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.

### Solving Single Variable Subtraction Equations

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 , then .

This means that if you add a number,

, to one side of an equation, you must add that same number, , to the other side, in order to keep both sides of the equation equal.Let’s look at an example.

Solve for

.

In the equation, 15 is subtracted from

. So, you use the addition property of equality to add 15 to both sides of the equation. This will isolate the variable, .

Notice that you can rewrite

as . This is a very useful strategy in solving equations. You can rewrite subtraction as adding a negative number.The answer is

.Here is another example.

Solve for

.

In the equation,

is subtracted from . So, you use the addition property of equality, and add to both sides of the equation. This isolates the variable .First, add

to both sides of the equation.

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

The answer is

.### Examples

#### Example 1

Earlier, you were given a problem about 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

, represent the amount of flour the group originally had. This amount, minus the .9 kg Sal took equals 2.7 kg.

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.

#### Example 2

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 forLet

be the amount of money Harry earned last week. Then you can write an equation.

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.

**Solve each equation.**

#### Example 3

Solve for

.

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

.#### Example 4

Solve for

.

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

.#### Example 5

Solve for

.

Use the addition property of equality and add

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}\end{align*} as the equivalent fraction and then add.

you need common denominators. Write\begin{align*}\begin{array}{rcl} y &=& \frac{2}{4}+\frac{1}{4} \\ y &=& \frac{3}{4} \end{array}\end{align*}

The answer is

.### Review

Solve each single-variable subtraction equation.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 7.8.