# Lesson 8

Reasoning about Solving Equations (Part 2)

Let’s use hangers to understand two different ways of solving equations with parentheses.

### 8.1: Equivalent to $2(x+3)$

Select all the expressions equivalent to $$2(x+3)$$.

1. $$2 \boldcdot (x+3)$$
2. $$(x + 3)2$$
3. $$2 \boldcdot x + 2 \boldcdot 3$$
4. $$2 \boldcdot x + 3$$
5. $$(2 \boldcdot x) + 3$$
6. $$(2 + x)3$$

### 8.2: Either Or

1. Explain why either of these equations could represent this hanger:

$$14=2(x+3)$$ or $$14=2x+6$$

2. Find the weight of one circle. Be prepared to explain your reasoning.

### 8.3: Use Hangers to Understand Equation Solving, Again

Here are some balanced hangers. Each piece is labeled with its weight.

For each diagram:

1. Assign one of these equations to each hanger:

$$2(x+5)=16$$

$$3(y+200)=3\!,000$$

$$20.8=4(z+1.1)$$

$$\frac{20}{3}=2\left(w+\frac23\right)$$

2. Explain how to figure out the weight of a piece labeled with a letter by reasoning about the diagram.
3. Explain how to figure out the weight of a piece labeled with a letter by reasoning about the equation.

### Summary

The balanced hanger shows 3 equal, unknown weights and 3 2-unit weights on the left and an 18-unit weight on the right.

There are 3 unknown weights plus 6 units of weight on the left. We could represent this balanced hanger with an equation and solve the equation the same way we did before.

\begin {align} 3x+6&=18 \\ 3x&=12 \\ x&=4 \\ \end{align}

Since there are 3 groups of $$x+2$$ on the left, we could represent this hanger with a different equation: $$3(x+2)=18$$.

The two sides of the hanger balance with these weights: 3 groups of $$x+2$$ on one side, and 18, or 3 groups of 6, on the other side.

The two sides of the hanger will balance with $$\frac13$$ of the weight on each side: $$\frac13 \boldcdot 3(x+2) = \frac13 \boldcdot 18$$.

We can remove 2 units of weight from each side, and the hanger will stay balanced. This is the same as subtracting 2 from each side of the equation.

An equation for the new balanced hanger is $$x=4$$. This gives the solution to the original equation.

Here is a concise way to write the steps above:

\begin{align} 3(x+2) &= 18 \\ x + 2 &= 6 & \text{after multiplying each side by } \tfrac13 \\ x &= 4 & \text{after subtracting 2 from each side} \\ \end{align}