<Parag>Here you can find pre-algebra lessons that will help you understand by explaining in detail and clarifying with some examples.</Parag>

Here you can find pre-algebra lessons that will help you understand by explaining in detail and clarifying with some examples.

Basic Algebra

# Math Order of Operations: BODMAS rule (i.e PEMDAS rule)

<Related>
<Link to="../basic-algebra/properties-of-addition-and-multiplication" as="/lessons/basic-algebra/properties-of-addition-and-multiplication">Properties of Addition and Multiplication: Commutative, Associative and Identity</Link> 
<Link to="../basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms" as="/lessons/basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms">Distributive Property of Multiplication over Addition and Subtraction and Combine Like Terms</Link> 
<Link to="../basic-algebra/evaluating-algebraic-expressions-with-exemples" as="/lessons/basic-algebra/evaluating-algebraic-expressions-with-exemples">Evaluating Algebraic Expressions with exemples</Link>
</Related>

The concept of order of operations in algebra lessons is one of the very basic concepts that the student must understand well, because this concept forms a basic base for all algebra lessons that will study later.

The arithmetic operations we encounter have addition, subtraction, multiplication and division (the four operations) combined or not in the same algebraic expression.

In other cases we find in the same algebraic expression in addition to the four operations, powers and grouping symbols(brackets), which will inevitably affect the change of priority of operations(BODMAS rule).

**So what is BODMAS rule?**

Bodmas' rule is a method for performing a series of arithmetic operations in an algebraic expression without errors. The meaning of these letters is:

- **B**: Brackets (Grouping symbols);
- **O**: Order (Exponent, square root, factorial);
- **D**: Division (/);
- **M**: Multiplication (*);
- **A**: Addition (+);
- **S**: Subtraction (-);

<Note>
Another synonym for PODMAS rule is **PEMDAS rule**, which means the same thing, the order of operations in math. The reason for changing some letters is due to several labels for the same concept. These are the meanings of the letters of the word PEMDAS:

- **P**: Parentheses (Grouping symbols);
- **E**: Exponent (Exponent, square root, factorial);
- **M**: Multiplication (*);
- **D**: Division (/);
- **A**: Addition (+);
- **S**: Subtraction (-);
</Note>

When evaluating expressions, proceed in the following rules order.

<Note>
1. Evaluate expressions contained in grouping symbols ﬁrst(**B**OMDAS). If grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols ﬁrst.
2. Evaluate all exponents that appear in the expression(B**O**MDAS).
3. Perform all multiplications and divisions in the order that they appear in the expression, moving left to right(BO**MD**AS).
4. Perform all additions and subtractions in the order that they appear in the expression, moving left to right(BOMD**AS**).
</Note>

These are some special cases in the order of operations.

<Note>
- Multiplication and division are the same importance.
- Addition and subtraction are the same importance.
</Note>

Let's take some examples to see the order in which the arithmetic operations are performed. These examples will be from simpler to complex.

------------

<Example type="example" num="1"/> 

calculate numerical expression below using the rules of Order of Operations.

<Tex block>A=2+15+10-12-1+6</Tex>

<Example type="solution"/>

This expression has only addition and subtraction (they have the same precedence), so we proceed from left to right.

<Tex block>
\begin{align*}
A&=17+10-12-1+6 \\\
&=27-12-1+6 \\\
&=15-1+6 \\\
&=14+6 \\\
&=20
\end{align*}
</Tex>

<Example type="example" num="2"/>

calculate numerical expression below using the rules of Order of Operations.

<Tex block>B=24\div8\times6\times2\div9</Tex>

<Example type="solution"/>

This expression has only multiplication and division (they have the same precedence), so we proceed from left to right.

<Tex block>
\begin{align*}
B&=3\times6\times2\div9 \\\
&=18\times2\div9 \\\
&=36\div9 \\\
&=18
\end{align*}
</Tex>

<Example type="example" num="3"/>

calculate numerical expression below using the rules of Order of Operations.

<Tex block>C=6+8\times5-10</Tex>

<Example type="solution"/>

This expression is combined by three operations: addition, subtraction and multiplication. So, we perform the multiplication first, Then, we accomplish the operation from left to right.

<Tex block>
\begin{align*}
C&=6+40-10 \\\
&=46-10 \\\
&=36
\end{align*}
</Tex>

<Example type="example" num="4"/>

calculate numerical expression below using the rules of Order of Operations.

<Tex block>D=(21\div7+3)+2\times3</Tex>

<Example type="solution"/>

This expression is combined by three operations: addition, division and multiplication, and parentheses So, we perform what is inside the parentheses first(Priority is for the division first).

<Tex block>
\begin{align*}
D&=(3+3)+2\times3 \\\
&=6+2\times3 \\\
&=6+6 \\\
&=12
\end{align*}
</Tex>

<Example type="example" num="5"/>

calculate numerical expression below using the rules of Order of Operations.

<Tex block>E=[50-(5+4\times4)]\times2-1</Tex>

<Example type="solution"/>

This expression is combined by three operations: addition, division and multiplication, and parentheses and barckets So, we perform what is inside the Brackets first(Priority is for what inside the parentheses).

<Tex block>
\begin{align*}
E&=[50-(5+16)]\times2-1 \\\
&=[50-21]\times2-1 \\\
&=29\times2-1 \\\
&=58-1 \\\
&=57
\end{align*}
</Tex>

<Example type="example" num="6"/>

calculate numerical expression below using the rules of Order of Operations.

<Tex block>F=3^2\times5+9^2-(7-2^2)</Tex>

<Example type="solution"/>

This expression is combined by three operations: addition, division and multiplication, and parentheses and powers So, we powers inside the parentheses first.

<Tex block>
\begin{align*}
F&=3^2\times5+9^2-(7-4) \\\
&=3^2\times5+9^2-3 \\\
&=9\times5+81-3 \\\
&=45+81-3 \\\
&=126-3 \\\
&=123
\end{align*}
</Tex>

<Worksheet>
<Link to="" as=""></Link>
</Worksheet>

Rules of math order of operation known as BODMAS rule or PEMDAS rule with evaluating some examples.

Math Order of Operations: BODMAS rule

# Properties of Addition and Multiplication: Commutative, Associative and Identity

<Related>
<Link to="../basic-algebra/math-order-of-operations-bodmas-rule" as="/lessons/basic-algebra/math-order-of-operations-bodmas-rule">Math Order of Operations: BODMAS rule (i.e PEMDAS rule)</Link> 
<Link to="../basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms" as="/lessons/basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms">Distributive Property of Multiplication over Addition and Subtraction and Combine Like Terms</Link> 
<Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms" as="/lessons/basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Simplifying algebraic expressions by Combining Like Terms</Link>
</Related>

## Commutative Property

Addition and multiplication are commutative in arithmetic operations. Commutative property means that the order of the terms can be changed so that the result does not change. This property is called **Commutative property over addition and mutiplication**.

Let's take an example of commutative over addition:

If <Tex>2+9=11</Tex> This means that <Tex>2+9=11</Tex>

Whether we add 9 to 2 or add 2 to 9 we get the same result

<Note>
Let <Tex>a</Tex> and <Tex>b</Tex> represent whole numbers. Then,
<Tex block>a+b=b+a</Tex>

</Note>

Let's take an example of commutative over multiplication:

If <Tex>3\times4=12</Tex> This means that <Tex>4\times3=12</Tex>

Whether we multiply 3 by 4 or multiply 4 by 3 we get the same result

<Note>
Let <Tex>a</Tex> and <Tex>b</Tex> represent whole numbers. Then,
<Tex block>a \times b=b \times a</Tex>

</Note>

But what about subtraction and division? Do you think they are commutative too? Let's try.

It is clear that <Tex>7 - 3 = 4</Tex>, but <Tex>3 - 7 \neq 4</Tex>

So the commutative property over subtraction has not been achieved.

It is also clear that <Tex>10\div2 = 5</Tex>, but <Tex>2\div10 \neq 5</Tex>.

So the commutative property was not achieved for the division as well.

<Note>
Let <Tex>a</Tex> and <Tex>b</Tex> represent whole numbers. Then, 
<Tex block>a-b \neq b-a</Tex>

<Tex block>a \div b \neq b \div a</Tex>

We say that subtraction and division are not commutative.

</Note>

-------------------------------

## Associative Property

Addition and multiplication are associative in arithmetic operations. Associative means in the case of an arithmetic operation in which there are 3 or more terms, we can add or multiply two different terms without paying attention to their order in the operation. This property is called the **associative property over addition and mutiplication**.

Let's take an example of associative over addition.

Obviously <Tex>(4+3)+5=7+5=12</Tex> means that <Tex>4+(3+5)=4+8=12</Tex>

Applying BODMAS rule, whether we add 5 to the sum of terms 4 and 3 or we add the sum of terms 5 and 3 to 4 we get the same result.

<Note>
Let <Tex>a, b</Tex> and <Tex>c</Tex> represent whole numbers. Then,
<Tex block>(a+b)+c=a+(b+c)</Tex>

</Note>

Let's take an example of associative over multiplication.

Obviously <Tex>(2\times6)\times5=12\times5=60</Tex> means that <Tex>2\times(6\times5)=2\times30=60</Tex>

Applying BODMAS rule, whether we multiply the product of 2 and 6 by 5 or we multiply 2 by the product of 6 and 5 we get the same result.

<Note>
Let <Tex>a, b</Tex> and <Tex>c</Tex> represent whole numbers. Then,
<Tex block>(a \centerdot b) \times c=a \times (b \centerdot c)</Tex>

</Note>

But what about subtraction and division? Do you think they are associative too? Let's try.

It is clear that <Tex>(8-3)-1=5-1=4</Tex>, but <Tex>8-(3-1)=8-2=6\neq4</Tex>

So the associative property over subtraction has not been achieved.

It is also clear that <Tex>(24 \div 6)\div2=4 \div 2=2</Tex>, but <Tex>24 \div (6 \div 2)=24 \div 3=8</Tex>

So the associative property was not achieved for the division as well.

<Note>
Let <Tex>a</Tex> and <Tex>b</Tex> represent whole numbers. Then, 
<Tex block>(a-b)-c \neq a-(b-c)</Tex>

<Tex block>(a \div b) \div c \neq a \div (b \div c)</Tex>

We say that subtraction and division are not associative.

</Note>

-------------------------------

## Identity Property (Neutral Element)

The identity element is every number that does not affect an arithmetic operation. That is, it does not affect the value of the algebraic expression, but not for all operations.

**For addition**:

In addition, the identity element is 0, because whether or not we add 0 to the expression, nothing changes.

For example: <Tex>2+0=0+2=2</Tex>

<Note>
Let <Tex>a</Tex> represent whole number. Then,
<Tex block>a+0=0+a=a</Tex>

</Note>

**For multiplication**:

In multiplication, the identity element is 1, because whether or not we multiply the expression by 1, nothing changes.

For example: <Tex>6 \times 1=1 \times 6=6</Tex>

<Note>
Let <Tex>a</Tex> represent whole number. Then,
<Tex block>a \times 1 = 1 \times a = a</Tex>

</Note>

The most important math rules for addition and multiplication, this article explain the commutative associative and identity rules.

Properties of Addition and Multiplication

# Distributive Property of Multiplication over Addition and Subtraction and Combine Like Terms

In this lesson, I will show you how to apply distributive property of multiplication over both operations addition and subtraction and simplifying the result using combine like terms rule.

<Related>
<Link to="../basic-algebra/math-order-of-operations-bodmas-rule" as="/lessons/basic-algebra/math-order-of-operations-bodmas-rule">Math Order of Operations: BODMAS rule (i.e PEMDAS rule)</Link> 
<Link to="../basic-algebra/properties-of-addition-and-multiplication" as="/lessons/basic-algebra/properties-of-addition-and-multiplication">Properties of Addition and Multiplication: Commutative, Associative and Identity</Link> 
<Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms" as="/lessons/basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Simplifying algebraic expressions by Combining Like Terms</Link>
</Related>

## Simple rule of distributive property of multiplication 

Let’s take the rectangle below and calculate its area using two different ways.

<Image src="/imgs/lessons/basicalgebra/Illustration_of_distributive_property_with_rectangles.png" alt="distributive property" width='500px' height='127px'/>

### **Method 1:**

We take the rectangle on the right (the two colored rectangles together). Its length is <Tex>b+c</Tex> and its width is <Tex>a</Tex>, so its area is:

<Tex block>S=a(b+c)</Tex>

### **Method 2:**

The second method consists to calculate the area of the two rectangles separated at left in the figure above.

The dimensions of the green rectangle are <Tex>a</Tex> and <Tex>b</Tex>, then this area is <Tex>S'=ab</Tex>

The dimensions of the purple rectangle are <Tex>a</Tex> and <Tex>c</Tex>, then this area is <Tex>S''=ac</Tex>

These two rectangles are the same parts as the rectangle on the right.

So, the area of the rectangle on the right is equal to the sum of the areas of the two rectangles on the left. We can express this result in the following expression:

<Text block>S=S'+S''</Tex>

<Tex block> a(b+c) = ab+ac</Tex>

This result we got is called **the distributive property of multiplication over addition**.

<Note>

Let <Tex>a, b</Tex> and <Tex>c</Tex> real numbers:

<Tex>a(b+c)=ab+ac</Tex>

</Note>

By similarity, we can achieve to **the distributive property of multiplication over subtraction**.

<Note>

Let <Tex>a, b</Tex> and <Tex>c</Tex> real numbers:

<Tex>a(b-c)=ab-ac</Tex>

</Note>

So how do we mathematically understand the distributive property?

Simply put, we multiply the number outside the parentheses by all the numbers inside the parentheses (distributive).

In case we have more than two numbers in parentheses, we apply the same rule as shown in the next note.

<Note>

Let <Tex>a, b, c</Tex> and <Tex>d</Tex> real numbers:

<Tex>a(b+c+d)=ab+ac+ad</Tex>

</Note>

When we factor, we get the same number of terms as inside the parentheses.

<Example type="example" num="1"/>

Distribute then simplify the algebraic expression below:

<Tex block>A=2(x+y)</Tex>

<Example type="solution"/>

This example is very simple, we multiply 2 by <Tex>x</Tex> and <Tex>y</Tex>.

<Tex block>A=2x+2y</Tex>

<Example type="example" num="2"/>

Distribute then simplify the algebraic expression below:

<Tex block>B=3(x-y+5)</Tex>

<Example type="solution"/>

by similarity, we multiply 3 by <Tex>x</Tex> and <Tex>y</Tex> and 5.

<Tex block>B=3 \times x-3 \times y+3 \times 5</Tex>

<Tex block>B=3x-3y+15</Tex>

<Example type="example" num="3"/>

Distribute then simplify the algebraic expression below:

<Tex block>C=-5(x-2y+1)</Tex>

<Example type="solution"/>

In this case, we need to pay attention to the sign of the number outside the bracket.

Here we multiply -5 by <Tex>x</Tex>, <Tex>2y</Tex> and 1

<Tex block>C=(-5) \times x-(-5) \times 2y+(-5) \times 1</Tex>

<Tex block>C=-5x+10y-5</Tex>

<Note>
We don't apply the distributive property if what is inside the parentheses is a product. 
<Tex>a(b \bullet c) \neq ab \bullet ac</Tex>

</Note>

------------

## Generalized rule of distributive property of multiplication

In this part I will try to show you another case of distributive in a simplified way. It is more general than the first property.

