Remember Marc and his lunch from the Single Variable Addition Equations Concept?

Previously we figured out that Marc spent $6.15 on lunch. After Marc had paid for his lunch, he bought an ice cream cone. It cost $3.15. If Marc received $5.85 back as change, how much did he give the cashier?

You can write a single variable subtraction equation to represent this dilemma.

Solving the equation will give you a solution to the problem presented.

**Pay attention and you will learn how to do this in this Concept.**

### Guidance

**To solve an equation in which a number is** *subtracted***from a variable, we can use the inverse of subtraction––addition. We can add that number to** *both***sides of the equation to solve it.**

**You can think about this as working backwards from the operation. If we have a problem with addition, we subtract. If we have a problem with subtraction, we add.**

We must add the number to *both* sides of the equation because of the ** Addition Property of Equality**, which states:

if \begin{align*}a=b\end{align*}, then \begin{align*}a+c=b+c\end{align*}.

So, if you add a number, \begin{align*}c\end{align*}, to one side of an equation, you must add that same number, \begin{align*}c\end{align*}, to the other side, too, to keep the values on both sides equal.

Let's apply this information to a problem.

Solve for \begin{align*}a. \ a-15=18\end{align*}.

**In the equation, 15 is** *subtracted***from \begin{align*}a\end{align*}. So, we can** *add***15 to both sides of the equation to solve for \begin{align*}a\end{align*}.**

\begin{align*}a-15 &= 18\\ a-15+15 &= 18+15\\ a+(-15+15) &= 33\\ a+0 &= 33\\ a &= 33\end{align*}

*Notice how we rewrote the subtraction as adding a negative integer.*

**The value of \begin{align*}a\end{align*} is 33.**

Here is another one.

Solve for \begin{align*}k. \ k-\frac{1}{3}=\frac{2}{3}\end{align*}.

**In the equation, \begin{align*}\frac{1}{3}\end{align*} is** *subtracted***from \begin{align*}k\end{align*}. So, we can** *add***\begin{align*}\frac{1}{3}\end{align*} to both sides of the equation to solve for \begin{align*}k\end{align*}.**

\begin{align*}k-\frac{1}{3} &= \frac{2}{3}\\ k-\frac{1}{3}+\frac{1}{3} &= \frac{2}{3}+\frac{1}{3}\\ k+\left(-\frac{1}{3}+\frac{1}{3}\right) &= \frac{3}{3}\\ k+0 &= \frac{3}{3}\\ k &= \frac{3}{3}=1\end{align*}

**The value of \begin{align*}k\end{align*} is 1.**

Again, we are using a property. **The Subtraction Property of Equality states that as long as you subtract the same quantity to both sides of an equation, that the equation will remain equal.**

Each of these properties makes use of an inverse operation. If the operation in the equation is addition, then you use the Subtraction Property of Equality. If the operation in the equation is subtraction, then you use the Addition Property of Equality.

Solve each equation.

#### Example A

\begin{align*}x-44=22\end{align*}

**Solution: \begin{align*}66\end{align*}**

#### Example B

\begin{align*}x-1.3=5.6\end{align*}

**Solution: \begin{align*}6.9\end{align*}**

#### Example C

\begin{align*}y-\frac{1}{4}=\frac{2}{4}\end{align*}

**Solution: \begin{align*}\frac{3}{4}\end{align*}**

Here is the original problem once again.

Previously we figured out that Marc spent $6.15 on lunch. After Marc had paid for his lunch, he bought an ice cream cone. It cost $3.15. If Marc received $5.85 back as change, how much did he give the cashier?

You can write a single variable subtraction equation to represent this dilemma.

Solving the equation will give you a solution to the problem presented.

First, let's write the equation.

Our unknown is the amount of money Marc gave the cashier. Let's call that \begin{align*}x\end{align*}.

\begin{align*}x\end{align*}

Then we know that the ice cream cone cost $3.15.

\begin{align*}x - 3.15\end{align*}

Marc received $5.85 in change.

\begin{align*}x - 3.15 = 5.85\end{align*}

Now we can solve this equation.

\begin{align*}x = 5.85 + 3.15\end{align*}

\begin{align*}x = \$9.00\end{align*}

**This is our answer.**

### Vocabulary

- Isolate the variable
- an explanation used to describe the action of getting the variable alone on one side of the equal sign.

- Inverse Operation
- the opposite operation.

- Subtraction Property of Equality
- states that you can subtract the same quantity from both sides of an equation and have the equation still balance.

- Addition Property of Equality
- states that you can add the same quantity to both sides of an equation and have the equation still balance.

### Guided Practice

Here is one for you to try on your own.

Harry earned $19.50 this week. That is $6.50 less than he earned last week.

a. Write an equation to represent \begin{align*}m\end{align*}, the amount of money, in dollars, that he earned last week.

b. Determine how much money Harry earned last week.

**Answer**

Consider part \begin{align*}a\end{align*} first.

Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem.

\begin{align*}& \text{Harry earned} \ \underline{\$19.50 \ \text{this week}}. \ \text{That} \ \underline{\text{is}} \ \underline{\$6.50} \ \underline{\text{less than}} \ldots \underline{\text{last week}.}\\ & \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \ \downarrow \quad \Box \quad \quad \downarrow \qquad \qquad \qquad \Box\\ & \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \ \downarrow \quad \Box \quad \quad \downarrow \qquad \qquad \qquad \Box\\ & \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \ \downarrow \quad \Box \quad \quad \downarrow \qquad \qquad \qquad \Box\\ & \qquad \qquad \qquad \qquad \ 19.50 \qquad \qquad \quad \ \ = \quad m \quad \quad - \qquad \qquad \quad 6.50\end{align*}

This equation, \begin{align*}19.50=m-6.50\end{align*}, represents \begin{align*}m\end{align*}, the number of dollars earned last week.

Next, consider part \begin{align*}b\end{align*}.

Solve the equation to find the number of blue tiles in the bag.

\begin{align*}19.50 &= m-6.50\\ 19.50+6.50 &= m-6.50+6.50\\ 26.00 &= m+(-6.50+6.50)\\ 26 &= m+0\\ 26 &= m\end{align*}

**Harry earned $26.00 last week.**

### Video Review

This is a James Sousa video on solving single variable subtraction equations.

### Practice

Directions: Solve each single-variable subtraction equation.

1. \begin{align*}x-8=9\end{align*}

2. \begin{align*}x-18=29\end{align*}

3. \begin{align*}a-9=29\end{align*}

4. \begin{align*}a-4=30\end{align*}

5. \begin{align*}b-14=27\end{align*}

6. \begin{align*}b-13=50\end{align*}

7. \begin{align*}y-23=57\end{align*}

8. \begin{align*}y-15=27\end{align*}

9. \begin{align*}x-9=32\end{align*}

10. \begin{align*}c-19=32\end{align*}

11. \begin{align*}x-1.9=3.2\end{align*}

12. \begin{align*}y-2.9=4.5\end{align*}

13. \begin{align*}c-6.7=8.9\end{align*}

14. \begin{align*}c-1.23=3.54\end{align*}

15. \begin{align*}c-5.67=8.97\end{align*}