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

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

Advanced Algebra

# Arithmetic Series: Sum of the first n terms of an arithmetic sequence

<Related>
<Link to="../intermediate-algebra/arithmetic-sequence-definition-and-examples" as="/lessons/intermediate-algebra/arithmetic-sequence-definition-and-examples">Arithmetic Sequence (Progression): Definition and Examples</Link> 
<Link to="../intermediate-algebra/arithmetic-sequence-formula-general-term" as="/lessons/intermediate-algebra/arithmetic-sequence-formula-general-term">Arithmetic Sequence Formula: How to Find General Term</Link> 
<Link to="../intermediate-algebra/arithmetic-sequence-formula-advanced-general-term" as="/lessons/intermediate-algebra/arithmetic-sequence-formula-advanced-general-term">Arithmetic Sequence Formula: Advanced General Term</Link> 
<Link to="../intermediate-algebra/arithmetic-sequence-mean-formula" as="/lessons/intermediate-algebra/arithmetic-sequence-mean-formula">Arithmetic Sequence Formula: Arithmetic Mean</Link>

</Related>

If we ask a student to add numbers from 1 to 1000, for example, how long will it take him to complete this operation?

Perhaps it would be easy if we asked him to add numbers from 1 to 10. But when a series of operations is long, accomplishing them becomes arduous and requires time and effort.

Can math shorten this task? In other words, is there a mathematical formula that directly calculates the sum from 1 to 1000?

Note with me carefully that the expression: 

<Tex block>1,\\; 2,\\; 3,\\; ......\\; ,\\; 999,\\; 1000</Tex>

It is an arithmetic sequence whose first term is 1 and its last term is 1000 and the common difference is 1. (The points mean that the sequence continues in the same pattern).

The goal of this lesson is to find how to add these terms in this way: 

<Tex block>1+2+3+......+999+1000</Tex>

This is what we will try to answer in this section of the Arithmetic Series lesson.

## Arithmetic Sequence vs Arithmetic Series

The difference between Arithmetic Sequence and Arithmetic Series is:

<Note>
- An arithmetic sequence is a set of terms in which there is a relation between the term and the term before or after it <Tex>a_{n+1}=a_n+d</Tex>
- In the expression, we separate these terms with a comma: <Tex>a_1; a_2; a_3; a_4; .....; a_n</Tex>
- Terms of an arithmetic sequence are arranged (either increasing or decreasing)
- An arithmetic series is the sum of the terms of an arithmetic series: <Tex>a_1+a_2+a_3+.....+a_n</Tex>
</Note>

The sum of the first n terms of this arithmetic sequence denoted by <Tex>S_n</Tex> and it is called the partial sum and we write <Tex>S_n=a_1+a_2+a_3+...+a_n</Tex>

Let's take the following example:

Let <Tex>(a_n)</Tex> be an arithmetic sequence where: <Tex>\\{1; \\;2;\\;3;\\;4;\\;5\\}</Tex>

Then,

<Tex block>
\begin{align*}
S_1 &=1 \\\
S_2 &=1+2 \\\
S_3 &=1+2+3 \\\
S_4 &=1+2+3+4 \\\
S_5 &=1+2+2+4+5
\end{align*}
</Tex>

But how do we find <Tex>S_{1000}</Tex>? Adding <Tex>1000</Tex> terms in this way, as we said before, requires a lot of time and effort. So we need a mathematical formula that sums it all up.

we know that:

<Tex block>S\_n=a\_1+a\_2+a\_3+......+a\_{n-1}+a\_n \tag{1}</Tex>

We rewrite this equation, but this time we start descending from the last term to the first term.

<Tex block>S\_n=a\_n+a\_{n-1}+......+a\_3+a\_2+a\_1 \tag{2}</Tex>

Adding the equations (1) and (2) side by side:

<Tex block>
\begin{align*}
S\_n+S\_n &= (a\_1+a\_n)+(a\_2+a\_{n-1})+......+(a\_{n-1}+a\_2)+(a\_n+a\_1) \\\
2S_n &= (a_1+a_n)+(a_1+d+a_n-d)+......+(a_n-d+a_1+d)+(a_n+a_1) \\\
2S\_n &= \underbrace{(a\_1+a\_n)+(a\_1+a\_n)+......+(a\_n+a\_1)+(a\_n+a\_1)}\_{\text{n terms}} \\\
2S_n &=n \cdot (a_1+a_n) \\\
S_n &=\frac{n \cdot (a_1+a_n)}{2}
\end{align*}
</Tex>

<Note>

Let <Tex>(a_n)</Tex> an arithmetic sequence.

<Tex block>\\{a\_1; \\;a\_2; \\ a\_3; \\; ......\\; ; \\; a\_{n-1}; \\; a\_n \\}</Tex>

The sum of n first terms of the sequence is:

<Tex block>S_n =\frac{n \cdot (a_1+a_n)}{2}</Tex> 

- <Tex>a_1</Tex> : first term 
- <Tex>a_n</Tex> : nth term 
- <Tex>n</Tex> : number of terms to add
</Note>

The expression <Tex>S_n</Tex> is written in terms of the general term <Tex>a_n</Tex>:

<Tex block>a_n=a_1+(n-1)d</Tex>

We substitute <Tex>a_n</Tex> into the <Tex>S_n</Tex> formula:

<Tex block>
\begin{align*}
S_n &= \frac{n \cdot (a_1+a_n)}{2} \\\
S_n &= \frac{n \cdot (a_1+a1+(n-1)d)}{2} \\\
S_n &= \frac{n \cdot (2a_1+(n-1)d)}{2} \\\
S_n &= na_1+\frac{nd(n-1)}{2}
\end{align*}
</Tex>

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

### Some illustrative examples of arithmetic series.

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

Let's go back to the example we started from in this lesson. What is the sum of the numbers from 1 to 1000?

<Example type="solution"/>

This example is very easy, then we will be able to calculate more complex sums.

The numbers from 1 to 1000 are an arithmetic sequence:

- Its first term is: <Tex>a_1=1</Tex>
- Its last term is: <Tex>a_{1000}=1000</Tex>
- The number of terms is: <Tex>1000</Tex>

therefore,

<Tex block>
\begin{align*}
S\_n &=\frac{n \cdot (a\_1+a\_n)}{2} \\\
S\_{1000} &= \frac{1000 \times (a\_1+a\_{1000})}{2} \\\
&= \frac{1000 \times (1+1000)}{2} \\\
&= \frac{1000 \times 1001}{2} \\\
&= 500500
\end{align*}
</Tex>

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

Find the sum of the terms of the following arithmetic series:

<Tex block>3;\\;8;\\;13;\\;18;\\;23;\\; \cdots; \\; 233</Tex>

<Example type="solution"/>

This example differs from the previous example in that the number of terms <Tex>n</Tex> in this example is unknown.

To calculate n we return to the formula that enables us to calculate the number of terms in an arithmetic sequence:

<Tex block>n = \frac{a_n-a_1}{d}+1</Tex>

- <Tex>a_n</Tex> is the nth term. (<Tex>a_n=233</Tex>)
- <Tex>a_1</Tex> is the first term. (<Tex>a_1=3</Tex>)
- <Tex>d</Tex> is the common difference. (<Tex>d=8-3=5</Tex>)

therefore:

<Tex block>
\begin{align*}
n &=\frac{233-3}{5}+1 \\\
n &=\frac{230}{5}+1 \\\
n &=47
\end{align*}
</Tex>

So the partial sum of <Tex>S_{47}</Tex> is:

<Tex block>
\begin{align*}
S_n &=\frac{n(a_1+a_n)}{2} \\\
S_{47} &=\frac{47(3+233)}{2} \\\
S_{47} &= \frac{47 \times 236}{2} \\\
S_{47} &= 472
\end{align*}
</Tex>

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

Find the sum of the first 69 terms of the following arithmetic sequence:

<Tex block>2; \\; 11; \\; 20; \\; 29; \\; 38; \\; \cdots</Tex>

<Example type="solution"/>

Always in the same way, we add the terms using the following formula:

<Tex block>
\begin{align*}
S\_{69} &= a\_1+a\_2+a\_3+\cdots +a\_{69} \\\
&= \frac{n(a\_1+a\_{69})}{2}
\end{align*}
</Tex>

- <Tex>a_1</Tex> is the first term. (<Tex>a_1=2</Tex>)
- <Tex>n</Tex> is the numbers of terms. (<Tex>n=69</Tex>)
- <Tex>a_{69}</Tex> is the 69th term. (we must calculate it using the term general formula)

<Tex block>a_{69}=a_1+(69-1)d</Tex> 

Here we need to know the common difference: <Tex>d=11-2=9</Tex>

Therefore,

<Tex block>
\begin{align*}
a_{69} &=a_1+(69-1)\times9 \\\
&=2+68\times9 \\\
&=614
\end{align*}
</Tex>

Thus,

<Tex block>
\begin{align*}
S_{69} &=\frac{69(2+614)}{2} \\\
&=\frac{69\times616}{2} \\\
&= 21252
\end{align*}
</Tex>

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

let <Tex>(a_n)</Tex> an arithmetic sequence in which <Tex>a_5=18</Tex> and <Tex>a_{13}=-14</Tex>

Calculate <Tex>S_{70}</Tex>.