We start from the next rule:

<Note>
Let <Tex>a, b, c</Tex> and <Tex>d</Tex> real numbers. 

<Tex block>
\begin{align*} 
(a+b)(c+d)&=a(c+d)+b(c+d) \\\
&=ac+ad+bc+bd 
\end{align*}
</Tex>

</Note>

So as you can see, distributive property in this case is basically based on the first rule and applying it twice.

Let's take some exemples.

<Example type="example" num="1"/> 

Distributive then simplify the algebraic expression below:

<Tex block>D=(x+2)(x+3)</Tex>

<Example type="solution"/>

<Tex block>
\begin{align*} 
D&=(x+2)(x+3) \\\
&=x(x+3)+2(x+3) \\\
&=x \times x+x \times 3+2 \times x+2 \times 3 \\\
&=x^2+3x+2x+6
\end{align*}
</Tex>

After removing the parentheses and simplifying each term separately, we now combine like terms.

<Tex block>
\begin{align*} 
D&=x^2+(3+2)x+6 \\\
&=x^2+5x+6
\end{align*}
</Tex>

<Example type="example" num="2"/>

Distributive then simplify the algebraic expression below:

<Tex block>E=(x+4)(x-3)</Tex>

<Example type="solution"/>

<Tex block>
\begin{align*} 
E&=(x+4)(x-3) \\\
&=x(x-3)+4(x-3) \\\
&=x \times x-x \times 3+4 \times x-4 \times 3 \\\
&=x^2-3x+4x-12
\end{align*}
</Tex>

After removing the parentheses and simplifying each term separately, we now combine like terms.

<Tex block>
\begin{align*} 
E&=x^2+(-3+4)x-12 \\\
&=x^2+x-12
\end{align*}
</Tex>

<Example type="example" num="3"/>

Distributive then simplify the algebraic expression below:

<Tex block>F=(x-5)(x-6)</Tex>

<Example type="solution"/>

<Tex block>
\begin{align*} 
F&=(x-5)(x-6) \\\
&=x(x-6)-5(x-6) \\\
&=x \times x-x \times 6-5 \times x-5 \times (-6) \\\
&=x^2-6x-5x+30
\end{align*}
</Tex>

After removing the parentheses and simplifying each term separately, we now combine like terms.

<Tex block>
\begin{align*} 
F&=x^2+(-6-5)x+30 \\\
&=x^2-11x+30
\end{align*}
</Tex>

<Example type="example" num="4"/>

Distributive then simplify the algebraic expression below:

<Tex block>G=(-x+1)(2x-3)+2x^2</Tex>

<Example type="solution"/>

<Tex block>
\begin{align*} 
G&=(-x+1)(2x-3)+2x^2 \\\
&=-x(2x-3)+1(2x-3)+2x^2 \\\
&=-x \times 2x-x \times (-3)+1 \times 2x-1 \times 3+2x^2\\\
&=-2x^2+3x+2x-3 +2x^2
\end{align*}
</Tex>

After removing the parentheses and simplifying each term separately, we now combine like terms.

<Tex block>
\begin{align*} 
G&=(-2+2)x^2+(3+2)x-3 \\\
&=0x^2+5x-3 \\\
&=5x-3
\end{align*}
</Tex>

Applying and simplifying distributive property of multiplication over addition and subtraction using combine like terms rule.

Distributive Property of Multiplication over Addition and subtraction and Combine Like Terms

# Evaluating Algebraic Expressions with exemples

Each algebraic expression consists of a series of numbers, variables, four operations, parentheses and other mathematical symbols.

Each variable can be replaced by a known number while keeping the same form of the expression. This process is called **evaluating an expression**, i.e. calculating its value by giving each unknown a certain value

In this section we will evaluate algebraic expressions for given values of the variables contained in the expressions. Here are some simple tips to help you be successful.

<Note>
1. Replace all occurrences of variables in the expression with open parentheses. Leave room between the parentheses to substitute the given value of the variable.
2. Substitute the given values of variables in the open parentheses prepared in the ﬁrst step.
3. Evaluate the resulting expression according to the Rules Guiding Order of Operations.
</NOte>

Let's take some examples to make the idea clear.

------------

<Example type="example" num="1"/> 

Evaluate the expression <Tex>A=2x+1</Tex> at <Tex>x=0</Tex>.

<Example type="solution"/>

Following the tips above, first we replace all occurrences of variable <Tex>x</Tex> in the expression <Tex>A</Tex> with open parentheses.

<Tex block>
\begin{align*} 
A&=2\times()+1 \\\
&=2\times(0)+1 \\\
&=0+1 \\\
&=1
\end{align*}
</Tex>

<Example type="example" num="2"/>

Evaluate the expression <Tex>B=20-3y</Tex> at <Tex>y=-5</Tex>.

<Example type="solution"/>

Following the tips above, first we replace all occurrences of variable <Tex>y</Tex> in the expression <Tex>B</Tex> with open parentheses.

<Tex block>
\begin{align*} 
B&=20-3\times() \\\
&=20-3\times(-5) \\\
&=20+15 \\\
&=35
\end{align*}
</Tex>

<Example type="example" num="3"/> 

Evaluate the expression <Tex>C=(2x-1)(x+6)-6x</Tex> at <Tex>x=3</Tex>.

<Example type="solution"/>

Following the tips above, first we replace all occurrences of variable <Tex>x</Tex> in the expression <Tex>C</Tex> with open parentheses.

<Tex block>
\begin{align*} 
C&=(2\times()-1)(()+6)-6\times() \\\
&=(2\times(3)-1)((3)+6)-6\times(3) \\\
&=(6-1)(9)-18 \\\
&=5\times9-18 \\\
&=45-18 \\\
&=27
\end{align*}
</Tex>


<Example type="example" num="4"/> 

Evaluate the expression <Tex>D=x^5-2</Tex> at <Tex>x=2</Tex>.

<Example type="solution"/> 

Following the tips above, first we replace all occurrences of variable <Tex>x</Tex> in the expression <Tex>D</Tex> with open parentheses.

<Tex block>
\begin{align*} 
D&=()^5-2 \\\
&=(2)^5-2 \\\
&=32-2 \\\
&=30
\end{align*}
</Tex>

<Example type="example" num="5"/> 

Evaluate the expression <Tex>E=x^3</Tex> at <Tex>x=-5</Tex>.

<Example type="solution"/>

Following the tips above, first we replace all occurrences of variable <Tex>x</Tex> in the expression <Tex>E</Tex> with open parentheses.

<Tex block>
\begin{align*} 
E&=()^3 \\\
&=(-5)^3 \\\
&=-125 \\\
\end{align*}
</Tex>

<Example type="example" num="6"/> 

Evaluate the expression <Tex>F=x^4</Tex> at <Tex>x=-5</Tex>.

<Example type="solution"/> 

Following the tips above, first we replace all occurrences of variable <Tex>x</Tex> in the expression <Tex>F</Tex> with open parentheses.

<Tex block>
\begin{align*} 
F&=()^4 \\\
&=(-5)^4 \\\
&=625 \\\
\end{align*}
</Tex>

<Example type="example" num="7"/> 

Evaluate the expression <Tex>G=a+b+2ab-b^3</Tex> at <Tex>a=2</Tex> and <Tex>b=3</Tex>.

<Example type="solution"/> 

Following the tips above, first we replace all occurrences of variable <Tex>x</Tex> in the expression <Tex>G</Tex> with open parentheses.

<Tex block>
\begin{align*} 
G&=(2)+(3)+2\times(2)\times(3)-(3)^3 \\\
&=(2)+(3)+2\times(2)\times(3)-27 \\\
&=(2)+(3)+12-27 \\\
&=5+12-27 \\\
&=17-27 \\\
&=-10
\end{align*}
</Tex>

Evaluating algebraic expressions with 7 examples.

Evaluating Algebraic Expressions with exemples

# Simplifying algebraic expressions by Combining Like Terms

<Related>
<Link to="../basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms" as="/lessons/basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms">Distributive Property of Multiplication over Addition and Subtraction and Combine Like Terms</Link> 
<Link to="../basic-algebra/evaluating-algebraic-expressions-with-exemples" as="/lessons/basic-algebra/evaluating-algebraic-expressions-with-exemples">Evaluating Algebraic Expressions with exemples</Link> 
<Link to="../basic-algebra/writing-algebraic-expressions-from-word-problems" as="/lessons/basic-algebra/writing-algebraic-expressions-from-word-problems">Writing Algebraic Expressions from Word Problems</Link>
</Related>


Sometimes, some algebraic expressions need to be simplified by adding (or subtracting) terms that have the same variable.

We encounter the necessity of simplifying these expressions when we develop an algebraic expression or solve an equation or an inequality. But it is also necessary to know the meaning of term in the algebraic expression.

<Note>
A term is every algebraic expression (simple or complex) between the (+) and (-) signs.

</Note>

To simplify any algebraic expression we apply the rule:

<Note>
Like terms with common variable part can be combined and simplified by adding (or substracting) the coefficients of this terms.

</Note>

### Simplifying a simple algebraic expression

Let's take the following example:

<Tex block>A=5x-3y+4x+2y-3x+8y.</Tex>

This expression consists of six terms:

Term 1 is : <Tex>5x</Tex>, the coefficient is <Tex>5</Tex>

Term 2 is : <Tex>-3y</Tex>, the coefficient is <Tex>-3</Tex>

Term 3 is : <Tex>4x</Tex>, the coefficient is <Tex>4</Tex>

Term 4 is : <Tex>2y</Tex>, the coefficient is <Tex>2</Tex>

Term 5 is : <Tex>-3x</Tex>, the coefficient is <Tex>-3</Tex>

Term 6 is : <Tex>8y</Tex>, the coefficient is <Tex>8</Tex>

We note that all terms have one of the variables <Tex>x</Tex> or <Tex>y</Tex>. To simplify this expression, now it's possible to combine like terms of the variable parts <Tex>x</Tex> and <Tex>y</Tex> (**combining like terms**).

- For <Tex>x</Tex> we have : <Tex>5x+4x-3x = (5+4-3)x = 6x</Tex>
- For <Tex>y</Tex> we have : <Tex>-3y+2y+8y=(-3+2+8)y=7y</Tex>

So the simplified expression is:

<Tex block>A=6x+7y</Tex>

### Simplifying an algebraic expression with powers

Let's take the following example:

<Tex block>B=10x-2xy-4x^2+7y+6xy+5x^2-3x-11y</Tex>

This expression consists of eight terms:

Term 1 is: <Tex>10x</Tex>, the coefficient is <Tex>10</Tex>

Term 2 is: <Tex>-2xy</Tex>, the coefficient is <Tex>-2</Tex>

Term 3 is: <Tex>-4x^2</Tex>, the coefficient is <Tex>-4</Tex>

Term 4 is: <Tex>7y</Tex>, the coefficient is <Tex>7</Tex>

Term 5 is: <Tex>6xy</Tex>, the coefficient is <Tex>6</Tex>

Term 6 is: <Tex>5x^2</Tex>, the coefficient is <Tex>5</Tex>

Term 7 is: <Tex>-3x</Tex>, the coefficient is <Tex>-3</Tex>

Term 8 is: <Tex>-11y</Tex>, the coefficient is <Tex>-11</Tex>

We note that all terms have one of the variables <Tex>x</Tex> , <Tex>xy</Tex>, <Tex>x^2</Tex> or <Tex>y</Tex>. To simplify this expression, now it's possible to combine like terms of the variable parts <Tex>x</Tex> , <Tex>xy</Tex>, <Tex>x^2</Tex> and <Tex>y</Tex>.

- For <Tex>x</Tex> we have : <Tex>10x-3x = (10-3)x = 7x</Tex>
- For <Tex>xy</Tex> we have : <Tex>-2xy+6xy=(-2+6)y=4xy</Tex>
- For <Tex>x^2</Tex> we have : <Tex>-4x^2+5x^2 = (-4+5)x^2 = -x^2</Tex>
- For <Tex>y</Tex> we have : <Tex>7y-11y = (7-11)y = -4y</Tex>

So the simplified expression is:

<Tex block>B=7x+4xy-x^2-4y</Tex>

### Simplifying an algebraic expression with symbol of grouping

Removal of symbols of grouping is governed by the following rules.

<Note>
1. If a <Tex>(+)</Tex> sign precedes a symbol of grouping, this symbol of grouping may be removed without affecting the terms contained.
2. If a <Tex>(-)</Tex> sign precedes a symbol of grouping, this symbol of grouping may be removed if each sign of the terms contained is changed.
3. If more than one symbol of grouping is present, the inner ones are to be removed first.
</Note>

In this example, we take a complex "term" (in parentheses) and try to simplify it using the same rule we saw earlier.

Let's take the following example:

<Tex block>C=2(x+y)-3(y+z)-5(x+y)+4(y+z)</Tex>

We note that all terms have one of the variables <Tex>(x+y)</Tex> or <Tex>(y+z)</Tex>. To simplify this expression, now it's possible to combine like terms of the variable parts <Tex>(x+y)</Tex> and <Tex>(y+z)</Tex> (**combine like terms**).

- For <Tex>(x+y)</Tex> we have : 

<Tex block>2(x+y)-5(x+y) = (2-5)(x+y)=-3(x+y)</Tex>

- For <Tex>(y+z)</Tex> we have : 

<Tex block>-3(y+z)+4(y+z) = (-3+4)(x+y)=-(y+z)</Tex>

So the simplified expression is:

<Tex block>C=-3(x+y)-(y+z)</Tex>

So as you can see, the most important thing in this simplification, **to applying combine like terms**, is finding the common variable in terms of the expression. As you also note, this terms can be simple and it can be complex.
I hope you have benefited from this explanation.

understand how to combine like terms in algebraic expressions to simplify it by examples.

Simplifying algebraic expressions by Combining Like Terms

# Writing Algebraic Expressions from Word Problems

<Related>
<Link to="../basic-algebra/properties-of-addition-and-multiplication" as="/lessons/basic-algebra/properties-of-addition-and-multiplication">Properties of Addition and Multiplication: Commutative, Associative and Identity</Link> 
<Link to="../basic-algebra/evaluating-algebraic-expressions-with-exemples" as="/lessons/basic-algebra/evaluating-algebraic-expressions-with-exemples">Evaluating Algebraic Expressions with exemples</Link> 
<Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms" as="/lessons/basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Simplifying algebraic expressions by Combining Like Terms</Link>
</Related>

Algebraic expressions are a series of arithmetic operations consisting of the four operations, numbers, variables, parentheses and other mathematical symbols (square roots, powers, factors...).

These algebraic expressions are abstracted from physical concepts so that we can see their applications in all fields.

As we said earlier, algebraic expressions mainly consist of the four operations, numbers, variables, parentheses and various mathematical symbols that we will discuss as follows:

- **Numbers:** They are fixed values such as <Tex>-3, 0, 2, \frac2{5}, \pi, \sqrt{2}...</Tex>
- **Operations Symbols:** These are addition(+), subtraction(-), multiplication(<Tex>\times</Tex>) and division(<Tex>\div</Tex>)
- **Variables:** These are the unknowns that we usually express with letters such as <Tex>x, y, z, t, a, b,\alpha, \beta, ...</Tex> 
- **Grouping Symbols:** It contains other expressions such as <Tex>(\cdots), [\cdots], \\{\cdots\\}</Tex>
- **Other Mathematical Symbols:** such as square roots, exponents, <Tex>\sum, \prod, \int</Tex>

--------------

## Convert texts to algebraic expressions

Let's take some exemples:

<Example type="example" num="1"/> 

Sum of two unknown numbers.

