# Two-Step Equations and Properties of Equality

## Maintain balance of an equation while solving using addition, subtraction, multiplication, or division.

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Two-Step Equations and Properties of Equality

### 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 \begin{align*}x\end{align*} to represent the number of marbles in each bag, then we can see that on the left side of the scale we have three bags (each containing \begin{align*}x\end{align*} marbles) plus two extra marbles, and on the right side of the scale we have 29 marbles. The balancing of the scales is similar to the balancing of the following equation.

\begin{align*}3x + 2 = 29 \end{align*}

“Three bags plus two marbles equals 29 marbles”

To solve for \begin{align*}x\end{align*}, we need to first get all the variables (terms containing an \begin{align*}x\end{align*}) alone on one side of the equation. We’ve already got all the \begin{align*}x\end{align*}’s on one side; now we just need to isolate them.

\begin{align*}3x + 2 &= 29\\ 3x + 2 - 2 &= 29 - 2 \qquad \text{Get rid of the 2 on the left by subtracting it from both sides.}\\ 3x &= 27\\ \frac{3x}{3} &= \frac{27}{3} \qquad \quad \ \text{Divide both sides by 3.}\\ x &= 9\end{align*}

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 \begin{align*}6(x + 4) = 12 \end{align*}.

This equation has the \begin{align*}x\end{align*} buried in parentheses. To dig it out, we can proceed in one of two ways: we can either distribute the six on the left, or divide both sides by six to remove it from the left. Since the right-hand side of the equation is a multiple of six, it makes sense to divide. That gives us \begin{align*} x + 4 = 2\end{align*}. Then we can subtract 4 from both sides to get \begin{align*}x = -2\end{align*}.

2. Solve \begin{align*}\frac{x - 3}{5} = 7\end{align*}.

It’s always a good idea to get rid of fractions first. Multiplying both sides by 5 gives us \begin{align*}x - 3 = 35\end{align*}, and then we can add 3 to both sides to get \begin{align*}x = 38\end{align*}.

### Example

#### Example 1

Solve \begin{align*}\frac{5}{9}(x + 1) =\frac{2}{7}\end{align*}.

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

\begin{align*}\frac{9}{5} \cdot \frac{5}{9}(x + 1) &= \frac{9}{5} \cdot \frac{2}{7}\\ (x + 1) &= \frac{18}{35}\end{align*}

Then we subtract 1 from both sides. (\begin{align*}\frac{35}{35}\end{align*} is equivalent to 1.)

\begin{align*}x + 1 &= \frac{18}{35}\\ x + 1 - 1 &= \frac{18}{35} - \frac{35}{35}\\ x &= \frac{18 - 35}{35} \\ x &= \frac{-17}{35}\end{align*}

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.

1. \begin{align*}6x - 1.3 = 3.2 \end{align*}
2. \begin{align*}4(x + 3) = 1 \end{align*}
3. \begin{align*}5q - 7 = \frac{2}{3} \end{align*}
4. \begin{align*} \frac{3}{5}x + \frac{5}{2} = \frac{2}{3}\end{align*}
5. \begin{align*}0.1y + 11 = 0\end{align*}
6. \begin{align*}\frac{5q - 7}{12} = \frac{2}{3}\end{align*}
7. \begin{align*}\frac{5(q - 7)}{12} = \frac{2}{3}\end{align*}
8. \begin{align*}33t - 99= 0 \end{align*}
9. \begin{align*}5p - 2 = 32 \end{align*}
10. \begin{align*}10y + 5 = 10 \end{align*}
11. \begin{align*}10(y + 5) = 10 \end{align*}

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