### Two Step Equations and Properties of Equality

We’ve seen how to solve for an unknown by isolating it on one side of an equation and then evaluating the other side. Now we’ll see how to solve equations where the variable takes more than one step to isolate.

#### Real-World Application: Marbles

Rebecca has three bags containing the same number of marbles, plus two marbles left over. She places them on one side of a balance. Chris, who has more marbles than Rebecca, adds marbles to the other side of the balance. He finds that with 29 marbles, the scales balance. How many marbles are in each bag? Assume the bags weigh nothing.

We know that the system balances, so the weights on each side must be equal. If we use

“Three bags plus two marbles **equals** 29 marbles”

To solve for

There are nine marbles in each bag.

We can do the same with the real objects as we did with the equation. Just as we subtracted 2 from both sides of the equals sign, we could remove two marbles from each side of the scale. Because we removed the same number of marbles from each side, we know the scales will still balance.

Then, because there are three bags of marbles on the left-hand side of the scale, we can divide the marbles on the right-hand side into three equal piles. You can see that there are nine marbles in each.

*Three bags of marbles* *balances**three piles of nine marbles.*

So each bag of marbles balances nine marbles, meaning that each bag contains nine marbles.

#### Solving for Unknown Values

1. Solve

This equation has the

2. Solve

It’s always a good idea to get rid of fractions first. Multiplying both sides by 5 gives us

### Example

#### Example 1

Solve

First, we’ll cancel the fraction on the left by multiplying by the reciprocal (the multiplicative inverse).

Then we subtract 1 from both sides. (

These examples are called **two-step equations**, because we need to perform two separate operations on the equation to isolate the variable.

### Review

Solve the following equations for the unknown variable.

6x−1.3=3.2 4(x+3)=1 5q−7=23 35x+52=23 0.1y+11=0 5q−712=23 5(q−7)12=23 33t−99=0 5p−2=32 10y+5=10 10(y+5)=10

### Review (Answers)

To view the Review answers, open this PDF file and look for section 3.4.