<Example type="solution"/>

Let the first number be <Tex>x</Tex> and the second number be <Tex>y</Tex>.

So their sum is <Tex>x + y</Tex>

<Example type="example" num="2"/> 

The sum of twice <Tex>x</Tex> and <Tex>5</Tex>

<Example type="solution"/>

Let the first number be <Tex>x</Tex>, the twice of <Tex>x</Tex> is <Tex>2x</Tex>.

So the sum is <Tex>2x + 5</Tex>

<Example type="example" num="3"/> 

The difference of the number <Tex>m</Tex> and the triple of the number <Tex>n</Tex>.

<Example type="solution"/>

The variables are already defined <Tex>m</Tex> and <Tex>n</Tex>. the triple of <Tex>n</Tex> is <Tex>3n</Tex>.

So the difference is <Tex>m-3n</Tex>

<Example type="example" num="4"/> 

The difference of the product of <Tex>3</Tex> and <Tex>x</Tex>, and the quotient of <Tex>y</Tex> and <Tex>17</Tex>

<Example type="solution"/>

The product of <Tex>3</Tex> and <Tex>x</Tex> is <Tex>3x</Tex>

The quotient of <Tex>y</Tex> and <Tex>17</Tex> is <Tex>\frac y{17}</Tex>

So the difference is <Tex>3x-\frac y{17}</Tex>

<Example type="example" num="5"/> 

The product of two consecutive numbers.

<Example type="solution"/>

Let <Tex>x</Tex> be the first number. then the second number is <Tex>x+1</Tex>.

So, the product of this two numbers is <Tex>x(x+1)</Tex>

<Example type="example" num="6"/> 

The product of the sum of the numbers <Tex>x</Tex> and <Tex>4</Tex> and the sum of the numbers <Tex>-1</Tex> and twice of <Tex>x</Tex>.

<Example type="solution"/>

The sum of <Tex>x</Tex> and <Tex>4</Tex> is <Tex>x+4</Tex>

The sum of <Tex>-1</Tex> and the twice of <Tex>x</Tex> is <Tex>-1+2x</Tex>

The product is <Tex>(x+4)(-1+2x)</Tex>

<Example type="example" num="7"/> 

The sum of the square root of a number and the number next of it

<Example type="solution"/>

Let <Tex>x</Tex> the unknown number. then the square root of <Tex>x</Tex> is <Tex>\sqrt{x}</Tex>.
The number next of <Tex>x</Tex> is <Tex>x+1</Tex>.
The sum is <Tex>\sqrt{x}+x+1.</Tex>

Learn how to convert a Word Problem into an algebraic expression with a few simple rules and various examples to illustrate.

Writing Algebraic Expressions from Word Problems

# Fractions Basics

## Slicing a Cake into Fractions

This is a simple example, if you cut a cake into two equal pieces, each piece represents half of the cake. In other words, the half of the cake you took is one of the two parts.
Mathematically, we express this division by <Tex>\frac{1}{2}</Tex> which is a fraction number where 1 represents the piece you took and 2 represents the number of portions the cake was divided into.

## Definition of Fraction

Every fraction number is a real number consisting of two integers separated by a "fraction bar", which can be written horizontally or diagonally. For example, we can write <Tex>\frac{1}{2}</Tex> or <Tex>1/2</Tex>.

A fraction is a "part" of a "whole": <Tex>\frac{PART}{WHOLE}</Tex>

* The upper part (**PART**) of a fraction is called a **numerator**. This number tells us how many parts we took;

* the lower part (**WHOLE**) is called a **denominator**. This number tells us the total number of parts;

Now suppose that if you divide this cake into 6 pieces and eat 5 pieces of it, what is the number that represents all the cake you ate? I think it has become easy now, we divided the cake into 6 parts which means the denominator is 6, and you ate five parts which means the numerator is 5, so what you ate is <Tex>\frac{5}{6}</Tex> of the cake.

<Note>

Let <Tex>a</Tex> and <Tex>b</Tex> two integers with <Tex>b\not=0</Tex>.

The expression <Tex>\frac{a}{b}</Tex> or <Tex>a/b</Tex> called a Fractional Number.

- The number <Tex>a</Tex> is called a Numerator;

- The number <Tex>b</Tex> is called a Denominator which is always non-zero number.

</Note>

## Types of Fractions

There are 3 types of fractions:

* **Proper Fractions**: The numerator is less than the denominator. (<Tex>a)

 Exemples: <Tex>\dfrac{5}{8}</Tex> ; <Tex>\dfrac{9}{10}</Tex> ; <Tex>\dfrac{1}{100}</Tex>

* **Improper Fractions**: The numerator is greater than, or equal to, the denominator. (<Tex>a\geqslant b </Tex>)

 Exemples: <Tex>\dfrac{7}{3}</Tex> ; <Tex>\dfrac{39}{7}</Tex> ; <Tex>\dfrac{8}{8}</Tex>

* **Mixed numbers**: There is a whole number and a fraction.

 Exemples: <Tex>2\dfrac{2}{5}</Tex> ; <Tex>4\dfrac{5}{6}</Tex> ; <Tex>-2\dfrac{1}{2}</Tex>

## Remainder of Fraction

Going back to the previous example, remember you ate 5 cookies out of 6. How many do you have left? This is very clear that the rest is only one piece. From this example, we can draw the following conclusion:

<Note>

**REMAINDER** part, quantity, or number left over after division

</Note>

## Fraction Word Problems

<Example type="example" num="1"/> 

In my fruit basket, there are 9 pieces of fruit, 5 of which are apples. 

How can we express the number of apples as a fraction?

<Example type="solution"/>

Total pieces: 9, corresponds to the Denominator

Apple pieces: 5, corresponds to the numerator

<Tex>\frac{5}{9}</Tex>

-------- 

<Example type="example" num="2"/> 

Maria spent <Tex>\frac{3}{4}</Tex> of the money her grandparents gave her on an adventure book.

What fraction number represents the remainder?

<Example type="solution"/>

Maria spent three quarters of her money, Then, she had only one quarter left.

<Tex>\frac{1}{4}</Tex>

-------- 

<Example type="example" num="3"/> 

What fraction of the colored part of each figure?

<Image src="/imgs/lessons/basicalgebra/fractions_word_problem.jpeg" alt="fractions of colored parts in cercle and shapes" width={500} height={200} />

<Example type="solution"/>

Figure 1: <Tex>\frac{3}{8}</Tex>

Figure 2: <Tex>\frac{9}{10}</Tex>

Figure 3: <Tex>\frac{4}{5}</Tex>

-------- 

<Example type="example" num="4"/> 

What fraction of the colored part of the two circles?

<Image src="/imgs/lessons/basicalgebra/mixed_fraction.jpeg" alt="mixed fractions of colored parts in cercle and shapes" width={242} height={114} />

<Example type="solution"/>

In this figure, we took all the eight parts of the left circle, and we also took 5 parts out of the eight from the right circle. Which means we took <Tex>\frac{13}{8}</Tex> of the parts.

<Related>
<Link to="../basic-algebra/adding-subtracting-fractions-like-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-like-denominators">Adding and Subtracting Fractions with Like Denominators</Link> 
<Link to="../basic-algebra/dividing-fractions-and-fraction-by-whole-number" as="/lessons/basic-algebra/dividing-fractions-and-fraction-by-whole-number">Dividing Fractions Divide Fractions and Whole Numbers</Link> 
<Link to="../basic-algebra/multiplying-fractions-by-whole-number-by-reciprocal" as="/lessons/basic-algebra/multiplying-fractions-by-whole-number-by-reciprocal">Multiplying Fractions Multiplying Fraction by Whole Number and by its Reciprocal</Link> 
<Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">Finding Greatest Common Factor (GCF)</Link>
</Related>

Learn what is the fraction and what is the part of the whole and how to write the fraction with the denominator and the numerator. Learn proper fraction, improper fraction and mixed fraction.

Types of Fratcions and Basics

# Converting Improper Fractions to Mixed Fractions and Vice Versa

In a previous lesson, we saw that there are three types of fractures:

* **Proper Fractions**: The numerator is less than the denominator. (<Tex>a)

 Exemples: <Tex>\dfrac{5}{8}</Tex> ; <Tex>\dfrac{9}{10}</Tex> ; <Tex>\dfrac{1}{100}</Tex>

* **Improper Fractions**: The numerator is greater than, or equal to, the denominator. (<Tex>a\geqslant b </Tex>)

 Exemples: <Tex>\dfrac{7}{3}</Tex> ; <Tex>\dfrac{39}{7}</Tex> ; <Tex>\dfrac{8}{8}</Tex>

* **Mixed numbers**: There is a whole number and a fraction.

 Exemples: <Tex>2\dfrac{2}{5}</Tex> ; <Tex>4\dfrac{5}{6}</Tex> ; <Tex>-2\dfrac{1}{2}</Tex>

In this lesson, we will focus on the last two types, Improper Fractions and Mixed Fractions, so that we will see how to convert a mixed fraction to an improper fraction, and how to convert an improper fraction to a mixed fraction using a very simple rule. All you have to do, dear student, is to continue reading the lesson to the end, and re-perform the examples that we will see in this lesson.

## Converting Improper Fractions to Mixed Fractions

<Image src="/imgs/lessons/basicalgebra/improper-mixed-fraction.jpeg" alt="convert an improper fraction to a mixed fraction" width='650px' height='290px'/>

There are three very easy steps to convert an improper fraction to a mixed fraction, which are as follows:

* Divide the denominator by the numerator;

* Keep the obtained quotient as the whole part;

* Write the fraction part which the numerator is the remainder and the denominator is the original denominator;

* Write down the final result by combining the parts **\[Whole Part][Fraction Part]**;

<Alert type='infos'>

The fraction part is always a **proper fraction**.

</Alert>

### Examples of how to convert an improper fraction to a mixed fraction

<Example type="example" num="1"/>

Convert <Tex>\dfrac{28}{5}</Tex> to a mixed fraction.

<Example type="solution"/>

The denominator (28) is greater than the numerator (5), so <Tex>\dfrac{28}{5}</Tex> is an improper fraction.

* Divide 28 by 5

* Quotient = 5 (then 5 is the whole part)

* Remainder = 3 (then <Tex>\dfrac{3}{5}</Tex> is the fraction part)

Thus, the final answer is: <Tex>\dfrac{28}{5}=5\dfrac{3}{5}</Tex>

<Example type="example" num="2"/>

Convert <Tex>\dfrac{39}{4}</Tex> to a mixed fraction.

<Example type="solution"/>

The denominator (39) is greater than the numerator (4), so <Tex>\dfrac{39}{4}</Tex> is an improper fraction.

* Divide 39 by 4

* Quotient = 9 (then 9 is the whole part)

* Remainder = 3 (then <Tex>\dfrac{3}{4}</Tex> is the fraction part)

Thus, the final answer is: <Tex>\dfrac{39}{4}=9\dfrac{3}{4}</Tex>

<Example type="example" num="3"/>

Convert <Tex>\dfrac{85}{19}</Tex> to a mixed fraction.

<Example type="solution"/>

The denominator (85) is greater than the numerator (19), so <Tex>\dfrac{85}{19}</Tex> is an improper fraction.

* Divide 85 by 19

* Quotient = 4 (then 4 is the whole part)

* Remainder = 9 (then <Tex>\dfrac{9}{19}</Tex> is the fraction part)

Thus, the final answer is: <Tex>\dfrac{85}{19}=4\dfrac{9}{19}</Tex>

## Converting Mixed Fractions to Improper Fractions

For the reverse process, Converting a mixed fraction to an improper fraction is also easy, you just follow this three steps below.

<Note>

Let <Tex>a\dfrac{b}{c}</Tex> is a mixed fraction. To convert the mixed fraction to improper fraction we use the following formula:

<Tex block>a\dfrac{b}{c}=\dfrac{a\times c+b}{c}</Tex>

</Note>

Lets take some examples of converting the mixed fractions to improper fractions.

### Examples of how to convert an mixed fraction to a improper fraction

<Example type="example" num="4"/>

Convert <Tex>6\dfrac{3}{7}</Tex> to an improper fraction.

<Example type="solution"/>

We can do the process in one go: 

<Tex block>6\dfrac{3}{7}=\dfrac{6\times7+3}{7}=\dfrac{45}{7}</Tex>

<Example type="example" num="5"/>

Convert <Tex>2\dfrac{2}{11}</Tex> to an improper fraction.

<Example type="solution"/>

We can do the process in one go: 

<Tex block>2\dfrac{2}{11}=\dfrac{2\times11+2}{11}=\dfrac{24}{11}</Tex>

<Example type="example" num="6"/>

Convert <Tex>10\dfrac{11}{13}</Tex> to an improper fraction.

<Example type="solution"/>

We can do the process in one go: 

<Tex block>10\dfrac{11}{13}=\dfrac{10\times13+11}{13}=\dfrac{141}{13}</Tex>

----------

<Related>

<Link to="../basic-algebra/types-fratcions-and-basics" as="/lessons/basic-algebra/types-fratcions-and-basics">Types of Fratcions and Basics</Link> 
<Link to="../basic-algebra/adding-subtracting-fractions-like-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-like-denominators">Adding and Subtracting Fractions with Like Denominators</Link> 
<Link to="../basic-algebra/adding-subtracting-fractions-unlike-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-unlike-denominators">Adding and Subtracting Fractions with Unlike Denominators</Link> 
<Link to="../basic-algebra/dividing-fractions-and-fraction-by-whole-number" as="/lessons/basic-algebra/dividing-fractions-and-fraction-by-whole-number">Dividing Fractions Divide Fractions and Whole Numbers</Link> 

</Related>

Learn how to convert improper fractions to mixed fractions and convert mixed fractions to improper fractions with detailed steps and many examples.

Converting Improper Fractions and Mixed Fractions

# Reducing Fractions to Lowest Terms - Find the Equivalent Fraction

Sometimes, the numerator and denominator will have common factors. This fraction is considered **Not Simplified**. We can **SIMPLIFY** it by dividing the numerator and the denominator by the greatest common factor.

As you can see, you must be familiar with finding the greatest common factor of two whole numbers. If not, I suggest you visit <Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">this lesson</Link> before you set out to simplify fractions.

Dividing the numerator and the denominator by one of their common divisors does not necessarily give us the reduced fraction to the lowest term, but rather another fraction called **equivalent fraction**.

But dividing the numerator and the denominator by their <Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">greatest common factor (GCF)</Link> does always give us the reduced fraction to the lowest term and also this fraction is the **equivalent fraction**.

Let's explain what we said above with a mathematical text:

<Note>

Let <Tex>\dfrac{a}{b}</Tex> represent a fraction, and <Tex>m</Tex> represent a whole number.

if <Tex>m</tex> is one of the common divisors of <Tex>a</tex> and <Tex>b</tex>, then we can simplify the fraction <Tex>\dfrac{a}{b}</Tex> using the following formula:

<Tex block>\dfrac{a}{b}=\dfrac{a\div m}{b \div m}</Tex>

if <Tex>m</tex> is the greatest common factor of <Tex>a</tex> and <Tex>b</tex> (i.e <Tex>gcf(a,b)=m</tex>), then the formula above give us the fration reduced to the lowest term.