<Example type="solution"/>

In this example, we only know the 5th term and the 13th term.

Let's calculate the common difference <Tex>d</Tex> from the general term formula:

<Tex block>
\begin{align*}
a_n &=a_p+(n-p)d \\\
a_{13} &= a_5 +(13-5)d \\\
-14 &= 18+8d \\\
8d &= -14-18 \\\
8d &=-32 \\\
d &= -4
\end{align*}
</Tex>

Therefor,

<Tex block>
\begin{align*}
S\_n &= \frac{n(a_1+a\_n)}{2} \\\
S\_{70} &= \frac{70(a\_1+a\_{70})}{2} 
\end{align*}
</Tex>

But we don't know <Tex>a_1</Tex> and <Tex>a_{70}</Tex>. Let's calculate it.

<Tex block>
\begin{align*}
a_{70} &= a_5+(70-5)\times(-4)\\\
&=18+65\times(-4)\\\
&=18+(-260)\\\
&=-242
\end{align*}
</Tex>

And,

<Tex block>
\begin{align*}
a_1 &=a_5+(1-5)(-4)\\\
&=18+(-4)\times (-4)\\\
&=18+16\\\
&=34
\end{align*}
</Tex>

Therefor,

<Tex block>
\begin{align*}
S\_{70} &=\frac{70\times(a\_1+a\_{70})}{2} \\\
&=\frac{70\times(34+(-242))}{2} \\\
&= 35\times(-208) \\\
&=-7280
\end{align*}
</Tex>

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

If <Tex>-5+\cdots \cdots +49=616</Tex> the sum of terms of an arithmetic sequence, How many terms are there in this sequence?

<Example type="solution"/>

In this example we know the sum of the terms <Tex>(S_n=616)</Tex>, but we do not know the number of these terms <Tex>(n=?)</Tex>

Let's turn the problem into an algebraic expression:

<Tex block>a_1=-5\\; ; \\;a_n=49\\; ; \\;S_n=616</Tex>

The partial sum formula is:

<Tex block>
\begin{align*}
S_n&=\frac{n(a_1+a_n)}{2}\\\
\frac{n(-5+49)}{2}&=616\\\
\frac{44n}{2}&=616\\\
22n&=616\\\
n&=\frac{616}{22}\\\
n&=28
\end{align*}
</Tex>

So, the arithmetic sequence <Tex>-5;\cdots \cdots ;49</Tex> has 28 terms (Now we can calculate the common difference).

learn how to calculate the sum of first n terms of an arithmetic series, and understand the main difference between arithmetic sequence and arithmetic series.

Arithmetic Series Formula

# Systems Of Linear Equations (Simultaneous Linear Equations)

Suppose you buy a book for an amount where you give the seller $20 and they return $8. Can we know the price of the book? Of course, the matter is very simple, we can express this problem with a simple equation <Tex>20-x=8</Tex>, where <Tex>x</Tex> is the price of the book (One Unknown).

But now suppose you bought 3 books and 9 notebooks for $120. Can we know the price of one book and the price of one notebook? Of course not, this time it became a little more complicated. We can express this problem by the equation <Tex>3x+9y=120</Tex> where <Tex>x</Tex> is the price of the book and <Tex>y</Tex> is the price of the book.

To solve this problem, we need another equation with the same unknowns as the first equation. For example <Tex>2x+5y=75</Tex>.

## Define System of Linear Equations

<Note>
Every algebraic expression is written as:

<Tex block>
\begin{cases}
 ax+by=c \\\
 dx+ey=f
\end{cases}
</Tex>

is called a **system of linear equations** (In some references it is also called **Simultaneous Linear Equations**).

Whereas <Tex>a,\\, b,\\, c,\\, d,\\, e</Tex> and <Tex>f</Tex> are real numbers. 

The solution of the system is the ordered pair <Tex>(x,\\, y)</Tex>.

</Note>

## Graphical Interpretation of a System of Linear Equations

the algebraic expression <Tex>ax+by=c</Tex> is an equation of a straight line that can be written in the form:<Tex>y=mx+p</Tex>. So, the system of two linear equations give us the equations of two straight lines (graphes) in a cartesian coordinate system.
the relative positions of these two lines express the solution of the system:

1. The graphs intersect at exactly one point, which gives the (single) ordered-pair solution of the system. The system is consistent and the equations are independent.

2. The graphs are parallel lines, so there is no solution and the solution set is ∅. The system is inconsistent and the equations are independent.

3. The graphs are identical (lines coincide), and there are an infinitely many solutions. The system is consistent and the equations are dependent.

## Solving Systems of Linear Equations

To solve a linear system of two equations, we have three different methods:

### Graphical Solution

Write the equations of linear systems in the form of <Tex>y=mx+p</Tex>, and then represent the lines in the coordinate plane. As we explained earlier, these two lines can be intersecting, parallel, or coincident.

We will try to take an example for each of the three cases.

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

Let's solve the following system of linear equations:

<Tex block>
\begin{cases}
 x+y &=5 \\\
 2x-y &=4
\end{cases}
</Tex>

We rewrite the two equations as follows in order to make it easier to represent them in the coordinate plane.

<Tex block>
\begin{cases}
 y &=-x+5 \\\
 y &=2x-4 
\end{cases}
</Tex>

<Image src="/imgs/lessons/advalgebra/system_example1.png" alt="graphical solution of linear system of two equations" width='500px' height='346px'/>

From the graph, we notice that the two lines intersect at the point with coordinates <Tex>(3\\;;\\; 2)</Tex>.

So the solution of the system is the ordered-pair <Tex>(3\\;;\\; 2)</Tex>.

The ordered-pair <Tex>(3\\;;\\; 2)</Tex> that means <Tex>x=3</Tex> and <Tex>y=2</Tex>.

**Check the solution**

We substitute <Tex>x</Tex> and <Tex>y</Tex> in the two equations:

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

<Tex block>2x-y=2\times3-2=6-2=4</Tex>

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

Let's solve the following system of linear equations:

<Tex block>
\begin{cases}
 2x+y &=-2 \\\
 2x+y &=3
\end{cases}
</Tex>

We rewrite the two equations as follows in order to make it easier to represent them in the coordinate plane.

<Tex block>
\begin{cases}
 y &=-2x-2 \\\
 y &=-2x+3 
\end{cases}
</Tex>

<Image src="/imgs/lessons/advalgebra/system_example2.png" alt="graphical solution of linear system of two equations" width='500px' height='421px'/>

From the graph, we notice that the two lines are parallel. Then there is no solution to the system.

<Note
>We say that two lines are parallel if they have the same slope.

</Note>

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

Let's solve the following system of linear equations:

<Tex block>
\begin{cases}
 8x-2y &=-4 \\\
 -4x+y &=2
\end{cases}
</Tex>

We rewrite the two equations as follows in order to make it easier to represent them in the coordinate plane.

<Tex block>
\begin{cases}
 y &=4x+2 \\\
 y &=4x+2 
\end{cases}
</Tex>

<Image src="/imgs/lessons/advalgebra/system_example3.png" alt="graphical solution of linear system of two equations" width='434px' height='411px'/>

From the graph, we notice that the two lines are identical. Then there is infinitely many solutions.

<Note>

We say that two lines are coincide if they have the same slope and the same ordinate at the origin.

</Note>

### Algebraic Solution: Substitution Method

In a system of two linear equations with two variables, the **substitution method** involves converting the system to one equation in one variable by an appropriate substitution.

**Solving Linear Systems by Substitution**

<Note>

1. Find an expression for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step.)

2. Substitute the expression found in step 1 into the other equation of the system. This will excludes one variable.

3. Solve the equation containing one variable found in step 2.

4. Back-substitute the value found in step 3 into one of the original equations of the system. Simplify and find the value of the remaining variable.

