# 7.8: Single Variable Subtraction Equations

**At Grade**Created by: CK-12

**Practice**Single Variable Subtraction Equations

### Let’s Think About It

The baking club realized they only have 2.7 kg of flour left from their original order. The only person to use the flour since they bought it was Sal who used .9 kg of flour to make cakes. Can you write an equation to solve for how much flour they had before Sal took some, and then solve that equation?

In this concept, you will learn to solve single variable subtraction equations.

### Guidance

To solve an equation in which a number is subtracted from a variable, you can use addition to isolate the variable.

You can add the same 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*}.

This means that 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, in order to keep both sides of the equation equal.

Let’s look at an example.

Solve for \begin{align*}a\end{align*}.

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

In the equation, 15 is subtracted from \begin{align*}a\end{align*}. So, you use the addition property of equality to add 15 to both sides of the equation. This will isolate the variable, \begin{align*}a\end{align*}.

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

Notice that you can rewrite \begin{align*}(-15)\end{align*} as \begin{align*}(+ -15)\end{align*}. This is a very useful strategy in solving equations. You can rewrite subtraction as adding a negative number.

The answer is \begin{align*}a = 33\end{align*}.

Here is another example.

Solve for \begin{align*}k\end{align*}.

\begin{align*}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, you use the addition property of equality, and add \begin{align*}\frac{1}{3}\end{align*} to both sides of the equation. This isolates the variable \begin{align*}k\end{align*}.

First, add \begin{align*}\frac{1}{3}\end{align*} to both sides of the equation.

\begin{align*}\begin{array}{rcl} k-\frac{1}{3} &=& \frac{2}{3}\\ k-\frac{1}{3}+\frac{1}{3} &=& \frac{2}{3}+\frac{1}{3} \end{array}\end{align*}

Since the fractions have the same denominators you can add them together.

\begin{align*}\begin{array}{rcl} k-\frac{1}{3}+\frac{1}{3} &=& \frac{2}{3}+\frac{1}{3}\\ k &=& \frac{3}{3}\\ k &=& 1 \end{array}\end{align*}

The answer is \begin{align*}k=1\end{align*}.

### Guided Practice

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

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

Let \begin{align*}m\end{align*} be the amount of money Harry earned last week. Then you can write an equation.

\begin{align*}19.50 = m-6.50.\end{align*}

Next, solve the equation. Use the addition property of equality to add 6.50 to each side of the equation.

\begin{align*}\begin{array}{rcl} 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{array}\end{align*}

The answer is Harry earned $26.00 last week.

### Examples

Solve each equation.

#### Example 1

Solve for \begin{align*}x\end{align*}.

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

Use the addition property of equality and add 44 to both sides of the equation.

\begin{align*}\begin{array}{rcl} x- 44 &=& 22 \\ x- 44+44 &=& 22+44 \\ x &=& 66 \end{array}\end{align*}

The answer is \begin{align*}x = 66\end{align*}.

#### Example 2

Solve for \begin{align*}x\end{align*}.

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

Use the addition property of equality and add 1.3 to both sides of the equation.

\begin{align*}\begin{array}{rcl} x- 1.3 &=& 5.6 \\ x- 1.3+1.3 &=& 5.6+1.3 \\ x &=& 6.9 \end{array}\end{align*}

The answer is \begin{align*}x = 6.9\end{align*}.

#### Example 3

Solve for \begin{align*}y\end{align*}.

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

Use the addition property of equality and add \begin{align*}\frac{1}{4}\end{align*} to both sides of the equation.

\begin{align*} \begin{array}{rcl} y-\frac{1}{4} &=& \frac{1}{2} \\ y-\frac{1}{4}+\frac{1}{4} &=& \frac{1}{2}+\frac{1}{4} \\ y &=& \frac{1}{2}+\frac{1}{4} \end{array}\end{align*}

To add \begin{align*}\frac{1}{2}+\frac{1}{4}\end{align*} you need common denominators. Write \begin{align*}\frac{1}{2}\end{align*} as the equivalent fraction \begin{align*}\frac{2}{4}\end{align*} and then add.

\begin{align*}\begin{array}{rcl} y &=& \frac{2}{4}+\frac{1}{4} \\ y &=& \frac{3}{4} \end{array}\end{align*}

The answer is \begin{align*}y=\frac{3}{4}\end{align*}.

### Follow Up

Remember the baking club and Sal who used their flour to make cakes? They need to know how much flour they originally had so they can pay their supplier.

The club started with some flour, but Sal took .9 kg of this flour to make cake. They had 2.7 kg left. You have to find how much they had to begin with.

First, you need to write an equation that represents this information. Let \begin{align*}x\end{align*}, represent the amount of flour the group originally had. This amount, minus the .9 kg Sal took equals 2.7 kg.

\begin{align*}x-.9=2.7\end{align*}

Use the addition property of equality and add .9 to both sides of the equation and solve.

\begin{align*}\begin{array}{rcl} x-.9+.9 &=& 2.7+.9 \\ x &=& 3.6 \end{array}\end{align*}

The answer is they originally had 3.6 kg of flour.

### Video Review

https://www.youtube.com/watch?v=yqdlj0lv7Cc&feature=youtu.be

### Explore More

Solve each single-variable subtraction equation.

- \begin{align*}x-8 = 9\end{align*}
- \begin{align*}x- 18 = 29\end{align*}
- \begin{align*}a- 9 = 29\end{align*}
- \begin{align*}a-4 = 30\end{align*}
- \begin{align*}b-14 = 27\end{align*}
- \begin{align*}b-13 = 50\end{align*}
- \begin{align*}y-23 = 57\end{align*}
- \begin{align*}y- 15 = 27\end{align*}
- \begin{align*}x-9 = 32\end{align*}
- \begin{align*}c- 19 = 32\end{align*}
- \begin{align*}x-1.9 = 3.2\end{align*}
- \begin{align*}y-2.9 = 4.5\end{align*}
- \begin{align*}c-6.7 = 8.9\end{align*}
- \begin{align*}c-1.23 = 3.54\end{align*}
- \begin{align*}c-5.67 = 8.97\end{align*}

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Addition Property of Equality

The addition property of equality states that you can add the same quantity to both sides of an equation without changing the relative truth of the statement. If , then .Inverse Operation

Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.Isolate the variable

To isolate the variable means to manipulate an equation so that the indicated variable is by itself on one side of the equals sign.Subtraction Property of Equality

The subtraction property of equality states that you can subtract the same quantity from both sides of an equation and it will still balance.### Image Attributions

In this concept, you will learn to solve single variable subtraction equations.