</Note>

This simple example shows us how we went from the unsimplified fraction to the reduced fraction to the lowest term, and that we always get an equivalent fraction.

<Image src="/imgs/lessons/basicalgebra/equivalent_fraction.jpeg" alt="equivalent- fraction" width={680} height={300} />

Another way to reduce a fraction is **repeated division**, where we divide the denominator and numerator of the fraction each time by the least common divisor of them until the least common divisor is 1. This method is good for beginners only and important to understand the basis of reducing fractions to an equivalent fractions.

------------------

## Examples of How to Reducing Fractions to the lowest terms

<Example type="example" num="1"/>

Reduce the fraction <Tex>\dfrac{6}{30}</Tex> to the lowest term.

<Example type="solution"/>

**Using Repeated Divisions Method:**

In this example, the numerator and the denominator are both divisibles by 2.

- Divide 6 and 30 by 2 we get:

<Tex block>\dfrac{6}{30}=\dfrac{6\div2}{30\div2}=\dfrac{3}{15}</Tex>

The resulting fraction is still not reduced to the lowest term, because its numerator and its denominator are both divisible by 3.

- Divide 3 and 15 by 3 we get:

<Tex block>\dfrac{3}{15}=\dfrac{3\div3}{15\div3}=\dfrac{1}{5}</Tex>

The least common divisor of 1 and 5 is 1, so, this resulting fraction is in its simplest form. Finally:

<Tex block>\dfrac{6}{30}=\dfrac{1}{5}</Tex>

**Using Greatest Common Factor Method:**

In the above method, we have simplified the fraction <Tex>\dfrac{6}{30}</Tex> by two successive divisions, the first one is by 2 and the second one is by 3. Is there any shortcut method without repeating the division? Sure, the shortcut method is dividing the denominator and the numerator by the greatest common factor which is the product of the divisors obtained in the previous method (i.e <Tex>gcf(6,30)=2\times3=6</Tex>). Thus,

<Tex block>\dfrac{6}{30}=\dfrac{6\div6}{30\div6}=\dfrac{1}{5}</Tex>

<Example type="example" num="2"/>

Reduce the fraction <Tex>\dfrac{8}{24}</Tex> to the lowest term.

<Example type="solution"/>

**Using Repeated Divisions Method:**

In this example, the numerator and the denominator are both divisibles by 2.

- Divide 8 and 24 by 2 we get:

<Tex block>\dfrac{8}{24}=\dfrac{8\div2}{24\div2}=\dfrac{4}{12}</Tex>

The resulting fraction is still not reduced to the lowest term, because its numerator and its denominator are both divisible by 2.

- Divide 4 and 12 by 2 we get:

<Tex block>\dfrac{4}{12}=\dfrac{4\div2}{12\div2}=\dfrac{2}{6}</Tex>

The resulting fraction is still not reduced to the lowest term, because its numerator and its denominator are both divisible by 2.

- Divide 2 and 6 by 2 we get:

<Tex block>\dfrac{2}{6}=\dfrac{2\div2}{6\div2}=\dfrac{1}{3}</Tex>

The least common divisor of 1 and 5 is 1, so, this resulting fraction is in its simplest form. Finally:

<Tex block>\dfrac{8}{24}=\dfrac{1}{3}</Tex>

**Using Greatest Common Factor Method:**

The greatest common factor of 8 and 24 is 8 (the product of the divisors obtained in the above method. i.e <Tex>gcf(8,24)=2\times2\times2=8</Tex>). Thus,

<Tex block>\dfrac{8}{24}=\dfrac{8\div8}{24\div8}=\dfrac{1}{3}</Tex>

<Example type="example" num="3"/>

Reduce the fraction <Tex>\dfrac{12}{28}</Tex> to the lowest term.

<Example type="solution"/>

**Using Repeated Divisions Method:**

In this example, the numerator and the denominator are both divisibles by 2.

- Divide 12 and 28 by 2 we get:

<Tex block>\dfrac{12}{28}=\dfrac{12\div2}{28\div2}=\dfrac{6}{14}</Tex>

The resulting fraction is still not reduced to the lowest term, because its numerator and its denominator are both divisible by 2.

- Divide 6 and 14 by 2 we get:

<Tex block>\dfrac{6}{14}=\dfrac{6\div2}{14\div2}=\dfrac{3}{7}</Tex>

The least common divisor of 3 and 7 is 1, so, this resulting fraction is in its simplest form. Finally:

<Tex block>\dfrac{12}{28}=\dfrac{3}{7}</Tex>

**Using Greatest Common Factor Method:**

The greatest common factor of 12 and 28 is 4 (the product of the divisors obtained in the above method. i.e <Tex>gcf(12,28)=2\times2=4</Tex>). Thus,

<Tex block>\dfrac{12}{28}=\dfrac{12\div4}{28\div4}=\dfrac{3}{7}</Tex>

<Example type="example" num="5"/>

Reduce the fraction <Tex>\dfrac{28}{49}</Tex> to the lowest term.

<Example type="solution"/>

**Using Greatest Common Factor Method:**

The greatest common factor of 28 and 49 is 7. Thus,

<Tex block>\dfrac{28}{49}=\dfrac{28\div7}{49\div7}=\dfrac{4}{7}</Tex>

The resulting fraction is reduced to the lowest term.

<Example type="example" num="6"/>

Reduce the fraction <Tex>\dfrac{6}{39}</Tex> to the lowest term.

<Example type="solution"/>

**Using Greatest Common Factor Method:**

The greatest common factor of 6 and 39 is 3. Thus,

<Tex block>\dfrac{6}{39}=\dfrac{6\div3}{39\div3}=\dfrac{2}{13}</Tex>

The resulting fraction is reduced to the lowest term.

<Related>

<Link to="../basic-algebra/types-fratcions-and-basics" as="/lessons/basic-algebra/types-fratcions-and-basics">Types of Fratcions and Basics</Link> 
<Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">Finding Greatest Common Factor (GCF)</Link>

</Related>

Learn how to reduce a fraction to the lowest term and find the equivalent fraction, using repeated divisions and the greatest common factor method with examples

Reducing Fractions to Lowest Terms

# Adding and Subtracting Fractions with Like Denominators

Alice and Bob order a cake which has been cut into six equal slices. Thus, each slice is <Tex>\frac{1}{6}</Tex> of the whole cake. Alice eats two slices (shaded in light green in Figure below), or <Tex>\frac{2}{6}</Tex> of the whole cake. Tony eats three slices (shaded in light red in the figure below), or <Tex>\frac{3}{6}</Tex> of the whole cake.

<Image src="/imgs/lessons/basicalgebra/add_fraction.webp" alt="add two fraction like denominators" width={96} height={96} />

Now it should be clear that together Alice and Bob eat ﬁve slices or <Tex>\frac{5}{6}</Tex> of the whole cake. This reﬂects the fact that:

<Tex block>\frac{2}{6}+\frac{3}{6}=\frac{5}{6}</Tex>

This demonstrates how to add two fractions with like (same) denominator. Keep the common denominator and add the numerators. That is,

<Tex block>\frac{2}{6}+\frac{3}{6}=\frac{2+3}{6}=\frac{5}{6}</Tex>

**Adding Fractions with Like Denominators**

<Note>

Let <Tex>\frac{a}{c}</Tex> and <Tex>\frac{b}{c}</Tex> be two fractions with like (same) denominator. Their sum is deﬁned as:

<Tex block>\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}</Tex>

</Note>
 
 **Subtracting Fractions with Like Denominators**

 A similar rule holds for subtraction.

 <Note>

Let <Tex>\frac{a}{c}</Tex> and <Tex>\frac{b}{c}</Tex> be two fractions with like (same) denominator. Their difference is deﬁned as:

<Tex block>\frac{a}{c}-\frac{b}{c}=\frac{a-b}{c}</Tex>

</Note>

## Examples of Adding and Subtracting Fractions with Like Denominator

<Example type="example" num="1"/> 

Add the fractions and reduce to the lowest terms if possible: <Tex>\frac{2}{3}+\frac{5}{3}</Tex>

<Example type="solution"/>

Both denominators are the same. So, we can add the numerators 2 and 5, and keep the denominator which is 3.

<Tex block>\frac{2}{3}+\frac{5}{3}=\frac{2+5}{3}=\frac{7}{3}</Tex>

Th fraction <Tex>\frac{7}{3}</Tex> is in the lowest terms because <Tex>gcf(7,3)=1</Tex>

-------------

<Example type="example" num="2"/> 

Add the fractions and reduce to the lowest terms if possible: <Tex>\frac{5}{12}+\frac{1}{12}</Tex>

<Example type="solution"/>

Both denominators are the same. So, we can add the numerators 5 and 1, and keep the denominator which is 12.

<Tex block>\frac{5}{12}+\frac{1}{12}=\frac{5+1}{12}=\frac{6}{12}</Tex>

Th fraction <Tex>\frac{6}{12}</Tex> isn't in the lowest terms because <Tex>gcf(6,12)=6</Tex>.

<Tex>\frac{6}{12}=\frac{6\times1}{6\times2}=\frac{1}{2}</Tex>

-------------

<Example type="example" num="3"/> 

Add the fractions and reduce to the lowest terms if possible: <Tex>\frac{2}{7}+\frac{5}{7}</Tex>

<Example type="solution"/>

Both denominators are the same. So, we can add the numerators 2 and 5, and keep the denominator which is 7.

<Tex block>\frac{2}{7}+\frac{5}{7}=\frac{2+5}{7}=\frac{7}{7}</Tex>

Th fraction <Tex>\frac{7}{7}</Tex> isn't in the lowest terms because <Tex>gcf(7,7)=7</Tex>.

<Tex>\frac{7}{7}=\frac{7\times1}{7\times1}=1</Tex>

-------------

<Example type="example" num="4"/> 

Add the fractions and reduce to the lowest terms if possible: <Tex>\frac{2}{9}+\frac{-7}{9}</Tex>

<Example type="solution"/>

Both denominators are the same. So, we can add the numerators 2 and -7, and keep the denominator which is 9.

<Tex block>\frac{2}{9}+\frac{-7}{9}=\frac{2+(-7)}{9}=\frac{-5}{9}</Tex>

Th fraction <Tex>\frac{-5}{9}</Tex> is in the lowest terms because <Tex>gcf(5,9)=1</Tex>

-------------

<Example type="example" num="5"/> 

Add the fractions and reduce to the lowest terms if possible: <Tex>\frac{-2}{5}+\frac{-8}{5}</Tex>

<Example type="solution"/>

Both denominators are the same. So, we can add the numerators -2 and -8, and keep the denominator which is 5.

<Tex block>\frac{-2}{5}+\frac{-8}{5}=\frac{(-2)+(-8)}{5}=\frac{-10}{5}</Tex>

Th fraction <Tex>\frac{-10}{5}</Tex> isn't in the lowest terms because <Tex>gcf(10,5)=5</Tex>

<Tex>\frac{-10}{5}=\frac{-2\times5}{5\times1}=-2</Tex>

-------------

<Example type="example" num="6"/> 

Subtract the fractions and reduce to the lowest terms if possible: <Tex>\frac{8}{11}-\frac{3}{11}</Tex>

<Example type="solution"/>

Both denominators are the same. So, we can subtract the numerators 8 and 3, and keep the denominator which is 11.

<Tex block>\frac{8}{11}-\frac{3}{11}=\frac{8-3}{11}=\frac{5}{11}</Tex>

Th fraction <Tex>\frac{5}{11}</Tex> is in the lowest terms because <Tex>gcf(5,11)=1</Tex>

-------------

<Example type="example" num="7"/> 

Subtract the fractions and reduce to the lowest terms if possible: <Tex>\frac{7}{16}-\frac{-3}{16}</Tex>

<Example type="solution"/>

Both denominators are the same. So, we can subtract the numerators 7 and -3, and keep the denominator which is 16.

<Tex block>\frac{7}{16}-\frac{-3}{16}=\frac{7-(-3)}{16}=\frac{10}{16}</Tex>

Th fraction <Tex>\frac{10}{16}</Tex> isn't in the lowest terms because <Tex>gcf(10,16)=2</Tex>

<Tex>\frac{10}{16}=\frac{2\times5}{2\times8}=\frac{5}{8}</Tex>

-------------

<Example type="example" num="8"/> 

Subtract the fractions and reduce to the lowest terms if possible: <Tex>\frac{-11}{21}-\frac{9}{21}</Tex>

<Example type="solution"/>

Both denominators are the same. So, we can subtract the numerators -11 and 9, and keep the denominator which is 21.

<Tex block>\frac{-11}{21}-\frac{9}{21}=\frac{-11-9}{21}=\frac{-20}{21}</Tex>

Th fraction <Tex>\frac{-20}{21}</Tex> is in the lowest terms because <Tex>gcf(20,21)=1</Tex>

-------------

<Example type="example" num="9"/> 

Add the fractions and reduce to the lowest terms if possible: <Tex>\frac{2}{-13}+\frac{5}{-13}</Tex>

<Example type="solution"/>

Both denominators are the same. So, we can add the numerators 2 and 5, and keep the denominator which is -13.

<Tex block>\frac{2}{-13}+\frac{5}{-13}=\frac{2+5}{-13}=\frac{7}{-13}</Tex>

Th fraction <Tex>\frac{7}{-13}</Tex> is in the lowest terms because <Tex>gcf(7,13)=1</Tex>

-------------

<Example type="example" num="10"/> 

Add the fractions and reduce to the lowest terms if possible: <Tex>\frac{7}{4}+\frac{6}{-4}</Tex>

<Example type="solution"/>

Both denominators aren't the same(4 and -4), but we can make them with the same denominator. We multiply the denominator and numerator of the fraction <Tex>\frac{6}{-4}</Tex> by -1, we get:

<Tex>\frac{6}{-4}=\frac{-1\times6}{-1\times(-4)}=\frac{-6}{4}</Tex>

Now, we add the fractions <Tex>\frac{7}{4}+\frac{-6}{4}</Tex> with like denominators.

<Tex block>\frac{7}{4}+\frac{-6}{4}=\frac{7+(-6)}{4}=\frac{1}{4}</Tex>

Th fraction <Tex>\frac{1}{4}</Tex> is in the lowest terms because <Tex>gcf(1,4)=1</Tex>

-------------

<Example type="example" num="11"/> 

Add the fractions and reduce to the lowest terms if possible: <Tex>\frac{2}{3}+\frac{5}{3}+\frac{7}{3}</Tex>

<Example type="solution"/>

The previous property can be extended as a sum of three fractions.

<Note>

Let <Tex>\frac{a}{d}</Tex>, <Tex>\frac{b}{d}</Tex> and <Tex>\frac{c}{d}</Tex> be three fractions with like (same) denominator. Their sum is deﬁned as:

<Tex block>\frac{a}{d}+\frac{b}{d}+\frac{c}{d}=\frac{a+b+c}{d}</Tex>

</Note>

Both denominators are the same. So, we can add the numerators 2, 5 and 7, and keep the denominator which is 3.