5. Check the solution obtained in both of the system’s given equations.

</Note>

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

Solve the following system by the substitution method:

<Tex block>
\begin{cases}
 5x-4y &=9 \\\
 x-2y &=-3
\end{cases}
</Tex>

<Example type="solution"/>

Before solving, it is advisable to number the two equations:

<Tex block>
\begin{cases}
 5x-4y &=9 \hspace7ex &(1) \\\
 x-2y &=-3 \hspace7ex &(2)
\end{cases}
</Tex>

#### **Step1: Find an expression for one variable in terms of the other**

We choose one of the unknowns in one of the two equations, write it in terms of the other unknown. If possible, solve
for a variable whose coefficient is 1 or - 1 to avoid working with fractions as much as possible.

we choose the equation (2) and write <Tex>x</Tex> in terms of <Tex>y</Tex> because the coeffiecient of <Tex>x</Tex> is <Tex>1</Tex>.

<Tex block>x=2y-3</Tex>

#### **Step2: Substitute the expression found in step 1 into the other equation of the system**

Substitute <Tex>2y-3</Tex> in the equation (1):

<Tex block>
\begin{align*}
5x-4y &= 9 \\\
5(2y-3)-4y &= 9
\end{align*}
</Tex>

The variable <Tex>x</Tex> has been eliminated.

#### **Step3: Solve the equation containing one variable found in step 2**

We simplify and solve the equation (3):

<Tex block>
\begin{align*}
5(2y-3)-4y &= 9 \\\
5\times2y-5\times 3 -4y &= 9 \\\
10y-15-4y &= 9 \\\
6y &= 9+15 \\\
6y &= 24 \\\
y &= \frac{24}{6} \\\
y &= 4
\end{align*}
</Tex>

#### **Step4: Back-substitute**

Now in Equation (1), we substitute <Tex>y</Tex> into the value 4.

<Tex block>
\begin{align*}
5x-4y &=9 \\\
5x-4\times4 &=9 \\\
5x-16 &=9 \\\
5x &=9+16 \\\
5x &=25 \\\
x &= \frac{25}{5} \\\
x &= 5
\end{align*}
</Tex>

The solution of the system is the ordered pair <Tex>(5;4)</Tex>.

#### **Step5: Check the solution**

Chek the solution in the both equations (1) and (2).

For equation (1):

<Tex block>
5x-4y=5\times5-4\times4=25-16=9
</Tex>

For equation (2):

<Tex block>
x-2y=5-2\times4=5-8=-3
</Tex>

True statements result when the solution is substituted in both equations, confirming that the solution set is <Tex>\\{(5;4)\\}</Tex>.

### Algebraic Solution: Elimination Method

In a system of two linear equations with two variables, the **Elimination Method** involves finding the equivalent system which the coefficients of one of the two variables must be additive inverses.

**Solving Linear Systems by Elimination**

<Note>
1. Choose the variable to be eliminated. (The coefficient is preferred 1 or -1 if exists)
2. We multiply all the terms of one equation (or both in some cases) by a constant that will get the variable chosen in step 1 to add up to zero.
3. Add the two equations together, in order to eliminate a variable, and then solve the simultaneous equation.
4. Back-substitute the value found in step 3 into one of the original equations of the system. Simplify and find the value of the remaining variable.
5. Check the solution obtained in both of the system’s given equations.
</Note>

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

Solve the following system by the elimination method:

<Tex block>
\begin{cases}
 4x-y &=-7 \\\
 -3x+2y &=9
\end{cases}
</Tex>

<Example type="solution"/>

Before solving, it is advisable to number the two equations:

<Tex block>
\begin{cases}
 4x-y &=-7 \hspace7ex &(1) \\\
 -3x+2y &=9 \hspace7ex &(2)
\end{cases}
</Tex>

#### **Step1: Choose the variable to be eliminated**

For example, we can choose to eliminat the variable <Tex>y</Tex> because the coefficient is <Tex>-1</Tex> in the equation(1).

#### **Step2: Find the eequivalent system**

The coefficient of the variable <Tex>y</Tex> in the equation (2) is 2, then we must multiply all terms of the equation (1) by **2** to get the additive inverses(<Tex>2y</Tex> and <Tex>-2y</Tex>).

The equivalent system is:

<Tex block>
\begin{cases}
 8x-2y &=-14 \hspace7ex &(3) \\\
 -3x+2y &=9 \hspace7ex &(4)
\end{cases}
</Tex>

#### **Step3: Adding the equations of the equivalent system**

We add the equations (3) and (4) side by side:

<Tex block>
\begin{align*}
(8x-2y)+(-3x+2y) &=-14+9 \\\
8x-2y-3x+2y &=-5 \\\
8x-3x &=-5 \\\
5x &=-5 \\\
x &=-1
\end{align*}
</Tex>

#### **Step4: Back-substitute**

Now in Equation (1), we substitute <Tex>x</Tex> into the value -1.

<Tex block>
\begin{align*}
4x-y &= -7 \\\
4\times (-1) -y &= -7 \\\
-4-y &= -7 \\\
-y &= -7+4 \\\
-y &= -3 \\\
y &= 3
\end{align*}
</Tex>

The solution of the system is the ordered pair <Tex>(-1;3)</Tex>.

#### **Step5: Check the solution**

Chek the solution in the both equations (1) and (2).

For equation (1):

<Tex block>
4x-y=4\times(-1)-3=-4-3=-7
</Tex>

For equation (2):

<Tex block>
-3x+2y=-3\times(-1)+2\times3=3+6=9
</Tex>

True statements result when the solution is substituted in both equations, confirming that the solution set is <Tex>\\{(-1;3)\\}</Tex>.

Solving Systems of Linear Equations (Simultaneous Linear Equation) with two variables using  three methods: graphic, substitution and elimination

Systems Of Linear Equations

# Solving Systems of Equations in 3 Unknowns Using Substitution Method

System of linear equations with three equations in three unknowns can be solved by making two substitutions.

Here is the detailed strategy for how to solve systems of equations with three variables.

1. **Step 1:** First, isolate one unknown (<Tex>x,\\; y,\\; or\\; z</Tex>) in one equation.

2. **Step 2:** Then substitute the expression for that unknown into the two other equations (not the equation that you already used in step 1.)

3. **Step 3:** Now isolate a different unknown in one of the two new equations (the two equations used in step 2).

4. **Step 4:** Substitute this expression into the last equation.

5. **Step 5:** Now, you should have one equation with just one unknown. Isolate that unknown to solve for it.

6. **Step 6:** Once you solve for one unknown, plug it back into any equation to find the other unknowns.

## Examples of how to solve system of equations with 3 variables using substitution method

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

Solve the following system of linear equations.

<Tex block>
\begin{cases}
 2x-3y+4z &=4 \\\
 3x+2y+3z &=7 \\\
 5x+8y-2z &=-7
\end{cases}
</Tex>

<Example type="solution"/>

Before proceeding to solve the systems, it is preferable to number the three equations.

<Tex block>
\begin{cases}
 2x-3y+4z &=4 \hspace7ex &(1) \\\
 3x+2y+3z &=7 \hspace7ex &(2) \\\
 5x+8y-2z &=-7 \hspace7ex &(3) 
\end{cases}
</Tex>

**Step 1:** Isolate <Tex>x</Tex> in the equation <Tex>(1)</Tex>

Doing this by moving <Tex>-3y</Tex> and <Tex>4z</Tex> with change of sign.

<Tex block>
\begin{align*}
2x &=3y-4z+4 \\\
x &=\dfrac{3y}{2}-\dfrac{4z}{2}+\dfrac{4}{2} \\\
x &=\dfrac{3}{2}y-2z+2
\end{align*}
</Tex>

Therefor: <Tex>\boxed{x =\dfrac{3}{2}y-2z+2} \\; \colorbox{aqua}{(*)}</Tex>

**Step 2:** Plug this expression in for <Tex>x</Tex> in the equations <Tex>(2)</Tex> and <Tex>(3)</Tex>.

<Tex block>
\begin{cases}
 3(\dfrac{3}{2}y-2z+2)+2y+3z &=7 \\\
 5(\dfrac{3}{2}y-2z+2)+8y-2z &=-7 
\end{cases}
</Tex>

Simplify these equations. First distribute the coefficient.

<Tex block>
\begin{cases}
 \dfrac{9}{2}y-6z+6+2y+3z &=7 \\\
 \dfrac{15}{2}y-10z+10+8y-2z &=-7 
\end{cases}
</Tex>

Now <Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms" as="/lessons/basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">group and combine like terms</Link> together for each equation.

