use addition and subtraction equations to solve problems
math on the spot for problems 1–3, solve each equation. check your answer.
1. $s - 5 = 12$
2. $35 = y - 6$
3. $x - 22 = 44$
4. model with mathematics mary compares the heights of two trees. their heights are shown. write and solve an equation to find the height $h$ of the taller tree.
5. open ended write an equation with a solution of 25. use the variable $x$ and addition.
6. each small square on this scale weighs 1 unit. each larger square weighs 10 units. the weight of the triangle $x$ is unknown. the scale is balanced. what is the weight of the triangle?
a. model with mathematics write an equation to represent the weights shown on the scale.
b. cross out the same weights on each side until $x$ is alone. what is the total weight that you crossed out on each side?
c. write the simplified equation represented by the scale now. what is the weight of the triangle?

Question

Accepted Answer

### Problem 1: Solve $ s - 5 = 12 $
# Explanation:
## Step 1: Isolate the variable $ s $
To solve for $ s $, we add 5 to both sides of the equation to undo the subtraction.
$ s - 5 + 5 = 12 + 5 $
## Step 2: Simplify both sides
Simplifying the left side, $ -5 + 5 = 0 $, so we have $ s = 12 + 5 $. Simplifying the right side, $ 12 + 5 = 17 $.
# Answer:
$ s = 17 $

### Problem 2: Solve $ 35 = y - 6 $
# Explanation:
## Step 1: Isolate the variable $ y $
To solve for $ y $, we add 6 to both sides of the equation to undo the subtraction.
$ 35 + 6 = y - 6 + 6 $
## Step 2: Simplify both sides
Simplifying the left side, $ 35 + 6 = 41 $. Simplifying the right side, $ -6 + 6 = 0 $, so we have $ 41 = y $ or $ y = 41 $.
# Answer:
$ y = 41 $

### Problem 3: Solve $ x - 22 = 44 $
# Explanation:
## Step 1: Isolate the variable $ x $
To solve for $ x $, we add 22 to both sides of the equation to undo the subtraction.
$ x - 22 + 22 = 44 + 22 $
## Step 2: Simplify both sides
Simplifying the left side, $ -22 + 22 = 0 $, so we have $ x = 44 + 22 $. Simplifying the right side, $ 44 + 22 = 66 $.
# Answer:
$ x = 66 $

### Problem 4: (Assuming the shorter tree is, for example, 8 feet and the taller tree is $ h $ feet with a difference, say, the shorter tree is 4 feet shorter. So equation: $ h - 4 = 8 $)
# Explanation:
## Step 1: Write the equation (example)
Suppose the shorter tree is 8 feet and the taller tree $ h $ is 4 feet taller. So $ h - 4 = 8 $.
## Step 2: Solve for $ h $
Add 4 to both sides: $ h - 4 + 4 = 8 + 4 $, so $ h = 12 $. (Note: Actual values depend on the diagram, but the process is adding to isolate $ h $.)
# Answer:
Equation: $ h - 4 = 8 $ (example), Solution: $ h = 12 $ (example, adjust based on diagram)

### Problem 5: Write an equation with solution 25 using $ x $ and addition
# Explanation:
## Step 1: Choose an addition equation
Let's use $ x + a = b $, where $ x = 25 $. Let $ a = 10 $, then $ 25 + 10 = 35 $, so equation: $ x + 10 = 35 $.
## Step 2: Verify
Substitute $ x = 25 $: $ 25 + 10 = 35 $, which is true.
# Answer:
$ x + 10 = 35 $ (or other valid, e.g., $ x + 5 = 30 $)

### Problem 6: (Assuming left side: triangle $ x $ + 2 small squares (1 unit each) + 1 large square (10 units); right side: 3 large squares (10 units each) + 1 small square (1 unit))
#### Part A: Write the equation
# Explanation:
## Step 1: Calculate weights
Left: $ x + 2(1) + 1(10) = x + 2 + 10 = x + 12 $
Right: $ 3(10) + 1(1) = 30 + 1 = 31 $
Equation: $ x + 12 = 31 $
# Answer:
$ x + 12 = 31 $ (adjust based on actual diagram)

#### Part B: Cross out same weights
# Explanation:
## Step 1: Identify common weights
Both sides have small squares (1 unit) and large squares? Wait, left: 2 small, 1 large; right: 3 large, 1 small. Cross out 1 small and 1 large from both.
Left: $ x + 1(1) + 0(10) = x + 1 $
Right: $ 2(10) + 0(1) = 20 $
Wait, better: If left has $ x $, 2 small (1), 1 large (10); right has 3 large (10), 1 small (1). Cross out 1 small and 1 large:
Left: $ x + 1(1) + 0(10) = x + 1 $
Right: $ 2(10) + 0(1) = 20 $
Wait, maybe cross out 1 small: left $ x + 1 + 10 $, right $ 30 + 1 $. Cross out 1 small (1 unit): left $ x + 10 $, right $ 30 $. Then cross out 1 large (10 units): left $ x $, right $ 20 $? Wait, process is to get $ x $ alone.
## Step 2: Cross out
Cross out 1 small square (1 unit) from both sides: left $ x + 2 + 10 - 1 = x + 11 $, right $ 30 + 1 - 1 = 30 $. Then cross out 1 large square (10 units): left $ x + 11 - 10 = x + 1 $, right $ 30 - 10 = 20 $. Wait, maybe my initial assumption is wrong. Let's assume left: $ x $ + 1 small (1) + 2 large (10); right: 3 large (10) + 2 small (1). Then equation: $ x + 1 + 20 = 30 + 2 $ → $ x + 21 = 32 $. Cross out 1 small and 2 large? No, better: cross out same number of small and large. Suppose both have 1 small and 2 large: cross out 1 small (1) and 2 large (10 each) from both. Left: $ x + 1 - 1 + 20 - 20 = x $. Right: $ 30 - 20 + 2 - 1 = 11 $. Wait, this is variable, but the idea is to subtract same weights from both sides to isolate $ x $.
# Answer:
After crossing out, left: $ x $, right: 20 (example, adjust based on diagram). Total weight crossed out: 12 (example).

#### Part C: Simplified equation
# Explanation:
## Step 1: From Part A equation $ x + 12 = 31 $, solve: $ x = 31 - 12 = 19 $ (adjust based on Part A)
# Answer:
Simplified equation: $ x = 19 $ (adjust based on Part A's equation)

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QUESTION IMAGE

Question

Explanation:

Problem 1: Solve \( s - 5 = 12 \)

Step 1: Isolate the variable \( s \)

Step 2: Simplify both sides

Step 1: Isolate the variable \( y \)

Step 2: Simplify both sides

Step 1: Isolate the variable \( x \)

Step 2: Simplify both sides

Answer:

Problem 2: Solve \( 35 = y - 6 \)