<Tex block>\frac{2}{3}+\frac{5}{3}+\frac{7}{3}=\frac{2+5+7}{3}=\frac{14}{3}</Tex>

Th fraction <Tex>\frac{14}{3}</Tex> is in the lowest terms because <Tex>gcf(14,3)=1</Tex>

<Related>

<Link to="../basic-algebra/types-fratcions-and-basics" as="/lessons/basic-algebra/types-fratcions-and-basics">Types of Fratcions and Basics</Link>

</Related>

Learn how to add and subtract two (or more) fractions with like or same denominators. This lesson is illustrate by many examples with solution.

Adding and Subtracting Fractions with Like Denominators

# Adding and Subtracting Fractions with Unlike (Different) Denominators

Adding and subtracting two fractions with different denominators takes some extra steps compared to adding and subtracting two fractions with the same denominator. In this lesson, we will try to explain how to add and subtract two fractions with different denominators by following three basic steps, and you can then skip them and complete the procedure directly.

Adding and subtracting two fractions with different denominators requires prior knowledge of the least common multiple, so if you are not familiar with this algebraic concept, I suggest you visit <Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">this lesson</Link> first.

Now we will learn how to add fractions with different denominators. When adding fractions, the denominators of those fractions must be the same. But they are not always the same.

For example, the two fractions <Tex>\dfrac{2}{5}</Tex> and <Tex>\dfrac{7}{5}</Tex> can be added because they have the same denominators.

But the fractions <Tex>\dfrac{2}{7}</Tex> and <Tex>\dfrac{7}{11}</Tex> cannot be added at once, because these fractions have different denominators. In such cases, rewrite each fraction to its **equivalent fraction** with a denominator which is equal to the Least Common Multiple of the two deniminators. As a result of these actions, fractions that had different denominators turn into fractions that have the same denominators. And we already know how to add such fractions.

By the same way, we subtract two fractions.

<Note>

Let <Tex>\dfrac{a}{b}</Tex> and <Tex>\dfrac{c}{d}</Tex> represent two fractions and, <Tex>a'</Tex>, <Tex>c'</Tex> represent whole numbers.

And let <Tex>m=LCM(b,d)</Tex>.

Convert the two fractions to its equivalent fractions:

<Tex block>\dfrac{a}{b}=\dfrac{a'}{m}</Tex>

<Tex block>\dfrac{c}{d}=\dfrac{c'}{m}</Tex>

Add the two fraction by add the two equivalent fractions.

<Tex block>\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{a'}{m}+\dfrac{c'}{m}=\dfrac{a'+c'}{m}</Tex>

Subtract the two fraction by add the two equivalent fractions.

<Tex block>\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{a'}{m}-\dfrac{c'}{m}=\dfrac{a'-c'}{m}</Tex>

</Note>

To make it easier to add fractions with different denominators, you can use the following step-by-step instructions:

1. Find the LCM of the denominators of fractions;

2. Divide the LCM by the denominator of each fraction and get an additional multiplier for each fraction;

3. Multiply the numerators and denominators of fractions by their additional factors;

4. Add fractions that have the same denominators;

5. Always make sure that the resulting fraction is reduced to the lowest term. That is, the gcf of the numerator and denominator is 1.

----------

## Examples of Adding and Subtracting Fractions with Different Denominators

<Example type="example" num="1"/>

Add and subtract the fractions <Tex>\dfrac{5}{6}</Tex> and <Tex>\dfrac{3}{4}</Tex>

<Example type="solution"/>

The two fractions <Tex>\dfrac{5}{6}</Tex> and <Tex>\dfrac{3}{4}</Tex> with unlike denominators. Therefore, we must convert each fraction to its equivalent fraction. But, first we must find the least common multiple of 6 and 4. that is, finding the least number that divides 6 and 4 together

<Tex>LCM(6,4)=12</Tex>

Thus, <Tex>\dfrac{5}{6}=\dfrac{10}{12}</Tex> and <Tex>\dfrac{3}{4}=\dfrac{9}{12}</Tex>

Once their denominators are equal, add (subtract) the fractions by adding (subtracting) their numerators and then keeping the common denominator.

<Tex block>
\begin{align*}
\dfrac{5}{6}+\dfrac{3}{4}&=\dfrac{10}{12}+\dfrac{9}{12}\\\
&=\dfrac{10+9}{12}\\\
&=\dfrac{19}{12}
\end{align*}
</Tex>

The fraction <Tex>\dfrac{19}{12}</Tex> is in the lowest term because <Tex>gcf(19,12)=1</Tex>

<Tex block>
\begin{align*}
\dfrac{5}{6}-\dfrac{3}{4}&=\dfrac{10}{12}-\dfrac{9}{12}\\\
&=\dfrac{10-9}{12}\\\
&=\dfrac{1}{12}
\end{align*}
</Tex>

The fraction <Tex>\dfrac{1}{12}</Tex> is in the lowest term because <Tex>gcf(1,12)=1</Tex>

<Example type="example" num="2"/>

Add and subtract the two fractions <Tex>\dfrac{9}{16}</Tex> and <Tex>\dfrac{1}{4}</Tex>

<Example type="solution"/>

The two fractions <Tex>\dfrac{9}{16}</Tex> and <Tex>\dfrac{1}{4}</Tex> are unlike denominators. Therefore, we must convert each fraction to its equivalent fraction. But, first we must find the least common multiple of 16 and 4. that is, finding the least number that divides 16 and 4 together

<Tex>LCM(16,4)=16</Tex>

In this case, we notice that one of the denominators is a multiple of the other, here we only convert the fraction with the smallest denominator to a fraction whose denominator is the denominator of the fraction with the largest denominator. Which, we convert only the fraction <Tex>\dfrac{1}{4}</Tex> to its equivalent fraction.

<Tex>\dfrac{1}{4}=\dfrac{4}{16}</Tex>

Once their denominators are equal, add (subtract) the fractions by adding (subtracting) their numerators and then keeping the common denominator.

<Tex block>
\begin{align*}
\dfrac{9}{16}+\dfrac{1}{4}&=\dfrac{9}{16}+\dfrac{4}{16}\\\
&=\dfrac{9+4}{16}\\\
&=\dfrac{13}{16}
\end{align*}
</Tex>

The fraction <Tex>\dfrac{13}{16}</Tex> is in the lowest term because <Tex>gcf(13,16)=1</Tex>

<Tex block>
\begin{align*}
\dfrac{9}{16}-\dfrac{1}{4}&=\dfrac{9}{16}-\dfrac{4}{16}\\\
&=\dfrac{9-4}{16}\\\
&=\dfrac{5}{16}
\end{align*}
</Tex>

The fraction <Tex>\dfrac{5}{16}</Tex> is in the lowest term because <Tex>gcf(5,16)=1</Tex>

<Example type="example" num="3"/>

Find the value of the expression: <Tex>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}</Tex>

<Example type="solution"/>

To add three fractions with different denominators we apply the same rule to add two fractions with different denominators, we first find the LCM of 3, 4 and 5.

<Tex>LCM(3, 4, 5)=60</Tex>

Now we can convert each fraction to its equivalent fraction with denominator is 60.

Thus, <Tex>\dfrac{1}{3}=\dfrac{20}{60}</Tex> and <Tex>\dfrac{1}{4}=\dfrac{15}{20}</Tex> and <Tex>\dfrac{1}{5}=\dfrac{12}{30}</Tex>

Once their denominators are equal, add the fractions by adding their numerators and then keeping the common denominator.

<Tex block>
\begin{align*}
\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}&=\dfrac{20}{60}+\dfrac{15}{60}+\dfrac{12}{60}\\\
&=\dfrac{20+15+12}{60}\\\
&=\dfrac{47}{60}
\end{align*}
</Tex>

The fraction <Tex>\dfrac{47}{60}</Tex> is in the lowest term because <Tex>gcf(47,60)=1</Tex>

<Example type="example" num="4"/>

Add and subtract the two fractions <Tex>\dfrac{9}{11}</Tex> and <Tex>\dfrac{2}{7}</Tex>

<Example type="solution"/>

The denominators are <Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">coprime</Link>, meaning that the lcm is the product of the two denominators. This means that:

<Tex>LCM(7,11)=7\times11=77</Tex>

Once their denominators are equal, add (subtract) the fractions by adding (subtracting) their numerators and then keeping the common denominator.

<Tex block>
\begin{align*}
\dfrac{9}{11}+\dfrac{2}{7}&=\dfrac{63}{77}+\dfrac{22}{77}\\\
&=\dfrac{63+22}{77}\\\
&=\dfrac{95}{77}
\end{align*}
</Tex>

The fraction <Tex>\dfrac{95}{77}</Tex> is in the lowest term because <Tex>gcf(95,77)=1</Tex>

<Tex block>
\begin{align*}
\dfrac{9}{11}-\dfrac{2}{7}&=\dfrac{63}{77}-\dfrac{22}{77}\\\
&=\dfrac{63-22}{77}\\\
&=\dfrac{41}{77}
\end{align*}
</Tex>

The fraction <Tex>\dfrac{41}{77}</Tex> is in the lowest term because <Tex>gcf(41,77)=1</Tex>

----------

<Related>

<Link to="../basic-algebra/types-fratcions-and-basics" as="/lessons/basic-algebra/types-fratcions-and-basics">Types of Fratcions and Basics</Link> 
<Link to="../basic-algebra/adding-subtracting-fractions-like-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-like-denominators">Adding and Subtracting Fractions with Like Denominators</Link> 
<Link to="../basic-algebra/dividing-fractions-and-fraction-by-whole-number" as="/lessons/basic-algebra/dividing-fractions-and-fraction-by-whole-number">Dividing Fractions Divide Fractions and Whole Numbers</Link> 
<Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">Finding Greatest Common Factor (GCF)</Link>

</Related>

Learn how to add and subtract two (or more) fractions with unlike or different denominators. This lesson is illustrated by many examples with solution.

Adding and Subtracting Fractions with Unlike Denominators

# Adding and Subtracting Mixed Fractions

In two of previous lessons, we saw how to add and subtract two fraction with like or unlike denominators. But what about two mixed fractions? How to add and subtract it?

Before we see how we do it, you first need to know how to convert a mixed fraction into an improper fraction. Then I advise you to go to <Link to="../basic-algebra/converting-improper-fractions-mixed-fractions" as="/lessons/basic-algebra/converting-improper-fractions-mixed-fractions">this lesson</Link> first if you are not familiar with how to convert mixed fraction to improper fraction.

There are two different methods to add and subtract mixed fractions.

**Method 1: by converting all mixed fractions to improper fractions**

* **Step 1:** Convert all mixed numbers into improper fractions.

* **Step 2:** Check! Do they have a common denominator? If not, find a common denominator.

* **Step 3:** When necessary, create equivalent fractions.

* **Step 4:** Add or subtract the numerators and keep the denominator the same.

* **Step 5:** If the answer is an improper form, reduce the fraction into a mixed number.

The following examples showing you hwo to apply the steps above.

<Example type="example" num="1"/>

Add the mixed frations: <Tex>2\dfrac{3}{4}+3\dfrac{1}{7}</Tex>

<Example type="solution"/>

* **Step 1:** Convert both mixed fractions into improper fractions

For the mixed fraction: <Tex>2\dfrac{3}{4}</Tex> :

<Tex block>2\dfrac{3}{4}=\dfrac{2\times4+3}{4}=\dfrac{11}{4}</Tex>

For the mixed fraction: <Tex>3\dfrac{1}{7}</Tex> :

<Tex block>3\dfrac{1}{7}=\dfrac{3\times7+1}{7}=\dfrac{22}{7}</Tex>

* **Step 2:** Find a common denominator

The common denominator is the lowest common multiple of 7 and 4.

By listing the multiples of 7 and 4, we fin that <Tex>LCM(7,4)=28</Tex>.

* **Step 3:** Create equivalent fraction

<Tex block>\dfrac{11}{4}=\dfrac{77}{28}</Tex>

<Tex block>\dfrac{22}{7}=\dfrac{88}{28}</Tex>

* **Step 4:** Add the fractions obtained in the step 3 (Fractions with like denominators)

<Tex block>
\begin{align*}
2\dfrac{3}{4}+3\dfrac{1}{7}&=\dfrac{11}{4}+\dfrac{22}{7}\\\
&=\dfrac{77}{28}+\dfrac{88}{28}\\\
&=\dfrac{77+88}{28}\\\
&=\dfrac{165}{28}
\end{align*}
</Tex>

* **Step 5:** Reduce the resulting fraction into a mixed fraction

 - Divide 165 by 28

 - Quotient = 5 (then 5 is the whole part)

 - Remainder = 25 (then <Tex>\dfrac{25}{28}</Tex> is the fraction part)

Thus, the final answer is: <Tex>\dfrac{165}{28}=5\dfrac{25}{28}</Tex>

hence the answer is final is: <Tex>2\dfrac{3}{4}+3\dfrac{1}{7}=5\dfrac{25}{28}</Tex>

<Example type="example" num="2"/>

Add the mixed frations: <Tex>7\dfrac{2}{3}+8\dfrac{1}{4}</Tex>

<Example type="solution"/>

* **Step 1:** Convert both mixed fractions into improper fractions

For the mixed fraction: <Tex>7\dfrac{2}{3}</Tex> :

<Tex block>7\dfrac{2}{3}=\dfrac{7\times3+2}{3}=\dfrac{23}{3}</Tex>

For the mixed fraction: <Tex>8\dfrac{1}{4}</Tex> :

<Tex block>8\dfrac{1}{4}=\dfrac{8\times4+1}{4}=\dfrac{33}{4}</Tex>

* **Step 2:** Find a common denominator

The common denominator is the lowest common multiple of 3 and 4.

By listing the multiples of 7 and 4, we fin that <Tex>LCM(3,4)=12</Tex>.

* **Step 3:** Create equivalent fraction

<Tex block>\dfrac{23}{3}=\dfrac{92}{12}</Tex>

<Tex block>\dfrac{33}{4}=\dfrac{99}{12}</Tex>

* **Step 4:** Add the fractions obtained in the step 3 (Fractions with like denominators)

<Tex block>
\begin{align*}
7\dfrac{2}{3}+8\dfrac{1}{4}&=\dfrac{23}{3}+\dfrac{33}{4}\\\
&=\dfrac{92}{12}+\dfrac{99}{12}\\\
&=\dfrac{92+99}{12}\\\
&=\dfrac{191}{12}
\end{align*}
</Tex>

* **Step 5:** Reduce the resulting fraction into a mixed fraction

 - Divide 191 by 12

 - Quotient = 15 (then 15 is the whole part)

 - Remainder = 11 (then <Tex>\dfrac{11}{12}</Tex> is the fraction part)

Thus, the final answer is: <Tex>\dfrac{191}{12}=15\dfrac{11}{12}</Tex>

hence the answer is final is: <Tex>7\dfrac{2}{3}+8\dfrac{1}{4}=15\dfrac{11}{12}</Tex>

___________

**Method 2: by adding or subtracting the whole number parts.**

* **Step 1:** Add or subtract the whole number part.

* **Step 2:** Check! Does the fraction part share a common denominator? If not, find one.

* **Step 3:** When necessary, create equivalent fractions.

* **Step 4:** Add or subtract the numerators of the fraction part and keep the denominator
the same.

* **Step 5:** If the answer is an improper form, reduce the fraction into a mixed number.