<Tex block>
\begin{cases}
 \dfrac{9}{2}y+2y-6z+3z &=7-6 \\\
 \dfrac{15}{2}y+8y-10z-2z &=-7-10 
\end{cases}
</Tex>

When one term is a fraction, this requires first finding a common denominator.

<Tex block>
\begin{cases}
 \frac{13}{2}y-3z &=1 \hspace7ex &(4) \\\
 \frac{31}{2}y-12z &=-17 \hspace7ex &(5)
\end{cases}
</Tex>

**Step 3:** Isolate <Tex>y</Tex> in the equation <Tex>(4)</Tex>.

Doing this by moving <Tex>-3z</Tex> with change of sign.

<Tex block>
\begin{align*}
\dfrac{13}{2}y &=3z+1 \\\
y &=\dfrac{2}{13}(3z+1) \\\
y &=\dfrac{6}{13}z+\dfrac{2}{13}
\end{align*}
</Tex>

Therefor: <Tex>\boxed{y =\dfrac{6}{13}z+\dfrac{2}{13}} \\; \colorbox{aqua}{(\**)}</Tex>

**Step 4:** Plug this expression for <Tex>y</Tex> into the equation <Tex>(5)</Tex>

<Tex block>
\begin{align*}
\frac{31}{2}y-12z &=-17 \\\
\frac{31}{2}(\dfrac{6}{13}z+\dfrac{2}{13})-12z &=-17
\end{align*}
</Tex>

**Step 5:** Solve for <Tex>z</Tex> in this equation.

Simplify these equation. First distribute the coefficient <Tex>\frac{31}{2}</Tex>. 

<Tex block>
\begin{align*}
\frac{31}{2}(\dfrac{6}{13}z+\dfrac{2}{13})-12z &=-17 \\\
\dfrac{186}{26}z+\dfrac{62}{26}-12z &=-17 \\\
\dfrac{93}{13}z+\dfrac{31}{13}-12z &=-17 \\\
\end{align*}
</Tex>

Now group and combine like terms together for the obtained equation.

<Tex block>
\begin{align*}
\dfrac{93}{13}z+\dfrac{31}{13}-12z &=-17 \\\
\dfrac{93}{13}z-12z &=-17-\dfrac{31}{13} \\\
\dfrac{-63}{13}z &=\dfrac{-252}{13} \\\
\dfrac{-63}{\cancel{13}}z &=\dfrac{-252}{\cancel{13}} \\\
-63z&=-252 \\\
63z&=252 \\\
z&=\dfrac{252}{63} \\\
z&=4
\end{align*}
</Tex>

**Step 6:** Use this value of <Tex>z</Tex> to solve for <Tex>x</Tex> and <Tex>y</Tex>.

- Plug <Tex>z=4</Tex> into the equation <Tex>\colorbox{aqua}{(\**)}</Tex>.

<Tex block>
\begin{align*}
y &=\dfrac{6}{13}z+\dfrac{2}{13} \\\
y &=\dfrac{6}{13}\times4+\dfrac{2}{13} \\\
y &=\dfrac{24}{13}+\dfrac{2}{13} \\\
y &=\dfrac{24+2}{13} \\\
y &=\dfrac{26}{13}\\\
y &=2
\end{align*}
</Tex>

- Plug <Tex>z=4</Tex> and <Tex>y=2</Tex> into the equation <Tex>\colorbox{aqua}{(*)}</Tex>.

<Tex block>
\begin{align*}
x &=\dfrac{3}{2}y-2z+2 \\\
x &=\dfrac{3}{2}\times2-2\times4+2 \\\
x &=3-8+2 \\\
x &=-3
\end{align*}
</Tex>

The final answers are:

<Tex block>x=-3\\; , \\; y=2\\; , \\; z=4</Tex>

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

Solve the following system of linear equations.

<Tex block>
\begin{cases}
 x+2y-3z &=-3 \\\
 2x-5y+4z &=13 \\\
 5x+4y-z &=5
\end{cases}
</Tex>

<Example type="solution"/>

Before proceeding to solve the systems, it is preferable to number the three equations.

<Tex block>
\begin{cases}
 x+2y-3z &=-3 \hspace7ex &(1) \\\
 2x-5y+4z &=13 \hspace7ex &(2) \\\
 5x+4y-z &=5 \hspace7ex &(3) 
\end{cases}
</Tex>

**Step 1:** Isolate <Tex>x</Tex> in the equation <Tex>(1)</Tex>

Doing this by moving <Tex>2y</Tex> and <Tex>-3z</Tex> with change of sign.

<Tex block>
\begin{align*}
x &=-2y+3z-3
\end{align*}
</Tex>

Therefor: <Tex>\boxed{x =-2y+3z-3} \\; \colorbox{aqua}{(*)}</Tex>

**Step 2:** Plug this expression in for <Tex>x</Tex> in the equations <Tex>(2)</Tex> and <Tex>(3)</Tex>.

<Tex block>
\begin{cases}
 2(-2y+3z-3)-5y+4z &=13 \\\
 5(-2y+3z-3)+4y-z &=5
\end{cases}
</Tex>

Simplify these equations. First distribute the coefficient.

<Tex block>
\begin{cases}
 -4y+6z-6-5y+4z &=13 \\\
 -10y+15z-15+4y-z &=5
\end{cases}
</Tex>

Now <Link to="../basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms" as="/lessons/basic-algebra/simplifying-algebraic-expressions-by-combining-like-terms">group and combine like terms</Link> together for each equation.

<Tex block>
\begin{cases}
 -4y-5y+6z+4z &=13+6 \\\
 -10y+4y+15z-z &=5+15
\end{cases}
</Tex>

Then,

<Tex block>
\begin{cases}
 -9y+10z &=19 \hspace7ex &(4) \\\
 -6y+14z &=20 \hspace7ex &(5)
\end{cases}
</Tex>

**Step 3:** Isolate <Tex>y</Tex> in the equation <Tex>(4)</Tex>.

Doing this by moving <Tex>-3z</Tex> with change of sign.

<Tex block>
\begin{align*}
-9y+10z &=19 \\\
-9y &=-10z+19 \\\
y &=\dfrac{-10}{-9}z+\dfrac{19}{-9}\\\
y &=\dfrac{10}{9}z-\dfrac{19}{9}
\end{align*}
</Tex>

Therefor: <Tex>\boxed{y =\dfrac{10}{9}z-\dfrac{19}{9}} \\; \colorbox{aqua}{(\**)}</Tex>

**Step 4:** Plug this expression for <Tex>y</Tex> into the equation <Tex>(5)</Tex>

<Tex block>
\begin{align*}
-6y+14z &=20 \\\
-6(\dfrac{10}{9}z-\dfrac{19}{9})+14z &=20
\end{align*}
</Tex>

**Step 5:** Solve for <Tex>z</Tex> in this equation.

Simplify these equation. First distribute the coefficient <Tex>-6</Tex>. 

<Tex block>
\begin{align*}
-6(\dfrac{10}{9}z-\dfrac{19}{9})+14z &=20 \\\
\dfrac{-60}{9}z+\dfrac{114}{9}+14z &=20\\\
\dfrac{-20}{3}z+\dfrac{38}{3}+14z &=20\\\
\end{align*}
</Tex>

Now group and combine like terms together for the obtained equation.

<Tex block>
\begin{align*}
\dfrac{-20}{3}z+\dfrac{38}{3}+14z &=20\\\
\dfrac{-20}{3}z+14z &=20-\dfrac{38}{3} \\\
\dfrac{22}{3}z &=\dfrac{22}{3} \\\
\dfrac{\cancel{22}}{\cancel{3}}z &=\dfrac{\cancel{22}}{\cancel{3}} \\\
z &=1
\end{align*}
</Tex>

**Step 6:** Use this value of <Tex>z</Tex> to solve for <Tex>x</Tex> and <Tex>y</Tex>.

- Plug <Tex>z=1</Tex> into the equation <Tex>\colorbox{aqua}{(\**)}</Tex>.

<Tex block>
\begin{align*}
y &=\dfrac{10}{9}z-\dfrac{19}{9} \\\
y &=\dfrac{10}{9}\times1-\dfrac{19}{9} \\\
y &=\dfrac{10}{9}-\dfrac{19}{9} \\\
y &=\dfrac{10-19}{9} \\\
y &=\dfrac{-9}{9}\\\
y &=-1
\end{align*}
</Tex>

- Plug <Tex>z=1</Tex> and <Tex>y=-1</Tex> into the equation <Tex>\colorbox{aqua}{(*)}</Tex>.

