Prealgebra

Module 9: Multi-Step Linear Equations

Solving Equations With Variables on Both Sides

Learning Outcomes

Identify the constant and variable terms in an equation
Solve linear equations by isolating constants and variables
Solve linear equations with variables on both sides that require several steps

The equations we solved in the last section simplified nicely so that we could use the division property to isolate the variable and solve the equation. Sometimes, after you simplify you may have a variable and a constant term on the same side of the equal sign.

Our strategy will involve choosing one side of the equation to be the variable side, and the other side of the equation to be the constant side. This will help us with organization. Then, we will use the Subtraction and Addition Properties of Equality, step by step, to isolate the variable terms on one side of the equation.

Read on to find out how to solve this kind of equation.

Examples

Solve:

4x+6=-14

.

Solution:

In this equation, the variable is only on the left side. It makes sense to call the left side the variable side. Therefore, the right side will be the constant side.

Since the left side is the variable side, the 6 is out of place. We must "undo" adding $6$ by subtracting $6$ , and to keep the equality we must subtract $6$ from both sides. Use the Subtraction Property of Equality.		$4x+6\color{red}{-6}=-14\color{red}{-6}$
Simplify.		$4x=-20$
Now all the $x$ s are on the left and the constant on the right.
Use the Division Property of Equality.		$\frac{4x}{\color{red}{4}}=\frac{-20}{\color{red}{4}}$
Simplify.		$x=-5$
Check:		$4x+6=-14$
Let $x=-5$ .		$4(\color{red}{-5})+6=-14$
		$-20+6=-14$
		$-14=-14\quad\checkmark$

Solve:

2y - 7=15

Show Solution

Solution:

Notice that the variable is only on the left side of the equation, so this will be the variable side and the right side will be the constant side. Since the left side is the variable side, the

7

is out of place. It is subtracted from the

2y

, so to "undo" subtraction, add

7

to both sides.

$2y-7$ is the side containing a variable. $15$ is the side containing only a constant.
Add $7$ to both sides.		$2y-7\color{red}{+7}=15\color{red}{+7}$
Simplify.		$2y=22$
The variables are now on one side and the constants on the other.
Divide both sides by $2$ .		$\frac{2y}{\color{red}{2}}=\frac{22}{\color{red}{2}}$
Simplify.		$y=11$
Check:		$2y-7=15$
Let $y=11$ .		$2\cdot\color{red}{11}-7\stackrel{\text{?}}{=}15$
		$22-7\stackrel{\text{?}}{=}15$
		$15=15\quad\checkmark$

Now you can try a similar problem.

Solve Equations with Variables on Both Sides

You may have noticed that in all the equations we have solved so far, we had variables on only one side of the equation. This does not happen all the time—so now we’ll see how to solve equations where there are variable terms on both sides of the equation. We will start like we did above—choosing a variable side and a constant side, and then use the Subtraction and Addition Properties of Equality to collect all variables on one side and all constants on the other side. Remember, what you do to the left side of the equation, you must do to the right side as well.

In the next example, the variable,

x

, is on both sides, but the constants appear only on the right side, so we'll make the right side the "constant" side. Then the left side will be the "variable" side.

ExampleS

Solve:

5x=4x+7

Show Solution

Solution:

$5x$ is the side containing only a variable. $4x+7$ is the side containing a constant.
We don't want any variables on the right, so subtract the $4x$ .		$5x\color{red}{-4x}=4x\color{red}{-4x}+7$
Simplify.		$x=7$
We have all the variables on one side and the constants on the other. We have solved the equation.
Check:		$5x=4x+7$
Substitute $7$ for $x$ .		$5(\color{red}{7})\stackrel{\text{?}}{=}4(\color{red}{7})+7$
		$35\stackrel{\text{?}}{=}28+7$
		$35=35\quad\checkmark$

Solve:

7x=-x+24

Show Solution

Solution:

The only constant,

24

, is on the right, so let the left side be the variable side.

$7x$ is the side containing only a variable. $-x+24$ is the side containing a constant.
Remove the $-x$ from the right side by adding $x$ to both sides.	$7x\color{red}{+x}=-x\color{red}{+x}+24$
Simplify.	$8x=24$
All the variables are on the left and the constants are on the right. Divide both sides by $8$ .	$\frac{8x}{\color{red}{8}}=\frac{24}{\color{red}{8}}$
Simplify.	$x=3$
Check:	$7x=-x+24$
Substitute $x=3$ .	$7(\color{red}{3})\stackrel{\text{?}}{=}-(\color{red}{3})+24$
	$21=21\quad\checkmark$

Did you see the subtle difference between the two equations? In the first, the right side looked like this:

2x+7

, and in the second, the right side looked like this:

-x+24

, even though they look different, we still used the same techniques to solve both.

Now you can try solving an equation with variables on both sides where it is beneficial to move the variable term to the left side.

In our last examples, we moved the variable term to the left side of the equation. In the next example, you will see that it is beneficial to move the variable term to the right side of the equation. There is no "correct" side to move the variable term, but the choice can help you avoid working with negative signs.

example

Solve:

5y - 8=7y

Show Solution

Solution:

The only constant,

-8

, is on the left side of the equation, and the variable,

y

, is on both sides. Let’s leave the constant on the left and collect the variables to the right.

