Printable worksheet for solving linear equations without negative numbers, divided into four sections with varying levels of difficulty.

Problem: Solve the following equations using inverse operations. Show your work.

#### Equations:
1. \( 10 \times y + 6 = 46 \)
2. \( 11 \times y - 6 = 104 \)
3. \( 4 \times y - y = 36 \)
4. \( 14 + y \times 10 = 144 \)
5. \( 1 + y \times 3 = 28 \)
6. \( 13 + 13 \times y = 91 \)
7. \( 4 \times y - 12 = 28 \)
8. \( 10^2 + y \times 8 = 164 \)
9. \( 6^2 - 15 \times y = -144 \)
10. \( 14 \times y + 15 = 57 \)

---

Solution:

#### 1. \( 10 \times y + 6 = 46 \)

1. Subtract 6 from both sides:
\[
10 \times y + 6 - 6 = 46 - 6
\]
\[
10 \times y = 40
\]

2. Divide both sides by 10:
\[
\frac{10 \times y}{10} = \frac{40}{10}
\]
\[
y = 4
\]

Answer: \( y = 4 \)

---

#### 2. \( 11 \times y - 6 = 104 \)

1. Add 6 to both sides:
\[
11 \times y - 6 + 6 = 104 + 6
\]
\[
11 \times y = 110
\]

2. Divide both sides by 11:
\[
\frac{11 \times y}{11} = \frac{110}{11}
\]
\[
y = 10
\]

Answer: \( y = 10 \)

---

#### 3. \( 4 \times y - y = 36 \)

1. Simplify the left side:
\[
4y - y = 3y
\]
So the equation becomes:
\[
3y = 36
\]

2. Divide both sides by 3:
\[
\frac{3y}{3} = \frac{36}{3}
\]
\[
y = 12
\]

Answer: \( y = 12 \)

---

#### 4. \( 14 + y \times 10 = 144 \)

1. Subtract 14 from both sides:
\[
14 + 10y - 14 = 144 - 14
\]
\[
10y = 130
\]

2. Divide both sides by 10:
\[
\frac{10y}{10} = \frac{130}{10}
\]
\[
y = 13
\]

Answer: \( y = 13 \)

---

#### 5. \( 1 + y \times 3 = 28 \)

1. Subtract 1 from both sides:
\[
1 + 3y - 1 = 28 - 1
\]
\[
3y = 27
\]

2. Divide both sides by 3:
\[
\frac{3y}{3} = \frac{27}{3}
\]
\[
y = 9
\]

Answer: \( y = 9 \)

---

#### 6. \( 13 + 13 \times y = 91 \)

1. Subtract 13 from both sides:
\[
13 + 13y - 13 = 91 - 13
\]
\[
13y = 78
\]

2. Divide both sides by 13:
\[
\frac{13y}{13} = \frac{78}{13}
\]
\[
y = 6
\]

Answer: \( y = 6 \)

---

#### 7. \( 4 \times y - 12 = 28 \)

1. Add 12 to both sides:
\[
4y - 12 + 12 = 28 + 12
\]
\[
4y = 40
\]

2. Divide both sides by 4:
\[
\frac{4y}{4} = \frac{40}{4}
\]
\[
y = 10
\]

Answer: \( y = 10 \)

---

#### 8. \( 10^2 + y \times 8 = 164 \)

1. Simplify \( 10^2 \):
\[
10^2 = 100
\]
So the equation becomes:
\[
100 + 8y = 164
\]

2. Subtract 100 from both sides:
\[
100 + 8y - 100 = 164 - 100
\]
\[
8y = 64
\]

3. Divide both sides by 8:
\[
\frac{8y}{8} = \frac{64}{8}
\]
\[
y = 8
\]

Answer: \( y = 8 \)

---

#### 9. \( 6^2 - 15 \times y = -144 \)

1. Simplify \( 6^2 \):
\[
6^2 = 36
\]
So the equation becomes:
\[
36 - 15y = -144
\]

2. Subtract 36 from both sides:
\[
36 - 15y - 36 = -144 - 36
\]
\[
-15y = -180
\]

3. Divide both sides by -15:
\[
\frac{-15y}{-15} = \frac{-180}{-15}
\]
\[
y = 12
\]

Answer: \( y = 12 \)

---

#### 10. \( 14 \times y + 15 = 57 \)

1. Subtract 15 from both sides:
\[
14y + 15 - 15 = 57 - 15
\]
\[
14y = 42
\]

2. Divide both sides by 14:
\[
\frac{14y}{14} = \frac{42}{14}
\]
\[
y = 3
\]

Answer: \( y = 3 \)

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \ y = 4 \\
2. & \ y = 10 \\
3. & \ y = 12 \\
4. & \ y = 13 \\
5. & \ y = 9 \\
6. & \ y = 6 \\
7. & \ y = 10 \\
8. & \ y = 8 \\
9. & \ y = 12 \\
10. & \ y = 3
\end{aligned}
}
\]