<Tex block>
\begin{align*}
x &=-2y+3z-3 \\\
x &=-2\times(-1)+3\times1-3 \\\
x &=2+3-3 \\\
x &=2
\end{align*}
</Tex>

The final answers are:

<Tex block>x=2\\; , \\; y=-1\\; , \\; z=1</Tex>

______________

<Related>

<Link to="../advanced-algebra/systems-of-linear-equations" as="/lessons/advanced-algebra/systems-of-linear-equations">Systems Of Linear Equations (Simultaneous Linear Equations</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> 
<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>

Solving Systems of Linear Equations (Simultaneous Linear Equation) with three variables using  the substitution method guided by many examples

Solve System of Equations in 3 Unknowns by Substitution Method

# Rational Expressions: Finding the Domain

In this lesson, we'll look at the mathematical definition of a rational expression, and then move on to determining the domain that fulfills this expression. By the way, if you are not familiar with the lesson "<Link to="../intermediate-algebra/polynomials-types-degree-properties" as="/lessons/intermediate-algebra/polynomials-types-degree-properties">Polynomials</Link>" and "<Link to="../intermediate-algebra/types-fratcions-and-basics" as="/lessons/basic-algebra/types-fratcions-and-basics">Fractions</Link>" I advise you to first read them to understand the basics that we will need in this lesson.

A **rational expression** is the quotient of two polynomials. Mathematically;

<Note>

Let <Tex>P(x)</Tex> and <Tex>Q(x)</Tex> two polynomials with <Tex>Q(x) \neq 0</Tex>. A rational expression is written as:

<Tex block>\dfrac{P(x)}{Q(x)}</Tex>

</Note>

**Examples of rational expression:**

* <Tex>\dfrac{x+4}{x+1}</Tex>


* <Tex>\dfrac{x-2}{x}</Tex>


* <Tex>\dfrac{x^2+3x-6}{x^2-2x+1}</Tex>


* <Tex>\dfrac{x^2+x+1}{x+1}</Tex>


* <Tex>\dfrac{x^2-5x-1}{(2x+3)(x+4)}</Tex>


* <Tex>\dfrac{x}{(x+1)(x-5)(x^2-4)}</Tex>

## Finding the Domain

As we said in the above property, the polynomial in the denominator must be different from 0. That is, the domain of the rational expression must **not include solutions** to the equation <Tex>Q(x)=0</Tex>, so, we must exclude numbers from a rational expression’s domain that make the denominator zero. Because if the denominator is equal to 0 then the rational expression becomes undefined.

To determine the domain of the rational expression <Tex>\dfrac{P(x)}{Q(x)}</Tex>, we follow these two steps:

1. We solve the equation <Tex>Q(x)=0</Tex>

2. We write the domain as the set of all real numbers excluding the solutions of the equation <Tex>Q(x)=0</Tex>.

We can write the domain in two different ways:

1. Write the domain is **all real numbers except ...**

2. Using intervals

-----------

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

Find the domain of the follwing rational expression:

<Tex block>\dfrac{6x+1}{x-1}</Tex>

<Example type="solution"/>

The denominator is <Tex>x-1</Tex> and must be not equal to 0.

Solving the equation <Tex>x-1=0</Tex>

<Tex block>
\begin{align*}
x-1&=0\\\
x&=1
\end{align*}
</Tex>

Hence, for the denominator to be different from 0, <Tex>x</Tex> must be different from 1.

The domain is "all real numbers except 1."

Or, with intervals:

<Tex>(-\infty;1)\cup(1;+\infty)</Tex>

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

Find the domain of the follwing rational expression:

<Tex block>\dfrac{-3x+1}{(x+2)(x-5)}</Tex>

<Example type="solution"/>

The denominator is <Tex>(x+2)(x-5)</Tex> and must be not equal to 0.

Solving the equation <Tex>(x+2)(x-5)=0</Tex>

<Tex block>
\begin{align*}
x+2=0\quad &or \quad x-5=0\\\
x=-2\quad &or \quad x=5
\end{align*}
</Tex>

Hence, for the denominator to be different from 0, <Tex>x</Tex> must be different from -2 and 5.

The domain is "all real numbers except -2 and 5."

Or, with intervals:

<Tex>(-\infty;-2)\cup(-2;5)\cup(5;+\infty)</Tex>

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

Find the domain of the follwing rational expression:

<Tex block>\dfrac{2x+5}{x(-x+2)}</Tex>

<Example type="solution"/>

The denominator is <Tex>x(-x+2)</Tex> and must be not equal to 0.

Solving the equation <Tex>x(-x+2)=0</Tex>

<Tex block>
\begin{align*}
x=0\quad &or \quad -x+2=0\\\
x=0\quad &or \quad x=2
\end{align*}
</Tex>

Hence, for the denominator to be different from 0, <Tex>x</Tex> must be different from 0 and 2.

The domain is "all real numbers except 0 and 2."

Or, with intervals:

<Tex>(-\infty;0)\cup(0;2)\cup(2;+\infty)</Tex>

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

Find the domain of the follwing rational expression:

<Tex block>\dfrac{x+1}{(x+1)(x-2)(2x+3)}</Tex>

<Example type="solution"/>

Solving the equation <Tex>(x+1)(x-2)(2x+3)=0</Tex>

<Tex block>
\begin{align*}
x+1=0\quad &or \quad x-2=0 \quad or \quad 2x+3=0\\\
x=-1\quad &or \quad x=2 \quad or \quad x=\dfrac{-3}{2}
\end{align*}
</Tex>

Hence, for the denominator to be different from 0, <Tex>x</Tex> must be different from -1 and 2 and <Tex>\dfrac{-3}{2}</Tex>.

The domain is "all real numbers except -1 and 2 and <Tex>\dfrac{-3}{2}</Tex>."

Or, with intervals:

<Tex>(-\infty;\dfrac{-3}{2})\cup(\dfrac{-3}{2};-1)\cup(-1;2)\cup(2;+\infty)</Tex>

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

Find the domain of the follwing rational expression:

<Tex block>\dfrac{6x^2-3x+9}{x^2-4}</Tex>

<Example type="solution"/>

To solve the equation <Tex>x^2-4=0</Tex>, we can factorize the expression as <Tex>(x-2)(x+2)=0</Tex>

The solutions of the equation are -2 and 2.

Hence, for the denominator to be different from -2 and 2.

The domain is "all real numbers except -2 and 2"

Or, with intervals:

<Tex>(-\infty;-2)\cup(-2;2)\cup(2;+\infty)</Tex>

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

Find the domain of the follwing rational expression:

<Tex block>\dfrac{6x^2-3x+9}{x^2-2x-15}</Tex>

<Example type="solution"/>

The equation <Tex>x^2-2x-15=0</Tex> is a <Link to="../intermediate-algebra/find-roots-quadratic-equation-quadratic-formula" as="/lessons/intermediate-algebra/find-roots-quadratic-equation-quadratic-formula">quadratic equation</Link>, to solve it, we must calculate the quadratic formula.

<Tex block>
\begin{align*}
\Delta &=b^2-4ac \\\
&=(-2)^2-4 \times 1 \times (-15) \\\
&=4+60\\\
&=64>0
\end{align*}
</Tex>

So the equation has two solutions:

<Tex block>
\begin{align*}
x&=\dfrac{-b-\sqrt{\Delta}}{2a} \quad or \quad x=\dfrac{-b+\sqrt{\Delta}}{2a} \\\
x&=\dfrac{2-\sqrt{64}}{2} \quad or \quad x=\dfrac{2+\sqrt{64}}{2} \\\
x&=\dfrac{2-8}{2} \quad or \quad x=\dfrac{2+8}{2} \\\
x&=\dfrac{-6}{2} \quad or \quad x=\dfrac{10}{2} \\\
x&=-3 \quad or \quad x=5
\end{align*}
</Tex>

Hence, for the denominator to be different from -3 and 5.

The domain is "all real numbers except -3 and 5"

Or, with intervals:

<Tex>(-\infty;-3)\cup(-3;5)\cup(5;+\infty)</Tex>

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

Find the domain of the follwing rational expression:

<Tex block>\dfrac{-x+1}{x^2+1}</Tex>

<Example type="solution"/>

The quadratic eqauation <Tex>x^2+1=0</Tex> has no solution because whatever <Tex>x</Tex> is, <Tex>x^2+1>0</Tex>.

In this case, the domaine is "all real numbers"

Or with intervals, <Tex>(-\infty; +\infty)</Tex>

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

Find the domain of the follwing rational expression:

<Tex block>\dfrac{1}{x}</Tex>

<Example type="solution"/>

The denominator is <Tex>x</Tex>, and must be different from 0.

