Lesson 3Staying in Balance

Learning Goal

Let’s use balanced hangers to help us solve equations.

Learning Targets

I can compare doing the same thing to the weights on each side of a balanced hanger to solving equations by subtracting the same amount from each side or dividing each side by the same number.
I can explain what a balanced hanger and a true equation have in common.
I can write equations that could represent the weights on a balanced hanger.

Lesson Terms

coefficient
interquartile range (IQR)
solution to an equation
variable

Warm Up: Hanging Around

Hanger A has a green triangle on the right hanging lower than a blue square on the left. Hanger B is hanging evenly with a green triangle on the left and 3 blue squares on right.

Problem 1

For diagram A, find:

One thing that must be true
One thing that could be true or false
One thing that cannot possibly be true

Problem 2

For diagram B, find:

One thing that must be true
One thing that could be true or false
One thing that cannot possibly be true

Activity 1: Match Equations and Hangers

Problem 1

Match each hanger to an equation.

$+ 3 = 6$
$3 \cdot = 6$
$6 = + 1$
$6 = 3 \cdot$

Problem 2

Using the hangers in the previous problem, complete the equation by writing $x$ , $y$ , $z$ , or $w$ in the empty box.

$+ 3 = 6$
$3 \cdot$ $= 6$
$6 =$ $+ 1$
$6 = 3 \cdot$

Problem 3

Find a solution to each equation. Use the hangers from problem 1 to explain what each solution means.

$+ 3 = 6$
$3 \cdot = 6$
$6 = + 1$
$6 = 3 \cdot$

Activity 2: Connecting Diagrams to Equations and Solutions

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

Problem 1

Balanced hanger, left side, 1 circle, x, 1 rectangle, 3, right side, 1 rectangle, 8.

Write an equation.
Explain how to reason with the diagram to find the weight of a piece with a letter.
Explain how to reason with the equation to find the weight of a piece with a letter.

Problem 2

Balanced hanger, left side, 1 rectangle, 12, right side, 2 identical pentagons, y.

Write an equation.
Explain how to reason with the diagram to find the weight of a piece with a letter.
Explain how to reason with the equation to find the weight of a piece with a letter.

Problem 3

Balanced hanger, left side, 1 rectangle, 11, right side, four identical triangles, z.

Write an equation.
Explain how to reason with the diagram to find the weight of a piece with a letter.
Explain how to reason with the diagram to find the weight of a piece with a letter.

Problem 4

Balanced hanger, left side, 1 rectangle, 13 and four fifths, right side, 1 crown, w, 1 rectangle, 3 and four fifths.

Write an equation.
Explain how to reason with the diagram to find the weight of a piece with a letter.
Explain how to reason with the diagram to find the weight of a piece with a letter.

Are you ready for more?

When you have the time, visit the site to solve some trickier puzzles that use hanger diagrams like the ones in this lesson. You can even build new ones. (If you want to do this during class, check with your teacher first!)

Lesson Summary

A hanger stays balanced when the weights on both sides are equal. We can change the weights and the hanger will stay balanced as long as both sides are changed in the same way. For example, adding 2 pounds to each side of a balanced hanger will keep it balanced. Removing half of the weight from each side will also keep it balanced.

An equation can be compared to a balanced hanger. We can change the equation, but for a true equation to remain true, the same thing must be done to both sides of the equal sign. If we add or subtract the same number on each side, or multiply or divide each side by the same number, the new equation will still be true.

This way of thinking can help us find solutions to equations. Instead of checking different values, we can think about subtracting the same amount from each side or dividing each side by the same number.

Balanced hanger A has 3 squares, x on left and rectangle labeled 11 on the right. Balanced Hanger B has a rectangle labeled 11 on left and on right is triangle, y and circle, 5.

Diagram A can be represented by the equation $3 x = 11$ .

If we break the 11 into 3 equal parts, each part will have the same weight as a block with an $x$ .

Splitting each side of the hanger into 3 equal parts is the same as dividing each side of the equation by 3.

$3 x$ divided by 3 is $x$ .
11 divided by 3 is $\frac{11}{3}$ .
If $3 x = 11$ is true, then $x = \frac{11}{3}$ is true.
The solution to $3 x = 11$ is $\frac{11}{3}$ .

Diagram B can be represented with the equation $11 = y + 5$ .

If we remove a weight of 5 from each side of the hanger, it will stay in balance.

Removing 5 from each side of the hanger is the same as subtracting 5 from each side of the equation.

$11 - 5$ is 6.
$y + 5 - 5$ is $y$ .
If $11 = y + 5$ is true, then $6 = y$ is true.
The solution to $11 = y + 5$ is 6.