The following examples showing you hwo to apply the steps above.

<Example type="example" num="3"/>

Add the mixed frations: <Tex>6\dfrac{2}{5}+4\dfrac{3}{8}</Tex> 

<Example type="solution"/>

* **Step 1:** Add the whole number part

<Tex block>6+4=10</Tex>

* **Step 2:** The fraction part withe unlike denominator, so we must find the common denominator

The lowest common multiple of 5 and 8 is 40. Hence, the common denominator is 40.

* **Step 3:** Creat equivalent fractions of the fraction part

<Tex block>\dfrac{2}{5}=\dfrac{16}{40}</Tex>

<Tex block>\dfrac{3}{8}=\dfrac{15}{40}</Tex>

* **Step 4:** Add the two fraction obtained in step 3. (Fractions with like denominatots)

<Tex block>
\begin{align*}
\dfrac{2}{5}+\dfrac{3}{8}&=\dfrac{16}{40}+\dfrac{15}{40}\\\
&=\dfrac{16+15}{40}\\\
&=\dfrac{31}{40}
\end{align*}
</Tex>

* **Step 5:**

The fraction <Tex>\dfrac{31}{40}</Tex> is reduced to its lowest term.

Hence the final answer is: <Tex>6\dfrac{2}{5}+4\dfrac{3}{8}=10\dfrac{31}{40}</Tex>

<Example type="example" num="4"/>

Add the mixed frations: <Tex>5\dfrac{2}{3}+2\dfrac{5}{6}</Tex> 

<Example type="solution"/>

* **Step 1:** Add the whole number part

<Tex block>5+2=7</Tex>

* **Step 2:** The fraction part withe unlike denominator, so we must find the common denominator

The lowest common multiple of 3 and 6 is 6. Hence, the common denominator is 6.

* **Step 3:** Creat equivalent fractions of the fraction part

<Tex block>\dfrac{2}{3}=\dfrac{4}{6}</Tex>

* **Step 4:** Add the two fraction obtained in step 3. (Fractions with like denominatots)

<Tex block>
\begin{align*}
\dfrac{2}{3}+\dfrac{5}{6}&=\dfrac{4}{6}+\dfrac{5}{6}\\\
&=\dfrac{4+5}{6}\\\
&=\dfrac{9}{6}
\end{align*}
</Tex>

* **Step 5:**

The fraction <Tex>\dfrac{9}{6}</Tex> isn't reduced to its lowest term. We must convert it to the mixed fraction.

By Applying the steps in <Link to="../basic-algebra/converting-improper-fractions-mixed-fractions" as="/lessons/basic-algebra/converting-improper-fractions-mixed-fractions">this lesson</Link>, we get:

<Tex>\dfrac{9}{6}=\dfrac{1\times6+3}{6}=1\dfrac{3}{6}</Tex>

The new whole part is : <Tex>7+1=8</Tex>

Hence the final answer is: <Tex>5\dfrac{2}{3}+2\dfrac{5}{6}=8\dfrac{3}{6}</Tex>

_____________

<Related>

<Link to="../basic-algebra/types-fratcions-and-basics" as="/lessons/basic-algebra/types-fratcions-and-basics">Types of Fratcions and Basics</Link> 
<Link to="../basic-algebra/adding-subtracting-fractions-like-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-like-denominators">Adding and Subtracting Fractions with Like Denominators</Link> 
<Link to="../basic-algebra/adding-subtracting-fractions-unlike-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-unlike-denominators">Adding and Subtracting Fractions with Unlike (Different) Denominators</Link> 
<Link to="../basic-algebra/converting-improper-fractions-mixed-fractions" as="/lessons/basic-algebra/converting-improper-fractions-mixed-fractions">Converting Improper Fractions to Mixed Fractions and Vice Versa</Link> 
<Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">Finding Greatest Common Factor (GCF)</Link>

</Related>

Learn how to add and subtract two mixed fractions using two different methods. This lesson is illustrated by many examples with solutions.

Adding and Subtracting Mixed Fractions

# Multiplying Fractions - Multiplying Fraction by Whole Number and by its Reciprocal

Multiplying two fractions is very easy, and the rule is simple. Just to remind you, the denominator of a fraction is always non-zero integer.

<Note>

Let <Tex>\dfrac{a}{b}</Tex> and <Tex>\dfrac{c}{d}</Tex> be two fractions.

<Tex block>\dfrac{a}{b} \times \dfrac{c}{d}=\dfrac{a \times c}{b \times d}</Tex>

To multiply two fractions, multiply the numerators and then multiply the denominators.

</Note>

Know that there are two other symbols of multiplication, using **Dot** or **Parentheses**.

<Note>

Let <Tex>\dfrac{a}{b}</Tex> and <Tex>\dfrac{c}{d}</Tex> be two fractions.

<Tex block>\dfrac{a}{b} \bullet \dfrac{c}{d}=\dfrac{a \bullet c}{b \bullet d}</Tex>

<Tex block>\left( \dfrac{a}{b} \right) \left( \dfrac{c}{d} \right)=\dfrac{(a) (c)}{(b) (d)}</Tex>

</Note>

<Alert type="infos">

* Note carefully that multiplying two fractions has nothing to do with whether the two fractions have the same or different denominators, as in the case of addition and subtraction.

* After multiplying two fractions, make sure your answer is reduced to lowest terms.

</Alert>

In some cases, the factor in the numerator of one of the two fractions is the same in the denominator of the other fraction. Here we resort to simplifying the operation before performing it.

**Best Procedure**:

- If possible, cancel the same factor from a numerator and a denominator.

- Multiply the remaining numerators and the remaining denominators.

These are some examples that show how to multiply two fractions, but before multiplying we always have to make sure that we cancel the common factor from the denominator and the numerator using reducing the factors rules.

<Example type="example" num="1"/> 

Multiply the following fractions:

1) <Tex>\dfrac{2}{5} \times \dfrac{3}{7}</Tex>

2) <Tex>\dfrac{7}{9} \times \dfrac{6}{5}</Tex>

3) <Tex>\dfrac{10}{15} \times \dfrac{9}{5}</Tex>

4) <Tex>\dfrac{-4}{3} \times \dfrac{1}{5}</Tex>

5) <Tex>\dfrac{11}{8} \times \dfrac{3}{5} \times \dfrac{1}{2}</Tex>

<Example type='solution' />

1) The two fractions <Tex>\dfrac{2}{5}</Tex> and <Tex>\dfrac{3}{7}</Tex> are in the lowest term.

<Tex block>\dfrac{2}{5} \times \dfrac{3}{7}=\dfrac{2\times3}{5\times7}=\dfrac{6}{35}</Tex>

The result of the multiplication is already in its reduced form (<Link to="../arithmetic/greates-common-factor-least-common-multiple">the greatest common factor</Link> of 6 and 35 is 1).

2) The two fractions <Tex>\dfrac{7}{9}</Tex> and <Tex>\dfrac{6}{5}</Tex> are in the lowest term, but the gcf of 9 and 6 is 3.

<Tex block>
\begin{align*}
\dfrac{7}{9} \times \dfrac{6}{5}&=\dfrac{7}{\cancel{3}\times3} \times \dfrac{\cancel{3}\times2}{5} \\\
&=\dfrac{7}{3} \times \dfrac{2}{5} \\\
&=\dfrac{7\times2}{3\times5} \\\
&=\dfrac{14}{15}
\end{align*}
</Tex>

The resulting fraction is in its reduced form since the greatest common factor of 14 and 15 is 1.

3) The fraction <Tex>\dfrac{10}{15}</Tex> should be reduced to the lowest term:

<Tex>\dfrac{10}{15}=\dfrac{\cancel{5}\times2}{\cancel{5}\times3}=\dfrac{2}{3}</Tex>

Thus,

<Tex block>
\begin{align*}
\dfrac{10}{15} \times \dfrac{9}{5}&=\dfrac{2}{3} \times \dfrac{9}{5} \\\
&=\dfrac{2}{\cancel {3}\times1} \times \dfrac{\cancel {3}\times3}{5} \\\
&=\dfrac{2}{1} \times \dfrac{3}{5} \\\
&=\dfrac{2\times3}{1\times5} \\\
&=\dfrac{6}{5}
\end{align*}
</Tex>

The resulting fraction is in its reduced form since the greatest common factor of 6 and 5 is 1.

4) The two fractions <Tex>\dfrac{-4}{3}</Tex> and <Tex>\dfrac{1}{5}</Tex> are in the lowest term.

<Tex block>\dfrac{-4}{3} \times \dfrac{1}{5}=\dfrac{-4\times1}{3\times5}=\dfrac{-4}{15}</Tex>

The result of the multiplication is already in its reduced form (the greatest common factor of 4 and 15 is 1).

5) In the same way, we multiple three fractions.

The three fractions <Tex>\dfrac{11}{8}</Tex>, <Tex>\dfrac{3}{5}</Tex> and <Tex>\dfrac{1}{2}</Tex> are in the lowest term.

<Tex block>
\begin{align*}
\dfrac{11}{8} \times \dfrac{3}{5} \times \dfrac{1}{2}&=\dfrac{11\times3\times1}{8\times5\times2} \\\
&=\dfrac{33}{80}
\end{align*}
</Tex>

The resulting fraction is in its reduced form since the greatest common factor of 33 and 88 is 1.

___________________

## Multiplying Fraction by Whole Number

In the previous paragraph, we saw how to multiply two fractions, but how do we multiply two numbers, one of which is a fraction and the other is a whole number? This is what we will see in this paragraph.

Before we get to know the rule, let's first derive it from a simple example.

We all agree that <Tex>3\times\dfrac{2}{7}</Tex> means: <Tex>\dfrac{2}{7}</Tex> goes 3 times, right?

We can express this mathematically as follows:

<Tex>3\times\dfrac{2}{7}=\dfrac{2}{7}+\dfrac{2}{7}+\dfrac{2}{7}</Tex>

Now applying the rule for adding fractions with the same denominator, we get:

<Tex block>
\begin{align*}
3\times\dfrac{2}{7}&=\dfrac{2}{7}+\dfrac{2}{7}+\dfrac{2}{7} \\\
&=\dfrac{2+2+2}{7}\\\
&=\dfrac{6}{7}
\end{align*}
</Tex>

After the result obtained, can you derive the property of multiplying a fraction by a whole number without going through the addition step? There is no doubt that you can, it is very clear that we multiplied 3 by the numerator of the fraction, that is, we multiplied 3 by 2 to get 6 in the numerator of the resulting fraction.

<Note>

Let <Tex>a</Tex> represent whole number and <Tex>\dfrac{b}{c}</Tex> be fraction.

<Tex block>a \times \dfrac{b}{c} = \dfrac{b}{c} \times a=\dfrac{a \times b}{c}</Tex>

To multiply a fraction by the whole number, multiply the numerator by the whole number and keep the denominator.

</Note>

<Example type="example" num="2"/> 

Multiply the following fractions:

1) <Tex>4\times\dfrac{3}{5}</Tex>

2) <Tex>12\times\dfrac{7}{4}</Tex>

3) <Tex>-3\times\dfrac{7}{2}</Tex>

4) <Tex>\dfrac{2}{25}\times 15</Tex>

<Example type='solution' />

1) The fraction <Tex>\dfrac{3}{5}</Tex> is in its reduced form. 

<Tex block>4\times\dfrac{3}{5}=\dfrac{4\times3}{5}=\dfrac{12}{5}</Tex>

The resulting fraction is in its reduced form since the greatest common factor of 12 and 5 is 1.

2) The fraction <Tex>\dfrac{7}{4}</Tex> is in its reduced form. But, the greatest common factor of 12 and 4 is 4.

<Tex block>
\begin{align*}
12\times\dfrac{7}{4}&=\dfrac{12\times7}{4} \\\
&=\dfrac{\cancel{4}\times3\times7}{\cancel{4}\times1} \\\
&=\dfrac{21}{1} \\\
&=21
\end{align*}
</Tex>

3) The fraction <Tex>\dfrac{7}{2}</Tex> is in its reduced form. 

<Tex block>-3\times\dfrac{7}{2}=\dfrac{-3\times7}{2}=\dfrac{-21}{2}</Tex>

The resulting fraction is in its reduced form since the greatest common factor of 21 and 2 is 1.

4) The fraction <Tex>\dfrac{2}{25}</Tex> is in its reduced form. But, the greatest common factor of 15 and 25 is 5.

<Tex block>
\begin{align*}
\dfrac{2}{25}\times 15&=\dfrac{2\times15}{25} \\\
&=\dfrac{2\times \cancel{5}\times 3}{\cancel{5}\times 5} \\\
&=\dfrac{6}{5}
\end{align*}
</Tex>

The resulting fraction is in its reduced form since the gcf of 6 and 5 is 1.

___________

## Multiplying Fraction by Its Reciprocal

Perhaps we'll need to mention the reciprocal of the fraction in the next property.

<Note>

Let <Tex>a</Tex> and <Tex>b</Tex> represent two non-zero integers.

- The reciprocal of <Tex>a</Tex> is <Tex>\dfrac{1}{a}</Tex>

- The reciprocal of <Tex>\dfrac{1}{a}</Tex> is <Tex>a</Tex>

- The reciprocal of <Tex>\dfrac{a}{b}</Tex> is <Tex>\dfrac{b}{a}</Tex>

The reciprocal of a fraction is switcing the numerator and the denominator.

</Note>

Multiplying a fraction by its reciprocal always gives us one as a result.

<Note>

Let <Tex>a</Tex> and <Tex>b</Tex> represent two non-zero integers.

<Tex block>a\times \dfrac{1}{a}=\dfrac{1}{a} \times a =1</Tex>

<Tex block>\dfrac{a}{b}\times \dfrac{b}{a}=1</Tex>

</Note>

Let's take some examples to further illustrate this important property that we will use often in dividing two fractions.

<Example type='example' num='3' />

Multiply the following number by its reciprocal.

<Tex block>3;\\quad-5;\\quad \dfrac{1}{6}; \\quad \dfrac{11}{9}; \\quad \dfrac{7}{-10}; \\quad 1; \\quad 0</Tex>

<Example type='solution' />

* The reciprocal of 3 is <Tex>\dfrac{1}{3}</Tex>.

<Tex block>3\times \dfrac{1}{3}=\dfrac{3}{3}=1</Tex>

* The reciprocal of -5 is <Tex>\dfrac{1}{-5}</Tex>.

<Tex block>-5\times \dfrac{1}{-5}=\dfrac{-5}{-5}=1</Tex>

* The reciprocal of <Tex>\dfrac{1}{6}</Tex> is 6.

<Tex block>\dfrac{1}{6}\times6=\dfrac{6}{6}=1</Tex>

* The reciprocal of <Tex>\dfrac{11}{9}</Tex> is <Tex>\dfrac{9}{11}</Tex>.

<Tex block>\dfrac{11}{9}\times \dfrac{9}{11}=\dfrac{99}{99}=1</Tex>

* The reciprocal of <Tex>\dfrac{7}{-10}</Tex> is <Tex>\dfrac{-10}{7}</Tex>.

<Tex block>\dfrac{7}{-10}\times \dfrac{-10}{7}=\dfrac{-70}{-70}=1</Tex>

* The reciprocal of 1 is itself.

* The number 0 has no reciprocal.