The domain of the rational expression is "all real numbers except 0".

With intervals, <Tex>(-\infty; 0)\cup(0;+\infty)</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="../intermediate-algebra/factoring-plynomials-greatest-common-factor" as="/lessons/intermediate-algebra/factoring-plynomials-greatest-common-factor">Factoring Polynomials Out the Greatest Common Factor</Link> 

<Link to="../intermediate-algebra/factoring-difference-two-perfect-squares" as="/lessons/intermediate-algebra/factoring-difference-two-perfect-squares">Factoring Difference of Two Perfect Squares</Link> 

<Link to="../intermediate-algebra/polynomials-types-degree-properties" as="/lessons/intermediate-algebra/polynomials-types-degree-properties">The polynomials: Definition, Types, Degree and Properties</Link> 

<Link to="../intermediate-algebra/find-roots-quadratic-equation-quadratic-formula" as="/lessons/intermediate-algebra/find-roots-quadratic-equation-quadratic-formula">Find Roots Of Quadratic Equation Using Quadratic Formula</Link> 

<Link to="../intermediate-algebra/factoring-simple-quadratic-trinomials" as="/lessons/intermediate-algebra/factoring-simple-quadratic-trinomials">Factoring a Simple Quadratic Trinomials</Link>

<Link to="../advanced-algebra/simplifying-rational-expressions" as="/lessons/advanced-algebra/simplifying-rational-expressions">Simplifying Rational Expressions</Link>

</Related>

You will learn how to find the domain of rational expression, which means finding all values of the variable that must be excluded.

Rational Expressions: Finding the Domain

# Simplifying Rational Expressions

A **rational expression** is an algebraic expression can be written as the quotient of two <Link to="../intermediate-algebra/polynomials-types-degree-properties" as="/lessons/intermediate-algebra/polynomials-types-degree-properties">polynomials</Link> (Hence any polynomial is also a rational expression). Rational expression domain is *set* of all real values of the variables except those that make the denominator equal to zero.

In this lesson, you'll learn how to simplify rational expressions using your skills of finding the greatest common factor and factoring the polynomials out this common factor.

## Lowest Terms of a Rational Expression

A rational expression is written in **lowest terms** if <Link to="../intermediate-algebra/factoring-plynomials-greatest-common-factor" as="/lessons/intermediate-algebra/factoring-plynomials-greatest-common-factor">the greatest common factor</Link> of its numerator and denominator is 1 or -1. To write a rational expression we must use the **Fondamental Principle of Fractions**, then reducing the rational expression to the lowest terms by dividing out the greatest common factor.

<Note>

**Fondamental Principle of Fractions:**

Let <Tex>a\\; , \\;b\\; and \\; k</Tex> are real numbers with <Tex>b\neq0 \\; and \\; k\neq0</Tex>

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

</Note>

## Procedure to Simplify Rational Expressions

A rational expression is **simplified** if it is reduced to the lowest terms (Its denominator and numerator have no common factor other than 1 or -1).

<Note>

**Simplify Rational Expressions:**

1. Factor the numerator and the denominator completely using the common factor.

2. Divide both the numerator and the denominator out the common factor.

</Note>

## Examples of how to Simplify Rational Expressions

This is a list of examples of how to reduce rational expressions to a lowest terms. The skill required in these examples is to be able to find the greatest common factor of the numerator and denominator.

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

Simplify the following rational expression:

<Tex block>\dfrac{x^2-2xy+y^2}{x^2-y^2}</Tex>

<Example type="solution"/>

**Factoring the denominator:** The expression at denominator is a <Link to="../intermediate-algebra/factoring-difference-two-perfect-squares" as="/lessons/intermediate-algebra/factoring-difference-two-perfect-squares">Factoring difference of two perfect squares</Link>, so:

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

**Factoring the numerator:** the expression at the numerator is a polynomial can be factored as product of two binomials:

<Tex block>x^2-2xy+y^2=(x-y)^2=(x-y)(x-y)</Tex>

Note carefully that there is a common factor between the two obtained expressions which is <Tex>(x-y)</Tex>.

Hence,

<Tex block>
\begin{align*}
\dfrac{x^2-2xy+y^2}{x^2-y^2}&=\dfrac{\bcancel{(x-y)}(x-y)}{\bcancel{(x-y)}(x+y)}\\\
&=\dfrac{x-y}{x+y}
\end{align*}
</Tex>

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

Simplify the following rational expression:

<Tex block>\dfrac{3x-6}{x^2-4}</Tex>

<Example type="solution"/>

**Factoring the denominator:** The expression at denominator can be written as a difference of two perfect squares:

<Tex block>x^2-4=x^2-2^2=(x-2)(x+2)</Tex>

**Factoring the numerator:** Factoring the numerator out the greatest common factor between the two terms <Tex>3x</Tex> and <Tex>-6</Tex> which is 3.

<Tex block>3x-6=3(x-2)</Tex>

Now, I assume that you noted the common factor wich is <Tex>(x-2)</Tex>

Hence,

<Tex block>
\begin{align*}
\dfrac{3x-6}{x^2-4}&=\dfrac{3\bcancel{(x-2)}}{\bcancel{(x-2)}(x+2)}\\\
&=\dfrac{3}{x+2}
\end{align*}
</Tex>

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

Simplify the following rational expression:

<Tex block>\dfrac{2x^2+7x-4}{5x^2+20x}</Tex>

<Example type="solution"/>

**Factoring the denominator:** The numerator can be factored using the greatest common factor of <Tex>5x^2</Tex> and <Tex>20x</Tex> which is <Tex>5x</Tex>.

<Tex block>
\begin{align*}
5x^2+20x&=5x\times x +5x \times 4 \\\
&=5x(x+4)
\end{align*}
</Tex>

**Factoring the numerator:** The numerator is a trinomial expression, we can factor it using <Link to="../intermediate-algebra/find-roots-quadratic-equation-quadratic-formula" as="/lessons/intermediate-algebra/find-roots-quadratic-equation-quadratic-formula">quadratic formula</Link>. So, we must calculate the descriminant:

<Tex block>\Delta=b^2-4ac=7^2-4\times2\times(-4)=81</Tex>

The quadratic expression has two roots:

<Tex block>
\begin{align*}
x&=\dfrac{-b-\sqrt{\Delta}}{2a}\\; or \\; x=\dfrac{-b+\sqrt{\Delta}}{2a}\\\
x&=\dfrac{-7-\sqrt{81}}{2\times2}\\; or \\; x=\dfrac{-7+\sqrt{81}}{2\times2}\\\
x&=\dfrac{-7-9}{4}\\; or \\; x=\dfrac{-7+9}{4}\\\
x&=\dfrac{-16}{4}\\; or \\; x=\dfrac{2}{4}\\\
x&=-4 \\; or \\; x=\dfrac{1}{2}
\end{align*}
</Tex>

Therefor;

<Tex block>
\begin{align*}
2x^2+7x-4&=2(x-\dfrac{1}{2})(x+4)\\\
&=(2x-1)(x+4)
\end{align*}
</Tex>

Note carefully that there is a common factor between the two obtained expressions which is <Tex>(x+4)</Tex>.

Hence;

<Tex block>
\begin{align*}
\dfrac{2x^2+7x-4}{5x^2+20x}&=\dfrac{(2x-1)\bcancel{(x+4)}}{5x\bcancel{(x+4)}}\\\
&=\dfrac{2x-1}{5x}\\\
\end{align*}
</Tex>

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

Simplify the following rational expression:

<Tex block>\dfrac{x^2-1}{x+1}</Tex>

<Example type="solution"/>

**Factoring the denominator:** The numerator is a difference of two perfect squares, so it can be factored as follow:

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

Note that there is already a common factor between the denominator and the numerator which is <Tex>(x+1)</Tex>.

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

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

Simplify the following rational expression:

<Tex block>\dfrac{6-3x}{x^2-4}</Tex>

<Example type="solution"/>

**Factoring the denominator:** The numerator is a difference of two perfect squares, so it can be factored as follow:

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

**Factoring the numerator:** The numerator is a binomial expression can be factored as follow:

<Tex>6-3x=3\times2-3\times x=3(2-x)</Tex>

But for a moment, the denominator and the numerator don't have a common factor, right? If you notice carefully when we factored by the number 3 we could also have factored with the number -3 because in this case, we will get a common factor.

