Solving for a Variable:

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

Not all problems only have one thing happening at a time, we often have multiplication and addition. We still add the additive inverse to both sides and multiply both sides by the multiplicative inverse, but which comes first? The following video will show this process:

Video Source (07:18 mins) | Transcript

Remember, when solving for a variable, we use the Order of Operations (PEDMAS), but we go backward instead. In your practice problems, you’re just dealing with Multiplication and Addition/Subtraction, so first you’ll add the additive inverse, then you multiply by the multiplicative inverse.

Additional Resources

Khan Academy: Introduction to Two-step Equations (05:11 mins, Transcript)
Khan Academy: Two-step Equations Intuition (08:40 mins, Transcript)

Practice Problems

Solve for the variable:

$5{\text{M}} + 2 = 12$

$-2{\text{Q}} + 1 = 5$

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

$8{\text{k}} {-} 7 = 17$

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

$-5{\text{q}} {-} 29 = 41$

$10{\text{f}} {-} 6 = 24$

Solutions

$2$ (Written Solution)
Where there are several operations within an equation, it is helpful to identify the operations and the order we should do them in.

In this example, there is addition and multiplication.

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

The order of operations is as follows:
1. Parentheses
2. Exponents
3. Multiplication & division
4. Addition and subtraction
Step 1: Going backward we start by undoing the addition and subtraction:

In the equation $5{\text{M}} + 2 = 12$ we need to remove the addition of the $+2$ from the left-hand side. We 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 we 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 ${\text{M}}$. We need to remove this to get ${\text{M}}$ all by itself. We do this by multiplying both sides of the equation by the multiplicative inverse of $5$.

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

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

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

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

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

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

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

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

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

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

$This is a picture of the equation 1(M)=(10)\frac{1}{5}. Next line is: M=\frac{10}{1} \times \frac{1}{5}=\frac{10 \times 1}{1 \times 5}=\frac{10}{5}=2$

Our final solution: ${\text{M}} = 2$

$-2$

$-7$

$3$ (Written Solution)

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

We need to unravel our equation in order to get the variable ${\text{k}}$ all by itself on one side of the equal sign. We 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 $8{\text{k}}$.

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

In order to remove the $-7$, we 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:

$17 + 7 = 24$

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

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

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

On the left-hand side of the equal sign:

$\left ( \frac{1}{8} \right )\left ( 8{\text{k}} \right ) = 1{\text{k}} = {\text{k}}$

On the right-hand side of the equal sign:

$\left ( \frac{1}{8} \right )24 = \left ( \frac{1}{8} \right )\left ( \frac{24}{1} \right ) = \left ( \frac{24}{8} \right ) = 3$

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

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

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

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

${\text{k}}=3$

Our final solution: ${\text{k}} = 3$

$7$

$-14$ (Solution Video | Transcript)

$3$ (Solution Video | Transcript)