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 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}