<Tex>6-3x=-3\times(-2)-3\times x=-3(-2+x)=-3(x-2)</Tex>

Now, I assume that you noted the common factor wich is <Tex>(x-2)</Tex>. Hence,

<Tex block>
\begin{align*}
\dfrac{6-3x}{x^2-4}&=\dfrac{-3\bcancel{(x-2)}}{(x+2)\bcancel{(x-2)}}\\\
&=\dfrac{-3}{x+2}
\end{align*}
</Tex>

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

Simplify the following rational expression:

<Tex block>\dfrac{x+1}{x^3+2x^2+x}</Tex>

<Example type="solution"/>

**Factoring the numerator:** No change will occur to the numerator because it is not factorizable as a product of two binomials. 

**Factoring the denominator:** Notice that the denominator is a trinomial which the three terms have a common factor which is <Tex>x</Tex>.

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

But up to here, we still don’t get a common factor of the numerator and denominator. If you notice well that the expression <Tex>x^2+2x+1</Tex> is a quadratic expression that can be factored as follows:

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

So, the denominator can factorized as follow:

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

Finally, the rational expression can be simplified as follow:

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

-----

<Related>

<Link to="../advanced-algebra/finding-domain-rational-expressions" as="/lessons/advanced-algebra/finding-domain-rational-expressions">Rational Expressions: Finding the Domain</Link> 

<Link to="../intermediate-algebra/factoring-plynomials-greatest-common-factor" as="/lessons/intermediate-algebra/factoring-plynomials-greatest-common-factor">Factoring Polynomials Out the Greatest Common Factor</Link> 

<Link to="../intermediate-algebra/factoring-difference-two-perfect-squares" as="/lessons/intermediate-algebra/factoring-difference-two-perfect-squares">Factoring Difference of Two Perfect Squares</Link> 

<Link to="../intermediate-algebra/factoring-quadratics-with-leading-coefficient" as="/lessons/intermediate-algebra/factoring-quadratics-with-leading-coefficient">Factoring Quadratics With a Leading Coefficient</Link> 

<Link to="../intermediate-algebra/find-roots-quadratic-equation-quadratic-formula" as="/lessons/intermediate-algebra/find-roots-quadratic-equation-quadratic-formula">Find Roots Of Quadratic Equation Using Quadratic Formula</Link> 

<Link to="../intermediate-algebra/factoring-simple-quadratic-trinomials" as="/lessons/intermediate-algebra/factoring-simple-quadratic-trinomials">Factoring a Simple Quadratic Trinomials</Link> 

<Link to="../intermediate-algebra/foil-method-for-multiplying-binomials" as="/lessons/intermediate-algebra/foil-method-for-multiplying-binomials">FOIL Method For Multiplying Two Binomials</Link> 

<Link to="../intermediate-algebra/factoring-trinomials-using-quadratic-equation-roots" as="/lessons/intermediate-algebra/factoring-trinomials-using-quadratic-equation-roots">Factoring Trinomials Using Quadratic Equation Roots</Link>

</Related>

learn how to simplify rational expression using the common factor. We use the fundamental principle of fractions to write a rational expression in lowest terms by dividing out common factors.

Simplifying Rational Expressions

# Multiplying Rational Expressions

As with fractions, the same rule we'll apply to multiplying rational expressions. If you want to first look at operations on fractions, I recommend these two lessons:

* <Link to="../basic-algebra/multiplying-fractions-by-whole-number-by-reciprocal" as="/lessons/basic-algebra/multiplying-fractions-by-whole-number-by-reciprocal">Multiplying Fractions</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</Link>

In this lesson, we will learn how to multiply two rational expressions. Very simple steps, and all you will need are some basic concepts such as simplifying a rational expression and finding the common factor of two expressions. Operations on rational expressions require some simplification and domain finding skills as detailed in the following two lessons:

* <Link to="../advanced-algebra/finding-domain-rational-expressions" as="/lessons/advanced-algebra/finding-domain-rational-expressions">Rational Expressions: Finding the Domain</Link>

* <Link to="../advanced-algebra/simplifying-rational-expressions" as="/lessons/advanced-algebra/simplifying-rational-expressions">Simplifying Rational Expressions</Link>

The basic rule for multiplying rational expressions is the same as the basic rule for fractions: the product of their numerators divided by the product of their denominators.

<Note>

Let <Tex>P(x),\\; Q(x),\\; R(x)\\; and \\; S(x)</Tex> four rational expressions whereas <Tex>R(x)\neq0</Tex> and <Tex>S(x)\neq0</Tex>.

<Tex block>\dfrac{P(x)}{R(x)}\times\dfrac{Q(x)}{S(x)}=\dfrac{P(x)\times Q(x)}{R(x)\times R(x)}</Tex>

</Note>

Here is a step-by-step procedure for multiplying rational expressions:

• Factor all numerators and denominators;

• Cancel any common factor that appears in both a numerator and a denominator;

• Multiply numerator by numerator and denominator by denominator;

<Alert>

After completing the simplification, be sure to keep the same domain for the original expression so that the entire answer is true.

</Alert>

## Examples of How to Multiply Rational Expressions

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

Multiply the rational expressions and give answer in simplest form:

<Tex block>A(x)=\dfrac{x-7}{x-1}\cdotp \dfrac{x^2-1}{3x-21}</Tex>

<Example type="solution"/>

**Step 1:** Always try to factor and find all the common factors in the expressions in the numerators and denominators.

<Tex block>
\begin{align*}
A(x)&=\dfrac{x-7}{x-1}\cdotp \dfrac{x^2-1}{3x-21}\\\
&=\dfrac{x-7}{x-1}\cdotp \dfrac{(x-1)(x+1)}{3(x-7)}
\end{align*}
</Tex>

The domain of this rational expression is "all real numbers except 1 and 7". i.e. <Tex>x\neq1,\\;7</Tex>

**Step 2:** Once the expression is factored, you can see all the common factors that appears in the numerators anddenominators.

<Tex block>
\begin{align*}
A(x)&=\dfrac{x-7}{x-1}\cdotp \dfrac{x^2-1}{3x-21}\\\
&=\dfrac{\colorbox{aqua}{$x-7$}}{\colorbox{aquamarine}{$x-1$}}\cdotp \dfrac{\colorbox{aquamarine}{$(x-1)$}(x+1)}{3\colorbox{aqua}{$(x-7)$}}
\end{align*}
</Tex>

Cancel them out, one from a numerator with one from a denominator (with same color).

<Tex block>
\begin{align*}
A(x)=\dfrac{x-7}{x-1}\cdotp \dfrac{x^2-1}{3x-21}=1\cdotp \dfrac{x+1}{3}\\\
\end{align*}
</Tex>

**Step 3:** Multiply the obtained expression.

<Tex block>
\begin{align*}
A(x)&=\dfrac{x-7}{x-1}\cdotp \dfrac{x^2-1}{3x-21}\\\
&=1\cdotp \dfrac{x+1}{3}\\\
&=\dfrac{x+1}{3}
\end{align*}
</Tex>

After canceling the common factors we got an expression that, logically, is not equal to the expression <Tex>A(x)</Tex> at the beginning, but why? In the step 1, we saw that <Tex>x\neq1,\\;7</Tex> which wasn't true of the expression <Tex>\dfrac{x+1}{3}</Tex>. So the final answer should exclude the two values of <Tex>x\neq1,\\;7</Tex>.

<Tex block>
\begin{align*}
A(x)&=\dfrac{x-7}{x-1}\cdotp \dfrac{x^2-1}{3x-21}\\\
&=\dfrac{x+1}{3} \quad for\quad x\neq1,\\;7
\end{align*}
</Tex>

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

Multiply the rational expressions and give answer in simplest form:

<Tex block>B(x)=\dfrac{6x-24}{9x+27}\cdotp \dfrac{3x+9}{2x-8}</Tex>

<Example type="solution"/>

**Step 1:** Factor all expressions in the numerators and denominators.

<Tex block>
\begin{align*}
B(x)&=\dfrac{6x-24}{9x+27}\cdotp \dfrac{3x+9}{2x-8}\\\
&=\dfrac{6(x-4)}{9(x+3)}\cdotp \dfrac{3(x+3)}{2(x-4)}
\end{align*}
</Tex>

The domain of this rational expression is "all real numbers except -3 and 4". i.e. <Tex>x\neq-3,\\;4</Tex>

**Step 2:** Now you can see all the common factors that appears more than once in the both top and bottom.

<Tex block>
\begin{align*}
B(x)&=\dfrac{6x-24}{9x+27}\cdotp \dfrac{3x+9}{2x-8}\\\
&=\dfrac{6\colorbox{aqua}{$(x-4)$}}{9\colorbox{aquamarine}{$(x+3)$}}\cdotp \dfrac{3\colorbox{aquamarine}{$(x+3)$}}{2\colorbox{aqua}{$(x-4)$}}
\end{align*}
</Tex>

Cancel them out, one from a numerator with one from a denominator (with same color).

