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 $2(x+3)$.
 $2 \boldcdot (x+3) $
 $(x + 3)2 $
 $2 \boldcdot x + 2 \boldcdot 3$
 $2 \boldcdot x + 3 $
 $(2 \boldcdot x) + 3$
 $(2 + x)3$
8.2: Either Or

Explain why either of these equations could represent this hanger:
\(14=2(x+3)\) or \(14=2x+6\)
 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:
 Assign one of these equations to each hanger:
\(2(x+5)=16\)
\(3(y+200)=3\!,000\)
\(20.8=4(z+1.1)\)
\(\frac{20}{3}=2\left(w+\frac23\right)\)
 Explain how to figure out the weight of a piece labeled with a letter by reasoning about the diagram.
 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 2unit weights on the left and an 18unit 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:
\(\begin{align} 3(x+2) &= 18 \\ x + 2 &= 6 & \text{after multiplying each side by } \tfrac13 \\ x &= 4 & \text{after subtracting 2 from each side} \\ \end{align} \)