<Related>
<Link to="../basic-algebra/types-fratcions-and-basics" as="/lessons/basic-algebra/types-fratcions-and-basics">Types of Fratcions and Basics</Link> 
<Link to="../basic-algebra/adding-subtracting-fractions-like-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-like-denominators">Adding and Subtracting Fractions with Like Denominators</Link> 
<Link to="../basic-algebra/dividing-fractions-and-fraction-by-whole-number" as="/lessons/basic-algebra/dividing-fractions-and-fraction-by-whole-number">Dividing Fractions Divide Fractions and Whole Numbers</Link> 
<Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">Finding Greatest Common Factor (GCF)</Link>
</Related>

Learn how to multiply two fractions and learn how to multiply a fraction by whole number and by its reciprocal. This lesson is illustrate by many examples.

Multiplying Fractions - Multiplying Fraction by Whole Number

# Dividing Fractions - Divide Fractions and Whole Numbers

As with multiplying fractions, we saw that it is very easy, we just multiply the denominators and multiply the numerators. In the process of division is not much different from multiplication.

To divide two fractions, we convert the division operation into a multiplication operation. but how? There are 3 basic steps in order to perform division.

- **First**, we convert the division symbol (<Tex>\div</Tex> or <Tex>/</Tex>) to the multiplication symbol (<Tex>\times</Tex>)

- **Second**: We keep the first number as it is, and flip the second fraction to its reciprocal.

- **Third**: We perform the multiplication process normally, as we explained in this lesson.

<Note>

Let <Tex>a</Tex>, <Tex>b</Tex>, <Tex>c</Tex> and <Tex>d</Tex> four non-zero integers.

<Tex block>\dfrac{a}{b} \div \dfrac{c}{d}=\cfrac{\dfrac{a}{b}}{{\dfrac{c}{d}}}=\dfrac{a}{b} \times \dfrac{d}{c}</Tex>

To divide <Tex>\dfrac{a}{b}</Tex> by <Tex>\dfrac{c}{d}</Tex> we multiply <Tex>\dfrac{a}{b}</Tex> by the reciprocal of <Tex>\dfrac{c}{d}</Tex>. i.e. we multiply by <Tex>\dfrac{d}{c}</Tex>

</Note>

<Alert type="infos">

* Note carefully that dividing two fractions has nothing to do with whether the two fractions have the same or different denominators, as in the case of addition and subtraction.

* After dividing two fractions, make sure your answer is reduced to lowest terms.

</Alert>

Let's take some examples of dividing two non-zero fractions.

<Example type="example" num="1"/>

Divide the fractions below.

1) <Tex>\dfrac{2}{9} \div \dfrac{7}{4}</Tex>

2) <Tex>\dfrac{-3}{8} \div \dfrac{11}{5}</Tex>

3) <Tex>\cfrac{\dfrac{6}{7}}{{\dfrac{10}{3}}}</Tex>

4) <Tex>\cfrac{\dfrac{-5}{4}}{{\dfrac{3}{-2}}}</Tex>

<Example type='solution' />

1) Let's calculate <Tex>\dfrac{2}{9} \div \dfrac{7}{4}</Tex>

The reciprocal of <Tex>\dfrac{7}{4}</Tex> is <Tex>\dfrac{4}{7}</Tex>. Thus,

<Tex block>
\begin{align*}
\dfrac{2}{9} \div \dfrac{7}{4}&=\dfrac{2}{9} \times \dfrac{4}{7} \\\
&=\dfrac{2\times4}{9\times7} \\\
&=\dfrac{8}{63}
\end{align*}
</Tex>

The resulting fraction is in its reduced form since <Link to="../arithmetic/greates-common-factor-least-common-multiple">the greatest common factor</Link> of 8 and 63 is 1.

2) Let's calculate <Tex>\dfrac{-3}{8} \div \dfrac{11}{5}</Tex>

The reciprocal of <Tex>\dfrac{11}{5}</Tex> is <Tex>\dfrac{5}{11}</Tex>. Thus,

<Tex block>
\begin{align*}
\dfrac{-3}{8} \div \dfrac{11}{5}&=\dfrac{-3}{8} \times \dfrac{5}{11} \\\
&=\dfrac{-3\times5}{8\times11} \\\
&=\dfrac{-15}{88}
\end{align*}
</Tex>

The resulting fraction is in its reduced form since the greatest common factor of 15 and 88 is 1.

3) Let's calculate <Tex>\cfrac{\dfrac{6}{7}}{{\dfrac{10}{3}}}</Tex>

We can rewrite the operation using the division symbol.

<Tex>\cfrac{\dfrac{6}{7}}{{\dfrac{10}{3}}} = \dfrac{6}{7} \div \dfrac{10}{3}</Tex>

The reciprocal of <Tex>\dfrac{10}{3}</Tex> is <Tex>\dfrac{3}{10}</Tex>. Thus,

<Tex block>
\begin{align*}
\dfrac{6}{7} \div \dfrac{10}{3}&=\dfrac{6}{7} \times \dfrac{3}{10} \\\
&=\dfrac{6\times3}{7\times10} \\\
&=\dfrac{\cancel{2} \times 3\times3}{7\times \cancel{2}\times 5} \\\
&=\dfrac{9}{35}
\end{align*}
</Tex>

The resulting fraction is in its reduced form since the greatest common factor of 9 and 35 is 1.

4) Let's calculate <Tex>\cfrac{\dfrac{-5}{4}}{{\dfrac{3}{-2}}}</Tex>

We can rewrite the operation using the division symbol.

<Tex>\cfrac{\dfrac{-5}{4}}{{\dfrac{3}{-2}}} = \dfrac{-5}{4} \div \dfrac{3}{-2}</Tex>

The reciprocal of <Tex>\dfrac{3}{-2}</Tex> is <Tex>\dfrac{-2}{3}</Tex>. Thus,

<Tex block>
\begin{align*}
\dfrac{-5}{4} \div \dfrac{3}{-2}&=\dfrac{-5}{4} \times \dfrac{-2}{3} \\\
&=\dfrac{-5\times(-2)}{4\times3} \\\
&=\dfrac{5\times2}{4\times3} \\\
&=\dfrac{5\times \cancel{2}}{2\times \cancel{2}\times 3} \\\
&=\dfrac{5}{6}
\end{align*}
</Tex>

The resulting fraction is in its reduced form since the greatest common factor of 5 and 6 is 1.

_________________

## Dividing Fraction by Whole Number - Dividing Whole Number by Fraction

A whole number can be considered as a fraction so that its denominator is 1, and thus we have to do the division of a fraction by an integer, or vice versa by applying the following property:

<Note>

Let <Tex>a</Tex>, <Tex>b</Tex> and <Tex>c</Tex> three non-zero integers.

<Tex block>\cfrac{a}{\dfrac{b}{c}}= a \div \dfrac{b}{c}=a \times \dfrac{c}{b}</Tex>

<Tex block>\cfrac{\dfrac{a}{b}}{c}=\dfrac{a}{b} \div c=\dfrac{a}{b} \div \dfrac{1}{c}</Tex>

</Note>

Let's take some examples of dividing two non-zero fraction by whole number, and whole number by fraction.

<Example type="example" num="2"/>

Divide the fractions below.

1) <Tex>5\div \dfrac{8}{3}</Tex>

2) <Tex>-9\div \dfrac{3}{4}</Tex>

3) <Tex>\dfrac{11}{6}\div 8</Tex>

4) <Tex>\dfrac{-7}{12} \div (-6)</Tex>

<Example type='solution' />

1) To divide a whole number by a fraction, we must find the reciprocal of the fraction and then performe the multiplication of whole number by fraction.

<Tex block>5\div \dfrac{8}{3}=5 \times \dfrac{3}{8}=\dfrac{5\times3}{8}=\dfrac{15}{8}</Tex>

2) By the same way as in the example above:

<Tex block>
\begin{align*}
-9\div \dfrac{3}{4}&=-9 \times \dfrac{4}{3}\\\
&=\dfrac{-9\times4}{3}\\\
&=\dfrac{-3 \times \cancel{3}\times4}{\cancel{3}\times1}\\\
&=\dfrac{-12}{1}\\\
&=-12
\end{align*}
</Tex>

3) To divide a fraction by a whole number, we must find the reciprocal of the whole number and then performe the multiplication of whole number by fraction.

<Tex block>
\begin{align*}
\dfrac{11}{6}\div 8 &=\dfrac{11}{6} \times \dfrac{1}{8}\\\
&=\dfrac{11\times1}{6\times8}\\\
&=\dfrac{11}{48}\\\
\end{align*}
</Tex>

4) By the same way as in the example above:

<Tex block>
\begin{align*}
\dfrac{-7}{12} \div (-6) &=\dfrac{-7}{12} \times \dfrac{1}{-6}\\\
&=\dfrac{-7\times1}{12\times(-6)}\\\
&=\dfrac{-7}{-72}\\\
&=\dfrac{7}{72}\\\
\end{align*}
</Tex>

<Related>
<Link to="../basic-algebra/types-fratcions-and-basics" as="/lessons/basic-algebra/types-fratcions-and-basics">Types of Fratcions and Basics</Link> 
<Link to="../basic-algebra/adding-subtracting-fractions-like-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-like-denominators">Adding and Subtracting Fractions with Like Denominators</Link> 
<Link to="../basic-algebra/multiplying-fractions-by-whole-number-by-reciprocal" as="/lessons/basic-algebra/multiplying-fractions-by-whole-number-by-reciprocal">Multiplying Fractions Multiplying Fraction by Whole Number and by its Reciprocal</Link> 
<Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">Finding Greatest Common Factor (GCF)</Link>
</Related>

Learn how to divide two fractions and learn how to divide a fraction by whole number and a whole number by fraction. This lesson is illustrate by many examples.

Dividing Fractions - Divide Fractions and Whole Numbers

# Comparing Fractions Using Three Different Methods

Undoubtedly you know how to compare two integers, if they have two different signs, the positive number is greater, if they have the same sign we have two cases:

If both numbers are positive, then the farthest from zero is greater
If both numbers are negative, then the farthest from zero is the smaller

But how do we compare two mixed numbers? For example, if you ate three-quarters of your pizza, and your brother ate two-thirds of his pizza, which one ate the most?
This is what we will try to answer in this lesson, comparing two fractions in three different ways.

To compare two fractions, we can use three different methods:

* Decimal Method

* The Same Denominator Method

* The Same Numerator Method

## Decimal Method

This method is based on dividing the numerator by the denominator and then comparing the two whole parts. If they are equal we compare the tenths place, if they are equal we compare the hundredths place, if they are equal we compare the thousandths place and so on.

<Example type="example" num="1"/>

Compare the frations: <Tex>\dfrac{3}{4}</Tex> and <Tex>\dfrac{2}{3}</Tex>

<Example type="solution"/>

Converting each fraction to a decimal:

<Tex>\dfrac{3}{4}=3\div4=0.75</Tex> 

<Tex>\dfrac{2}{3}=2\div3=0.6666...</Tex>

The two whole parts are equal which is 0. Move to the next.

The digit in tenth place of <Tex>.75</Tex> is 7, and the digit in tenth place of <Tex>.66669...</Tex> is 6

So, <Tex>\dfrac{3}{4} > \dfrac{2}{3}</Tex>

<Example type="example" num="2"/>

Compare the frations: <Tex>\dfrac{15}{8}</Tex> and <Tex>\dfrac{9}{5}</Tex>

<Example type="solution"/>

Converting each fraction to a decimal:

<Tex>\dfrac{15}{8}=15\div8=1.875</Tex> 

<Tex>\dfrac{9}{5}=9\div5=1.800</Tex>

The two whole parts are equal which is 1. Move to the next.

The digit in tenth place of <Tex>.875</Tex> is 8, and the digit in tenth place of <Tex>.800</Tex> is 8. Move to the next.

The digit in hundredths place of <Tex>.875</Tex> is 7, and the digit in tenth place of <Tex>.800</Tex> is 0.

So, <Tex>\dfrac{15}{8} > \dfrac{9}{5}</Tex>

_________

## The Same Denominator Method

When we have two fractions with **like denominators**, it's easy to compare them, The greatest fraction is the fraction with the greatest numerator.

<Note>

Let <Tex>\dfrac{a}{c}</Tex> and <Tex>\dfrac{b}{c}</Tex> represent two fractions.

If <Tex>a>c</Tex>, so <Tex>\dfrac{a}{c}>\dfrac{b}{c}</Tex>

</Note>

When we have two fractions with **unlike denominators**, we make them the same (check out <Link to="../basic-algebra/adding-subtracting-fractions-unlike-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-unlike-denominators">this lesson</Link>)

<Example type="example" num="3"/>

Compare the frations: <Tex>\dfrac{7}{18}</Tex> and <Tex>\dfrac{11}{18}</Tex>

<Example type="solution"/>

The two fractions have the same denominator, we compare the numerators. It's clear that <Tex> 7 < 11</Tex>.

Thus, <Tex>\dfrac{7}{18} < \dfrac{11}{18}</Tex>

<Example type="example" num="4"/>

Compare the frations: <Tex>\dfrac{9}{4}</Tex> and <Tex>\dfrac{15}{7}</Tex>

<Example type="solution"/>

The two fractions withy unlike denominators, we need to make the the same.

The Greatest Common Factor: <Tex>gcf(7,4)=28</Tex>.

The equivalent fractions:

<Tex>\dfrac{9}{4}=\dfrac{63}{28}</Tex> 

<Tex>\dfrac{15}{7}=\dfrac{60}{28}</Tex>

And since <Tex> 63 > 60</Tex>, therefore <Tex> \dfrac{63}{28} > \dfrac{60}{28}</Tex>

Hence, <Tex>\dfrac{9}{4} > \dfrac{15}{7}</Tex>

______

## The Same Numerator Method

When we have two fractions with **like numerators**, it's easy to compare them. But pay attention, the greatest fraction is the fraction with the smallest denominator. Because as we increase the denominator, the fraction decreases

<Example type="example" num="5"/>

Compare the frations: <Tex>\dfrac{17}{8}</Tex> and <Tex>\dfrac{17}{7}</Tex>

<Example type="solution"/>

The two fractions withy like numerators, we need to compare the denominators. It's clear that <Tex>8 > 7</Tex>. Thus,

<Tex>\dfrac{17}{8} < \dfrac{17}{7}</Tex>

<Example type="example" num="6"/>

Compare the frations: <Tex>\dfrac{5}{2}</Tex> and <Tex>\dfrac{5}{3}</Tex>

<Example type="solution"/>

The two fractions withy like numerators, we need to compare the denominators. It's clear that <Tex>2 < 3</Tex>. Thus,

<Tex>\dfrac{5}{2} > \dfrac{5}{3}</Tex>

## Comparing Mixed Fractions

To compare two mixed fractions, we need to convert them to improper fractions and apply the rules of comparing two fractions. (check out <Link to="../basic-algebra/converting-improper-fractions-mixed-fractions" as="/lessons/basic-algebra/converting-improper-fractions-mixed-fractions">this lesson</Link>)

<Example type="example" num="7"/>

Compare the mixed frations: <Tex>3\dfrac{4}{5}</Tex> and <Tex>2\dfrac{7}{8}</Tex>

<Example type="solution"/>

Firs, we need to convert the mixed fractions to improper fractions using this method:

<Note>

Let <Tex>a\dfrac{b}{c}</Tex> is a mixed fraction. To convert the mixed fraction to improper fraction we use the following formula:

<Tex block>a\dfrac{b}{c}=\dfrac{a\times c+b}{c}</Tex>

</Note>

<Tex>3\dfrac{4}{5}=\dfrac{3\times5+4}{5}=\dfrac{19}{5}</Tex> 

<Tex>2\dfrac{7}{8}=\dfrac{2\times8+7}{8}=\dfrac{23}{8}</Tex>

The improper fractions obtained are unlike denominators, we need to make them the same.

