lesson 17 | session 1
complete problems 3–5.
3 the hanger diagram models the equation $2b = 4$. use the diagram to find the value of $b$. show your reasoning.
hanger diagram with two $b$ circles on left and four 1 squares on right
4 the hanger diagram models $2b + 2 = 6$. you can use this diagram to find the value of $b$.
a. what can you remove from both sides of the hanger diagram so only variables are on the left side? cross off what you can remove.
b. fill in the blank to show the equation the hanger diagram now models.
$2b = \underline{\quad\quad}$
c. how can you use the diagram to find the value of $b$?
d. what does the diagram show is the value of $b$?
hanger diagram with two $b$ circles and two 1 squares on left, six 1 squares on right
5 think about the equation $3y - 9 = 24$.
a. the value of $3y$ is 33. how do you know this is true?
b. the value of $y$ is 11. how do you know this is true?

Question

Accepted Answer

### Problem 3
# Explanation:
## Step1: Analyze the hanger diagram
The hanger has 2 $ b $-weights on the left and 4 $ 1 $-weights on the right, modeling $ 2b = 4 $.
## Step2: Solve for $ b $
To find $ b $, we divide both sides of the equation $ 2b = 4 $ by 2.
$ b=\frac{4}{2} $
## Step3: Calculate the value
$ \frac{4}{2}=2 $
# Answer:
The value of $ b $ is 2.

### Problem 4
#### Part a
# Brief Explanations:
In the hanger diagram for $ 2b + 2 = 6 $, there are 2 $ 1 $-weights on the left (below the $ b $-weights) and 2 $ 1 $-weights on the right (the top two). To have only variables on the left, we can remove 2 $ 1 $-weights from both sides (cross off the two $ 1 $-weights from the left below the $ b $-weights and the top two $ 1 $-weights from the right).
# Answer:
Remove 2 (one - unit) weights from both sides (cross off the two 1 - weight blocks from the left side below the $ b $-weights and the top two 1 - weight blocks from the right side).

#### Part b
# Explanation:
## Step1: Analyze the original equation
The original equation is $ 2b + 2=6 $. When we remove 2 (1 - unit) weights from both sides, we subtract 2 from both sides of the equation.
$ 2b+2 - 2=6 - 2 $
## Step2: Simplify the equation
Simplifying both sides, we get $ 2b=4 $
# Answer:
4

#### Part c
# Brief Explanations:
After removing the 2 (1 - unit) weights from both sides, we have $ 2b = 4 $ (from part b). The hanger now has 2 $ b $-weights on the left and 4 $ 1 $-weights on the right. To find $ b $, we can divide the 4 (1 - unit) weights into two equal groups (since there are 2 $ b $-weights). Each group will have $ \frac{4}{2}=2 $ (1 - unit) weights, so each $ b $-weight is equal to 2 (1 - unit) weights, meaning $ b = 2 $.
# Answer:
After removing 2 (1 - unit) weights from both sides, we have $ 2b=4 $. We can divide the 4 (1 - unit) weights on the right into 2 equal groups (since there are 2 $ b $-weights on the left). Each group has 2 (1 - unit) weights, so $ b = 2 $.

#### Part d
# Brief Explanations:
From part c, after simplifying the hanger diagram, we found that when $ 2b = 4 $, dividing the 4 (1 - unit) weights into 2 equal parts (for the 2 $ b $-weights) gives us that each $ b $-weight is equal to 2 (1 - unit) weights. So the value of $ b $ is 2.
# Answer:
The value of $ b $ is 2.

### Problem 5
#### Part a
# Brief Explanations:
We know that in the equation $ 3y-9 = 24 $, we can use the addition property of equality. The addition property of equality states that if we add the same number to both sides of an equation, the equation remains true. If we add 9 to both sides of $ 3y - 9=24 $, we get $ 3y-9 + 9=24 + 9 $. Simplifying the left side ( $ - 9+9 = 0 $) and the right side ( $ 24 + 9=33 $), we find that $ 3y=33 $.
# Answer:
We know this is true by using the addition property of equality. Add 9 to both sides of the equation $ 3y - 9 = 24 $: $ 3y-9 + 9=24 + 9 $, which simplifies to $ 3y=33 $.

#### Part b
# Brief Explanations:
We know that in the equation $ 3y = 33 $ (from part a), we can use the division property of equality. The division property of equality states that if we divide both sides of an equation by the same non - zero number, the equation remains true. If we divide both sides of $ 3y = 33 $ by 3, we get $ \frac{3y}{3}=\frac{33}{3} $. Simplifying the left side ( $ \frac{3y}{3}=y $) and the right side ( $ \frac{33}{3}=11 $), we find that $ y = 11 $.
# Answer:
We know this is true by using the division property of equality. Divide both sides of the equation $ 3y = 33 $ by 3: $ \frac{3y}{3}=\frac{33}{3} $, which simplifies to $ y = 11 $.

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QUESTION IMAGE

Question

Explanation:

Problem 3

Step1: Analyze the hanger diagram

Step2: Solve for \( b \)

Step3: Calculate the value

Step1: Analyze the original equation

Step2: Simplify the equation

Answer:

Problem 4

Part a