<Tex block>
\begin{align*}
B(x)&=\dfrac{6x-24}{9x+27}\cdotp \dfrac{3x+9}{2x-8}\\\
&=\dfrac{6}{9}\cdotp \dfrac{3}{2}
\end{align*}
</Tex>

**Step 3:** We obtained a product of two fractions,

<Tex block>
\begin{align*}
B(x)&=\dfrac{6x-24}{9x+27}\cdotp \dfrac{3x+9}{2x-8}\\\
&=\dfrac{6}{9}\cdotp \dfrac{3}{2}\\\
&=\dfrac{18}{18}\\\
&=1 \quad for\quad x\neq-3,\\;4 \\\
\end{align*}
</Tex>

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

Multiply the rational expressions and give answer in simplest form:

<Tex block>C(x)=\dfrac{7x^2-7x}{4x^2}\cdotp \dfrac{4x-12}{x^2-4x+3}</Tex>

<Example type="solution"/>

**Step 1:** Factor all numerators and denominators. Note that the second denominator is a quadratic trinomial you can factor it using FOIL method or Simple case method.

<Tex block>
\begin{align*}
C(x)&=\dfrac{7x^2-7x}{4x^2}\cdotp \dfrac{4x-12}{x^2-4x+3}\\\
&=\dfrac{7x(x-1)}{4x\times x}\cdotp \dfrac{4(x-3)}{(x-1)(x-3)}
\end{align*}
</Tex>

The domain of this rational expression is "all real numbers except 0, 1 and 3". i.e. <Tex>x\neq0,\\;1,\\;3</Tex>

**Step 2:** Now you can see all the common factors that appears more than once in the both top and bottom.

<Tex block>
\begin{align*}
C(x)&=\dfrac{7x^2-7x}{4x^2}\cdotp \dfrac{4x-12}{x^2-4x+3}\\\
&=\dfrac{7\colorbox{aqua}{$x$}\colorbox{aquamarine}{$(x-1)$}}{\colorbox{lightgreen}{$4$}x\times \colorbox{aqua}{$x$}}\cdotp \dfrac{\colorbox{lightgreen}{$4$}\colorbox{bisque}{$(x-3)$}}{\colorbox{aquamarine}{$(x-1)$}\colorbox{bisque}{$(x-3)$}}
\end{align*}
</Tex>

Cancel them out, one from a numerator with one from a denominator (with same color).

<Tex block>
\begin{align*}
C(x)&=\dfrac{7x^2-7x}{4x^2}\cdotp \dfrac{4x-12}{x^2-4x+3}\\\
&=\dfrac{7}{x}\cdotp 1
\end{align*}
</Tex>

**Step 3:** The obtained expression is already in simplest form.

<Tex block>
\begin{align*}
C(x)&=\dfrac{7x^2-7x}{4x^2}\cdotp \dfrac{4x-12}{x^2-4x+3}\\\
&=\dfrac{7}{x} \quad for\quad x\neq0,\\;1,\\;3
\end{align*}
</Tex>

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

Multiply the rational expressions and give answer in simplest form:

<Tex block>D(x)=\dfrac{x^2-9}{x^2-6x+5}\cdotp \dfrac{x-5}{x^2+3x}</Tex>

<Example type="solution"/>

**Step 1:** Factor all numerators and denominators completly

<Tex block>
\begin{align*}
D(x)&=\dfrac{x^2-9}{x^2-6x+5}\cdotp \dfrac{x-5}{x^2+3x}\\\
&=\dfrac{(x-3)(x+3)}{(x-5)(x-1)}\cdotp \dfrac{x-5}{x(x+3)}
\end{align*}
</Tex>

The domain of this rational expression is "all real numbers except -3, 0, 1 and 5". i.e. <Tex>x\neq-3,\\;0,\\;1,\\;5</Tex>

**Step 2:** Once the expression is factored, you can see all the common factors that appears in the numerators anddenominators.

<Tex block>
\begin{align*}
D(x)&=\dfrac{x^2-9}{x^2-6x+5}\cdotp \dfrac{x-5}{x^2+3x}\\\
&=\dfrac{(x-3)\colorbox{aqua}{$(x+3)$}}{\colorbox{aquamarine}{$(x-5)$}(x-1)}\cdotp \dfrac{\colorbox{aquamarine}{$x-5$}}{x\colorbox{aqua}{$(x+3)$}}
\end{align*}
</Tex>

Cancel them out, one from a numerator with one from a denominator (with same color).

<Tex block>
\begin{align*}
D(x)&=\dfrac{x^2-9}{x^2-6x+5}\cdotp \dfrac{x-5}{x^2+3x}\\\
&=\dfrac{(x-3)}{(x-1)}\cdotp \dfrac{1}{x}
\end{align*}
</Tex>

**Step 3:** Multiply numerator by numerator and denominator by denominator of the obtained expression.

<Tex block>
\begin{align*}
D(x)&=\dfrac{(x-3)}{(x-1)}\cdotp \dfrac{1}{x}\\\
&=\dfrac{x-3}{x(x-1)} \quad for\quad x\neq-3,\\;0,\\;1,\\;5
\end{align*}
</Tex>

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

Simplify the following rational-expression multiplication

<Tex block>E(x)=\dfrac{8x^3}{3}\cdotp \dfrac{6}{2x^2} \cdotp \dfrac{x}{5}</Tex>

<Example type="solution"/>

Note that the domain is "all real numbers except 0"

Find all the common factors of denominators and numerators, then cancel them.

<Tex block>
\begin{align*}
D(x)&=\dfrac{8x^3}{3}\cdotp \dfrac{6}{2x^2} \cdotp \dfrac{x}{5}\\\
&=\dfrac{8x\times \colorbox{bisque}{$x^2$}}{\colorbox{aqua}{$3$}}\cdotp \dfrac{\colorbox{aqua}{$3$}\times\colorbox{aquamarine}{$2$}}{\colorbox{aquamarine}{$2$}\times \colorbox{bisque}{$x^2$}} \cdotp \dfrac{x}{5}\\\
&=\dfrac{8x}{1}\cdotp \dfrac{1}{1} \cdotp \dfrac{x}{5}
\end{align*}
</Tex>

After canceling the common factors, now we can multiply numerators by numerators and denominators by denominators as follow:

<Tex block>
\begin{align*}
D(x)&=\dfrac{8x^3}{3}\cdotp \dfrac{6}{2x^2} \cdotp \dfrac{x}{5}\\\
&=\dfrac{8x}{1}\cdotp \dfrac{1}{1} \cdotp \dfrac{x}{5}\\\
&=\dfrac{8x\times1\times x}{1\times1\times5}\\\
&=\dfrac{8x^2}{5}\quad for \quad x\neq0
\end{align*}
</Tex>

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

<Related>

<Link to="../advanced-algebra/finding-domain-rational-expressions" as="/lessons/advanced-algebra/finding-domain-rational-expressions">Rational Expressions: Finding the Domain</Link> 

<Link to="../advanced-algebra/simplifying-rational-expressions" as="/lessons/advanced-algebra/simplifying-rational-expressions">Simplifying Rational Expressions</Link> 

<Link to="../intermediate-algebra/factoring-plynomials-greatest-common-factor" as="/lessons/intermediate-algebra/factoring-plynomials-greatest-common-factor">Factoring Polynomials Out the Greatest Common Factor</Link> 

<Link to="../intermediate-algebra/factoring-difference-two-perfect-squares" as="/lessons/intermediate-algebra/factoring-difference-two-perfect-squares">Factoring Difference of Two Perfect Squares</Link> 

<Link to="../intermediate-algebra/find-roots-quadratic-equation-quadratic-formula" as="/lessons/intermediate-algebra/find-roots-quadratic-equation-quadratic-formula">Find Roots Of Quadratic Equation Using Quadratic Formula</Link> 

<Link to="../intermediate-algebra/factoring-simple-quadratic-trinomials" as="/lessons/intermediate-algebra/factoring-simple-quadratic-trinomials">Factoring a Simple Quadratic Trinomials</Link>

</Related>

learn how to multiply two rational expressions or more with a few steps. Rational expressions are multiplied and divided the same way numeric fractions are.

Explorer all lessons of Advanced Algebra