# Lesson 7: Reasoning about Solving Equations (Part 1)

Let’s see how a balanced hanger is like an equation and how moving its weights is like solving the equation.

## 7.1: Hanger Diagrams In the two diagrams, all the triangles weigh the same and all the squares weigh the same.

For each diagram, come up with . . .

1. One thing that must be true
2. One thing that could be true
3. One thing that cannot possibly be true

## 7.2: Hanger and Equation Matching

On each balanced hanger, figures with the same letter have the same weight. • $2 \boxed{\phantom{3}} + 3 = 5$
• $3 \boxed{\phantom{3}} + 2 = 3$
• $6 = 2 \boxed{\phantom{3}} + 3$
• $7 = 3 \boxed{\phantom{3}} + 1$
1. Match each hanger to an equation. Complete the equation by writing $x$, $y$, $z$, or $w$ in the empty box.
2. Find the solution to each equation. Use the hanger to explain what the solution means.

## 7.3: Use Hangers to Understand Equation Solving Here are some balanced hangers where each piece is labeled with its weight. For each diagram:

1. Write an equation.
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

In this lesson, we worked with two ways to show that two amounts are equal: a balanced hanger and an equation. We can use a balanced hanger to think about steps to finding an unknown amount in an associated equation.

The hanger shows a total weight of 7 units on one side that is balanced with 3 equal, unknown weights and a 1-unit weight on the other. An equation that represents the relationship is $7=3x+1$. We can remove a weight of 1 unit from each side and the hanger will stay balanced. This is the same as subtracting 1 from each side of the equation. An equation for the new balanced hanger is $6=3x$. So the hanger will balance with $\frac13$ of the weight on each side: $\frac13 \boldcdot 6 = \frac13 \boldcdot 3x$. The two sides of the hanger balance with these weights: 6 1-unit weights on one side and 3 weights of unknown size on the other side. Here is a concise way to write the steps above:

\begin {align} 7&=3x+1 & \\ 6&=3x & \text{after subtracting 1 from each side} \\ 2 &= x & \text{after multiplying each side by } \tfrac13 \\ \end{align}