Solving for a Variable on One Side Using Multiplication, Addition, and Subtraction

Introduction

In this lesson, you will practice using addition, subtraction, and multiplication to solve for a variable on one side of an equation.

This video illustrates the lesson material below. Watching the video is optional.

Solving for a Variable on One Side (Part 3): Multiplication, Addition, and Subtraction (07:18 mins) | Transcript

Solving for a Variable on One Side Using Addition, Subtraction, and Multiplication

Not all problems only have one thing happening at a time. Often multiplication and addition are happening. You must still add the additive inverse to both sides and multiply both sides by the multiplicative inverse, but which comes first?

When you solve to isolate a variable, work backwards or in reverse through the order of operations.

Add or Subtract from left to right.
Multiply or Divide from left to right.
Evaluate Exponents.
Simplify everything inside the parentheses.

Example 1

\begin{align*}3x + 5 &= 17 &\color{red}\small\text{Solve for x}\\\\3x + 5 \color{red}\mathbf{-5} &= 17\color{red}\mathbf{-5} &\color{red}\small\text{Additive inverse of 5 is -5}\\\\3x &= 12 &\color{red}\small\text{Simplify each side}\\\\3x \color{red}\mathbf{(\frac{1}{3})} &= 12\color{red}\mathbf{(\frac{1}{3})} &\color{red}\small\text{Multiplicative inverse of 3 is $\frac{1}{3}$}\\\\x &= 4 &\color{red}\small\text{Simplify each side}\\\\\end{align*}

Example 2

\begin{align*}5x -8 &= 27 &\color{red}\small\text{Solve for x}\\\\5x -8 \color{red}\mathbf{+8} &= 27\color{red}\mathbf{+8} &\color{red}\small\text{Additive inverse of -8 is +8}\\\\5x &= 35 &\color{red}\small\text{Simplify each side}\\\\5x \color{red}\mathbf{(\frac{1}{5})} &= 35\color{red}\mathbf{(\frac{1}{5})} &\color{red}\small\text{Multiplicative inverse of 5 is $\frac{1}{5}$}\\\\x &= 7 &\color{red}\small\text{Simplify each side}\\\\\end{align*}

Remember, when solving for a variable, use the Order of Operations (PEMDAS), but go backward instead. The practice problems above only deal with multiplication and addition/subtraction, so add the additive inverse first, then multiply by the multiplicative inverse.

Things to Remember

Work backwards through the order of operations to isolate a variable.
Make sure to rewrite the equation after performing a step. It will ensure that no steps are skipped or forgotten.

Practice Problems

$5{\text{M}} + 2 = 12$ (
Solution
x

Solution: $2$
Details:
Where there are several operations within an equation, it is helpful to identify the operations and the order you should do them in.

In this example, there is addition and multiplication.

When solving for a variable, do the order of operations backward.

The order of operations is as follows:
1. Parentheses
2. Exponents
3. Multiplication & Division
4. Addition & Subtraction

Step 1: Going backward you start by undoing the addition and subtraction:

In the equation $5{\text{M}} + 2 = 12$ you need to remove the addition of the $+2$ from the left-hand side. Do this by adding the additive inverse to both sides. The additive inverse of $+2$ is $-2$.

The $+2$ and $-2$ add to zero, so you are left with $5{\text{M}}$ on the left side.

The $12$ and $-2$ add to $10$ on the right side.

Step 2: Undo any multiplication or division:

The $5$ is currently being multiplied to the ${\text{M}}$. You need to remove this to get ${\text{M}}$ all by itself. Do this by multiplying both sides of the equation by the multiplicative inverse of $5$.

The multiplicative inverse of $5$ is $\dfrac{1}{5}$.

$5{\text{M}} = 10$

${\color{Cyan} \dfrac{1}{5}}\left ( 5{\text{M}} \right ) = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}$

Since $\dfrac{1}{5} \times 5 = 1$, you are left with $1{\text{M}}$. Anything multiplied by $1$ is still itself, so $1{\text{M}} = {\text{M}}$.

${\color{Cyan} \dfrac{1}{5}}\left ( {\color{DarkGreen} 5}{\text{M}} \right ) = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}$

${\color{Cyan} \dfrac{1}{5}}\cdot{\color{DarkGreen} \dfrac{5}{1}}\left ( {\text{M}} \right ) = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}$

$\dfrac{{\color{Cyan} 1}*{\color{DarkGreen} 5}}{{\color{Cyan} 5}*{\color{DarkGreen} 1}}\left ( {\text{M}} \right ) = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}$

$\dfrac{{\color{Orange} 5}}{{\color{Orange} 5}}\left ( {\text{M}} \right ) = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}$

${\color{Orange} 1}{\text{M}} = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}$

