# 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}