The Greatest Common Factor: <Tex>gcf(5,8)=40</Tex>.

The equivalent fractions:

<Tex>\dfrac{19}{5}=\dfrac{152}{40}</Tex> 

<Tex>\dfrac{23}{8}=\dfrac{115}{40}</Tex>

And since <Tex> 152 > 115</Tex>, therefore <Tex> \dfrac{152}{40} > \dfrac{115}{40}</Tex>

Hence, <Tex>3\dfrac{4}{5} > 2\dfrac{7}{8}</Tex>

Learn how to compare fractions using three ways: decimals, the same denominator, and the same numerator. This includes comparing mixed fractions.

Comparing Fractions Using Three Different Methods

# Solving Simple Linear Equation In One Variable

Before you got to know the concept of equations, you used them in your daily life without realizing it. do you believe that?

Yes, you used mental calculation to solve simple equations, perhaps before you entered school. For example, if I tell you what number if we add it to 8 we get 10. Here you do a mental calculation that includes addition and subtraction, but in a very simple way only, where you subtract 8 from 10 and you get 2, and to make sure of your solution you add the number 2 to 8 and you get 10. That's how simply all the children passed this stage.

The same thing if I told you what number that if we multiply it by 5 we get 20. You do a simple mental calculation that includes multiplication and division, where you divide 20 by 5 to get 4, and to make sure of your answer multiply 4 by 5 to get 20. This is what we will simply rely on in this lesson to solve very simple linear equations. Continue reading the article until the end if you are interested in this lesson.

## The concept of an equation

**An Equation** is a mathematical expression consisting of one or more unknown values, and a right and a left side. These two sides are always separated by the <Tex>"="</Tex> symbol.

**A Linear Equation** is an equation with one unknown. This unknown can be in one or several terms.

*Examples of Linear Equations:* 

<Tex>2x+6=0 \quad ; \quad-3x+1=-7 \quad ; \quad-x-4=3x-8</Tex>

**The root of the Equation** is the value of the unknown number that we get after solving the equation. When we plug this number back into the equation, we should get both sides equal.

## Standard Form Of A Linear Equation In One Variable

<Note>

Let <Tex>a</Tex> and <Tex>b</Tex> represent two real numbers with <Tex>a\neq0</Tex>

The standard form of a linear equation with one unknown <Tex>x</Tex> is:

<Tex block>ax+b=0</Tex>

The solution of this equation is <Tex>x=\dfrac{-b}{a}</Tex>

</Note>

In addition to linear equations, there are other types of equations, which we will get acquainted with next time:

- Linear Equations with two unknowns;

- Quadratic equations(one unknown with second degree);

- Rational equations

- Cubic equations

- etc.

## How To Solve Simple Linear Equation

To solve an equation with one unknown, the unknown must be isolated in the equation by following these four steps in this order.

* **Transfer Rule:** We shift the unknowns to the left side and shift the known numbers to the right side by changing the sign of the transferred terms;

* Simplify the expressions in the two sides obtained in the previous step;

* In this step, we multiply both sides by the reciprocal of the unknown coefficient;

* Simplify the expressions in the two sides;

___________

<Example type="example" num="1"/>

Solve the equation: <Tex>4x-1=7</Tex>

<Example type="solution"/>

The unknown of this equation is <Tex>x</Tex>, and its coefficient is 4.

* We must transfer -1 from the left side to the right side and changing the sign.

<Tex block>4x=7+1</Tex>

* Simplify the expression obtained.

<Tex block>4x=8</Tex>

* We must myltiply the both sides by the reciprocal of the coefficient of <Tex>x</Tex>. i.e, we multiply by <Tex>\dfrac{1}{4}</Tex>

<Tex block>\dfrac{1}{4}\times4x=8\times\dfrac{1}{4}</Tex>

<Tex block>\dfrac{4x}{4}=\dfrac{8}{4}</Tex>

* Simplify the expression to isolate <Tex>x</Tex>.

<Tex block>\dfrac{\cancel{4}\times x}{\cancel{4}}=\dfrac{\cancel{4}\times2}{\cancel{4}}</Tex>

<Tex block>x=2</Tex>

* The final answer is <Tex>x=2</Tex>. We substitute the value of <Tex>x</Tex> by 2 into the equation to make sure the answer is correct.

<Tex block>
\begin{align*}
4x-1=7\\\
4\times2-1=7\\\
8-1=7\\\
7=7
\end{align*}
</Tex>

Therefore, The answer is correct.

<Example type="example" num="2"/>

Solve the equation: <Tex>-3x+5=-10</Tex>

<Example type="solution"/>

The unknown of this equation is <Tex>x</Tex>, and its coefficient is -3.

* We must transfer 5 from the left side to the right side and changing the sign.

<Tex block>-3x=-10-5</Tex>

* Simplify the expression obtained.

<Tex block>-3x=-15</Tex>

* We must myltiply the both sides by the reciprocal of the coefficient of <Tex>x</Tex>. i.e, we multiply by <Tex>\dfrac{1}{-3}</Tex>

<Tex block>\dfrac{1}{-3}\times(-3)x=-15\times\dfrac{1}{-3}</Tex>

<Tex block>\dfrac{-3x}{-3}=\dfrac{-15}{-3}</Tex>

* Simplify the expression to isolate <Tex>x</Tex>.

<Tex block>\dfrac{\cancel{-3}\times x}{\cancel{-3}}=\dfrac{\cancel{-3}\times5}{\cancel{-3}}</Tex>

<Tex block>x=5</Tex>

* The final answer is <Tex>x=5</Tex>. We substitute the value of <Tex>x</Tex> by 5 into the equation to make sure the answer is correct.

<Tex block>
\begin{align*}
-3x+5=-10\\\
-3\times5+5=-10\\\
-15+5=-10\\\
-10=-10
\end{align*}
</Tex>

Therefore, The answer is correct.

<Example type="example" num="3"/>

Solve the equation: <Tex>8x+5=3x+4</Tex>

<Example type="solution"/>

In this expression, we have the unknown x found on both sides with different coefficients.

* We must transfer 5 from the left side to the right side and, transfer <Tex>3x</Tex> from the right side to the left side and changing the signs.

<Tex block>8x-3x=4-5</Tex>

* Simplify the expression obtained. In this case we simplify the left side by Combining Like Terms.

<Tex block>5x=-1</Tex>

* We must myltiply the both sides by the reciprocal of the coefficient of <Tex>x</Tex>. i.e, we multiply by <Tex>\dfrac{1}{5}</Tex>

<Tex block>\dfrac{1}{5}\times5x=-1\times\dfrac{1}{5}</Tex>

<Tex block>\dfrac{5x}{5}=\dfrac{-1}{5}</Tex>

* Simplify the expression to isolate <Tex>x</Tex>.

<Tex block>\dfrac{\cancel{5}\times x}{\cancel{5}}=\dfrac{-1}{5}</Tex>

<Tex block>x=\dfrac{-1}{5}</Tex>

* The final answer is <Tex>x=\dfrac{-1}{5}</Tex>. We substitute the value of <Tex>x</Tex> by <Tex>\dfrac{-1}{5}</Tex> into the equation to make sure the answer is correct.

<Tex block>
\begin{align*}
8\times \dfrac{-1}{5}+5 &=3 \times \dfrac{-1}{5}+4\\\
\dfrac{-8}{5}+5 &=\dfrac{-3}{5}+4\\\
\dfrac{-32}{20}+\dfrac{100}{20} &=\dfrac{-12}{20}+\dfrac{80}{20}\\\
\dfrac{-32+100}{20} &= \dfrac{-12+80}{20} \\\
\dfrac{68}{20} &= \dfrac{68}{20} \\\
\end{align*}
</Tex>

Therefore, The answer is correct.

<Example type="example" num="4"/>

Solve the equation: <Tex>-5(x-3)+2=3(x-4)+2x-1</Tex>

<Example type="solution"/>

This equation is also a linear equation, but it contains expressions in parentheses. So it is necessary to first get rid of these parentheses using the <Link to="../basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms" as="/lessons/basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms">distributive property</Link>.

* Applying distributive property;

<Tex block>
\begin{align*}
-5(x-3)+2 &=3(x-4)+2x-1 \\\
-5\times x-(-5)\times3+2 &=3\times x-3\times4+2x-1\\\
-5x+15+2 &= 3x-12+2x-1\\\
-5x+17 &=5x-13
\end{align*}
</Tex>

* We must transfer 17 from the left side to the right side and, transfer <Tex>5x</Tex> from the right side to the left side and changing the signs.

<Tex block>-5x-5x =-13-17</Tex>

* Simplify the expression obtained. In this case we simplify the left side by <Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms" as="/lessons/basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">Combining Like Terms</Link>.

<Tex block>-10x=-30</Tex>

* We must myltiply the both sides by the reciprocal of the coefficient of <Tex>x</Tex>. i.e, we multiply by <Tex>\dfrac{1}{-10}</Tex>

<Tex block>\dfrac{1}{-10}\times(-10)x=-30\times\dfrac{1}{-10}</Tex>

<Tex block>\dfrac{-10x}{-10}=\dfrac{-30}{-10}</Tex>

* Simplify the expression to isolate <Tex>x</Tex>.

<Tex block>\dfrac{\cancel{-10} \times x}{\cancel{-10}}=\dfrac{\cancel{-10} \times 3}{\cancel{-10}}</Tex>

<Tex block>x=3</Tex>

* The final answer is <Tex>x=3</Tex>. We substitute the value of <Tex>x</Tex> by 3 into the equation to make sure the answer is correct.

<Tex block>
\begin{align*}
-5(x-3)+2&=3(x-4)+2x-1\\\
-5(3-3)+2&=3(3-4)+2\times3-1\\\
-5\times0+2&=3\times(-1)+6-1\\\
2&=-3+6-1\\\
2&=2
\end{align*}
</Tex>

Therefore, The answer is correct.

<Example type="example" num="5"/>

Solve the equation: <Tex>\dfrac{2x}{3}-1=1</Tex>

<Example type="solution"/>

The unknown of this equation is <Tex>x</Tex>, and its coefficient is <Tex>\dfrac{2}{3}</Tex>.

* We must transfer -1 from the left side to the right side and changing the sign.

<Tex block>\dfrac{2x}{3}=1+1</Tex>

* Simplify the expression obtained.

<Tex block>\dfrac{2x}{3}=2</Tex>

* We must myltiply the both sides by the reciprocal of the coefficient of <Tex>x</Tex>. i.e, we multiply by <Tex>\dfrac{3}{2}</Tex>

<Tex block>\dfrac{3}{2}\times\dfrac{2x}{3}=2\times\dfrac{3}{2}</Tex>

<Tex block>\dfrac{6x}{6}=\dfrac{6}{2}</Tex>

* Simplify the expression to isolate <Tex>x</Tex>.

<Tex block>\dfrac{\cancel{6}\times x}{\cancel{6}}=\dfrac{\cancel{2}\times3}{\cancel{2}}</Tex>

<Tex block>x=3</Tex>

* The final answer is <Tex>x=3</Tex>. We substitute the value of <Tex>x</Tex> by 3 into the equation to make sure the answer is correct.

<Tex block>
\begin{align*}
\dfrac{2x}{3}-1=1\\\
\dfrac{2\times3}{3}-1=1\\\
\dfrac{6}{3}-1=1\\\
2-1=1\\\
1=1
\end{align*}
</Tex>

Therefore, The answer is correct.

<Example type="example" num="6"/>

Solve the equation: <Tex>\dfrac{2x-1}{3}-5=1-2x-\dfrac{x-1}{4}</Tex>

<Example type="solution"/>

The best way to solve such equations in my opinion is to make all terms have the same denominator. In order to reduce it and then make it easier for us to solve the equation.

Learn how to add and subtract two fractions with:

* <Link to="../basic-algebra/adding-subtracting-fractions-like-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-like-denominators">Like Denominators</Link>

* <Link to="../basic-algebra/adding-subtracting-fractions-unlike-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-unlike-denominators">Unlike Denominators</Link>

The same denominator is the <Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">Greatest Common Factor</Link> of all denominators in the equation. In our case, We must find the GCF of 3 and 4. It's clear that <Tex>GCF(3,4)=12</Tex>, so:

<Tex block>
\begin{align*}
\frac{2x-1}{3}-5 &=1-2x-\frac{x-1}{4} \\\
\frac{4(2x-1)}{12}-\frac{60}{12} &=\frac{12}{12}-\frac{24x}{12}-\frac{3(x-1)}{12} \\\
\frac{4(2x-1)}{\cancel{12}}-\frac{60}{\cancel{12}} &=\frac{12}{\cancel{12}}-\frac{24x}{\cancel{12}}-\frac{3(x-1)}{\cancel{12}}\\\
4(2x-1)-60 &= 12-24x-3(x-1)\\\
\end{align*}
</Tex>

After that, we must applying the distributive property.

<Tex block>
\begin{align*}
4(2x-1)-60 &= 12-24x-3(x-1)\\\
4\times2x-4\times1-60 &=12-24x-3\times x-3\times(-1)\\\
8x-4-60 &= 12-24x-3x+3\\\
8x-64 &=-27x-21
\end{align*}
</Tex>

Then, apllying the transfer rule.

<Tex block>
\begin{align*}
8x-64 &=-27x-21\\\
8x+27x=-21+64\\\
35x=43\\\
\end{align*}
</Tex>

And multiplying the both sides by <Tex>\dfrac{1}{35}</Tex>.

<Tex block>\dfrac{1}{35}\times35x=\dfrac{1}{35}\times43</Tex>

<Tex block>\dfrac{35x}{35}=\dfrac{43}{35}</Tex>

<Tex block>\dfrac{\cancel{35}\times x}{\cancel{35}}=\dfrac{43}{35}</Tex>

<Tex block>x=\dfrac{43}{35}</Tex>

The final answer is <Tex>x=\dfrac{43}{35}</Tex>

_______________

<Related>

<Link to="../basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms" as="/lessons/basic-algebra/distributive-property-multiplication-addition-substraction-combine-like-terms">distributive property</Link> 
<Link to="../basic-algebra/adding-subtracting-fractions-like-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-like-denominators">Adding and Subtracting Fractions with Like Denominators</Link> 
<Link to="../basic-algebra/adding-subtracting-fractions-unlike-denominators" as="/lessons/basic-algebra/adding-subtracting-fractions-unlike-denominators">Adding and Subtracting Fractions with Unlike (Different) Denominators</Link> 
<Link to="../arithmetic/greates-common-factor-least-common-multiple" as="/lessons/arithmetic/greates-common-factor-least-common-multiple">Finding Greatest Common Factor (GCF)</Link>

</Related>

Learn how to solve simple linear equations in one unknown withe the standard form ax+b=0. This lesson is illustrated by many examples with solutions.

Explorer all lessons of Basic Algebra