In the previous image, multiplication is written in several different ways.
A number and a variable right next to each other without any other operations between them means they are being multiplied together. Ex: $5{\text{M}}$
Parentheses are used to show multiplication between the numbers inside the parentheses and the numbers outside the parentheses.
A dot is used between the fractions showing they are being multiplied together.
The $*$ symbol is used within the fraction to show multiplication between the numbers in the fraction.
You now have ${\text{M}}$ all by itself on one side of the equal sign. The next step is to simplify the other side of the equal sign.
$This is a picture of the equation $1M=10\dfrac{1}{5}$. Next line is: $M=\dfrac{10}{1} \times \dfrac{1}{5}=\dfrac{10 \times 1}{1 \times 5}=\dfrac{10}{5}=2$$

$\left ( 10 \right )\left ( \dfrac{1}{5} \right )$ is the same as $\dfrac{10}{5}=2$.

The final solution is: ${\text{M}} = 2$.
)
$-2{\text{Q}} + 1 = 5$ (
Solution

x

Solution: $-2$

)
$2{\text{n}} + 9 = -5$ (
Solution

x

Solution: $-7$

)
$8{\text{k}} {-} 7 = 17$ (
Solution

x

Solution: $3$
Details:
In this example problem, there are two operations going on: multiplication and the addition of a negative number.

You need to unravel our equation in order to get the variable ${\text{k}}$ all by itself on one side of the equal sign. Do this by doing the order of operations backward.

Step 1: Do the inverse of any addition or subtraction:
In this example, $-7$ is being added to $8{\text{k}}$.

$8{\text{k}} {-} 7 = 8{\text{k}} + (-7)$

$This is a picture of the equation $8K-7=17$. There is a vertical dashed line through the equal sign. “Left-hand side” is written on the left side of the dashed line with “Right-hand side” written on the right. $8k+-7=17$ is written below the original equation, with a horizontal bracket indicating that subtracting 7 is equivalent to adding negative $-7$.$

In order to remove the $-7$, add the additive inverse to both sides.

The additive inverse of $-7$ is $+7$

On the left-hand side:
$8{\text{k}} {-} 7 + 7 = 8{\text{k}} + 0 = 8{\text{k}}$

On the right-hand side:
$17 + 7 = 24$

$This is a picture of the equation $8K-7+7=17+7$. There is a vertical dashed line through the equal sign. “Left-hand side” is written on the left side of the dashed line with “Right-hand side” written on the right. $8K+0=24$ is written on the next line. There are horizontal brackets between the two equations indicating that $-7+7=0$ and $17+7=24$.$

Step 2: Now the only operation on ${\text{k}}$ is the multiplication of $8$. Isolate ${\text{k}}$ by multiplying both sides of the equation by the multiplicative inverse of $8$.

$8\left ( \dfrac{1}{8} \right ) = 1$

Therefore, the multiplicative inverse of $8$ is $\dfrac{1}{8}$.

On the left-hand side of the equal sign:
$\left ( \dfrac{1}{8} \right )\left ( 8{\text{k}} \right ) = 1{\text{k}} = {\text{k}}$

On the right-hand side of the equal sign:
$\left ( \dfrac{1}{8} \right )24 = \left ( \dfrac{1}{8} \right )\left ( \dfrac{24}{1} \right ) = \left ( \dfrac{24}{8} \right ) = 3$

${\color{DarkGreen} 8}{\text{k}}={\color{Orange} 24}$

${\color{Cyan} \left ( \dfrac{1}{8} \right )} 8{\color{DarkGreen} {\text{k}}} = {\color{Cyan} \left ( \dfrac{1}{8} \right )}{\color{Orange} 24}$

$\left ( \dfrac{{\color{Cyan} 1}*{\color{DarkGreen} 8}}{{\color{Cyan} 8}*{\color{DarkGreen} 1}} \right ){\text{k}} = \dfrac{{\color{Cyan} 1}*{\color{Orange} 24}}{{\color{Cyan} 8}*{\color{Orange} 1}}$

$1{\text{k}}=\dfrac{{\color{DarkOrange} 24}}{{\color{Cyan} 8}}$

${\text{k}}=3$

The final solution is: ${\text{k}} = 3$.

)
$-3{\text{x}} + 14 = -7$ (
Solution

x

Solution: $7$

)
$-5{\text{q}} {-} 29 = 41$ (
Video Solution

x

Solution: $-14$
Details:

(Solving for a Variable on One Side Using Multiplication, Addition, and Subtraction #6 (02:51 mins) | Transcript)

| Transcript)
$10{\text{f}} {-} 6 = 24$ (
Video Solution

x

Solution: $3$
Details:

(Solving for a Variable on One Side Using Multiplication, Addition, and Subtraction #7 (02:44 mins) | Transcript)

| Transcript)

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