$5y-8$ is the side containing a constant. $7y$ is the side containing only a variable.
Subtract $5y$ from both sides.	$5y\color{red}{-5y}-8=7y\color{red}{-5y}$
Simplify.	$-8=2y$
We have the variables on the right and the constants on the left. Divide both sides by $2$ .	$\frac{-8}{\color{red}{2}}=\frac{2y}{\color{red}{2}}$
Simplify.	$-4=y$
Rewrite with the variable on the left.	$y=-4$
Check:	$5y-8=7y$
Let $y=-4$ .	$5(\color{red}{-4})-8\stackrel{\text{?}}{=}7(\color{red}{-4})$
	$-20-8\stackrel{\text{?}}{=}-28$
	$-28=-28\quad\checkmark$

Now you can try solving an equation where it is beneficial to move the variable term to the right side.

Solve Equations with Variables and Constants on Both Sides

The next example will be the first to have variables and constants on both sides of the equation. As we did before, we’ll collect the variable terms to one side and the constants to the other side. You will see that as the number of variable and constant terms increases, so do the number of steps it takes to solve the equation.

Examples

Solve:

7x+5=6x+2

Show Solution

Solution:

Start by choosing which side will be the variable side and which side will be the constant side. The variable terms are

7x

and

6x

. Since

7

is greater than

6

, make the left side the variable side and so the right side will be the constant side.

16=16\quad\checkmark

$7x+5=6x+2$
Collect the variable terms to the left side by subtracting $6x$ from both sides.	$7x\color{red}{-6x}+5=6x\color{red}{-6x}+2$
Simplify.	$x+5=2$
Now, collect the constants to the right side by subtracting $5$ from both sides.	$x+5\color{red}{-5}=2\color{red}{-5}$
Simplify.	$x=-3$
The solution is $x=-3$ .
Check:	$7x+5=6x+2$
Let $x=-3$ .	$7(\color{red}{-3})+5\stackrel{\text{?}}{=}6(\color{red}{-3})+2$
	$-21+5\stackrel{\text{?}}{=}-18+2$

Solve:

6n - 2=-3n+7

Show Solution

We have

6n

on the left and

-3n

on the right. Since

6>-3

, make the left side the "variable" side.

$6n-2=-3n+7$
We don't want variables on the right side—add $3n$ to both sides to leave only constants on the right.	$6n\color{red}{+3n}-2=-3n\color{red}{+3n}+7$
Combine like terms.	$9n-2=7$
We don't want any constants on the left side, so add $2$ to both sides.	$9n-2\color{red}{+2}=7\color{red}{+2}$
Simplify.	$9n=9$
The variable term is on the left and the constant term is on the right. To get the coefficient of $n$ to be one, divide both sides by $9$ .	$\frac{9n}{\color{red}{9}}=\frac{9}{\color{red}{9}}$
Simplify.	$n=1$
Check:	$6n-2=-3n+7$
Substitute $1$ for $n$ .	$6(\color{red}{1})-2\stackrel{\text{?}}{=}-3(\color{red}{1})+7$
	$4=4\quad\checkmark$

In the following video we show an example of how to solve a multi-step equation by moving the variable terms to one side and the constants to the other side. You will see that it doesn't matter which side you choose to be the variable side; you can get the correct answer either way.

In the next example, we move the variable terms to the right side to keep a positive coefficient on the variable.

EXAMPLE

Solve:

2a - 7=5a+8

Show Solution

Solution:

This equation has

2a

on the left and

5a

on the right. Since

5>2

, make the right side the variable side and the left side the constant side.

Let

a=-5

2(\color{red}{-5})-7\stackrel{\text{?}}{=}5(\color{red}{-5})+8

$2a-7=5a+8$
Subtract $2a$ from both sides to remove the variable term from the left.	$2a\color{red}{-2a}-7=5a\color{red}{-2a}+8$
Combine like terms.	$-7=3a+8$
Subtract $8$ from both sides to remove the constant from the right.	$-7\color{red}{-8}=3a+8\color{red}{-8}$
Simplify.	$-15=3a$
Divide both sides by $3$ to make 1[/latex] the coefficient of $a$ .	$\frac{-15}{\color{red}{3}}=\frac{3a}{\color{red}{3}}$
Simplify.	$-5=a$
Check:	$2a-7=5a+8$
	$-10-7\stackrel{\text{?}}{=}-25+8$
	$-17=-17\quad\checkmark$

The following video shows another example of solving a multi-step equation by moving the variable terms to one side and the constants to the other side.

Try these problems to see how well you understand how to solve linear equations with variables and constants on both sides of the equal sign.

We just showed a lot of examples of different kinds of linear equations you may encounter. There are some good habits to develop that will help you solve all kinds of linear equations. We’ll summarize the steps we took so you can easily refer to them.

Solve an equation with variables and constants on both sides

Choose one side to be the variable side and then the other will be the constant side.
Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.
Collect the constants to the other side, using the Addition or Subtraction Property of Equality.
Make the coefficient of the variable
$1$
, using the Multiplication or Division Property of Equality.
Check the solution by substituting it into the original equation.

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