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# 7.8: Single Variable Subtraction Equations

Difficulty Level: At Grade Created by: CK-12
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Practice Single Variable Subtraction Equations

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Remember Marc and his lunch?

In the last Concept, you 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 a=b\begin{align*}a=b\end{align*}, then a+c=b+c\begin{align*}a+c=b+c\end{align*}. So, if you add a number, c\begin{align*}c\end{align*}, to one side of an equation, you must add that same number, c\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 a a15=18\begin{align*}a \ a-15=18\end{align*}. In the equation, 15 is subtracted from a\begin{align*}a\end{align*}. So, we can add 15 to both sides of the equation to solve for a\begin{align*}a\end{align*}. a15a15+15a+(15+15)a+0a=18=18+15=33=33=33\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 a\begin{align*}a\end{align*} is 33. Here is another one. Solve for k k13=23\begin{align*}k \ k-\frac{1}{3}=\frac{2}{3}\end{align*}. In the equation, 13\begin{align*}\frac{1}{3}\end{align*} is subtracted from k\begin{align*}k\end{align*}. So, we can add 13\begin{align*}\frac{1}{3}\end{align*} to both sides of the equation to solve for k\begin{align*}k\end{align*}. k13k13+13k+(13+13)k+0k=23=23+13=33=33=33=1\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 k\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 x44=22\begin{align*}x-44=22\end{align*} Solution: 66\begin{align*}66\end{align*} #### Example B x1.3=5.6\begin{align*}x-1.3=5.6\end{align*} Solution: 6.9\begin{align*}6.9\end{align*} #### Example C y14=24\begin{align*}y-\frac{1}{4}=\frac{2}{4}\end{align*} Solution: 34\begin{align*}\frac{3}{4}\end{align*} Here is the original problem once again. In the last Concept, you figured out that Marc spent6.15 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 x\begin{align*}x\end{align*}.

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

Then we know that the ice cream cone cost 3.15. x3.15\begin{align*}x - 3.15\end{align*} Marc received5.85 in change.

x3.15=5.85\begin{align*}x - 3.15 = 5.85\end{align*}

Now we can solve this equation.

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

### Video Review

Here is a video for review.

### Practice

Directions: Solve each single-variable subtraction equation.

1. x8=9\begin{align*}x-8=9\end{align*}

2. x18=29\begin{align*}x-18=29\end{align*}

3. a9=29\begin{align*}a-9=29\end{align*}

4. a4=30\begin{align*}a-4=30\end{align*}

5. b14=27\begin{align*}b-14=27\end{align*}

6. b13=50\begin{align*}b-13=50\end{align*}

7. y23=57\begin{align*}y-23=57\end{align*}

8. y15=27\begin{align*}y-15=27\end{align*}

9. x9=32\begin{align*}x-9=32\end{align*}

10. c19=32\begin{align*}c-19=32\end{align*}

11. x1.9=3.2\begin{align*}x-1.9=3.2\end{align*}

12. y2.9=4.5\begin{align*}y-2.9=4.5\end{align*}

13. c6.7=8.9\begin{align*}c-6.7=8.9\end{align*}

14. c1.23=3.54\begin{align*}c-1.23=3.54\end{align*}

15. c5.67=8.97\begin{align*}c-5.67=8.97\end{align*}

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