Algebra equations worksheet with problems to solve for X and Y.

Problem: Solve the given algebra equations for \( x \).

#### Part 1: Solving for \( x \) in simple linear equations

We will solve each equation step by step.

---

Equation 1: \( 4x + 16 = 24 \)

1. Subtract 16 from both sides:
\[
4x + 16 - 16 = 24 - 16
\]
\[
4x = 8
\]

2. Divide both sides by 4:
\[
\frac{4x}{4} = \frac{8}{4}
\]
\[
x = 2
\]

Solution: \( x = 2 \)

---

Equation 2: \( 33 - 3x = 12 \)

1. Subtract 33 from both sides:
\[
33 - 3x - 33 = 12 - 33
\]
\[
-3x = -21
\]

2. Divide both sides by -3:
\[
\frac{-3x}{-3} = \frac{-21}{-3}
\]
\[
x = 7
\]

Solution: \( x = 7 \)

---

Equation 3: \( x - 6 = 17 \)

1. Add 6 to both sides:
\[
x - 6 + 6 = 17 + 6
\]
\[
x = 23
\]

Solution: \( x = 23 \)

---

Equation 4: \( 10x - 6 = 34 \)

1. Add 6 to both sides:
\[
10x - 6 + 6 = 34 + 6
\]
\[
10x = 40
\]

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

Solution: \( x = 4 \)

---

Equation 5: \( 23 + 6x = 29 \)

1. Subtract 23 from both sides:
\[
23 + 6x - 23 = 29 - 23
\]
\[
6x = 6
\]

2. Divide both sides by 6:
\[
\frac{6x}{6} = \frac{6}{6}
\]
\[
x = 1
\]

Solution: \( x = 1 \)

---

Equation 6: \( 2x + 8 = 16 \)

1. Subtract 8 from both sides:
\[
2x + 8 - 8 = 16 - 8
\]
\[
2x = 8
\]

2. Divide both sides by 2:
\[
\frac{2x}{2} = \frac{8}{2}
\]
\[
x = 4
\]

Solution: \( x = 4 \)

---

Equation 7: \( 4(x + 3) - 2x + 8 = 28 \)

1. Distribute the 4:
\[
4(x + 3) = 4x + 12
\]
So the equation becomes:
\[
4x + 12 - 2x + 8 = 28
\]

2. Combine like terms:
\[
(4x - 2x) + (12 + 8) = 28
\]
\[
2x + 20 = 28
\]

3. Subtract 20 from both sides:
\[
2x + 20 - 20 = 28 - 20
\]
\[
2x = 8
\]

4. Divide both sides by 2:
\[
\frac{2x}{2} = \frac{8}{2}
\]
\[
x = 4
\]

Solution: \( x = 4 \)

---

Equation 8: \( 34 - 5x + 2(x - 2) = 15 \)

1. Distribute the 2:
\[
2(x - 2) = 2x - 4
\]
So the equation becomes:
\[
34 - 5x + 2x - 4 = 15
\]

2. Combine like terms:
\[
(-5x + 2x) + (34 - 4) = 15
\]
\[
-3x + 30 = 15
\]

3. Subtract 30 from both sides:
\[
-3x + 30 - 30 = 15 - 30
\]
\[
-3x = -15
\]

4. Divide both sides by -3:
\[
\frac{-3x}{-3} = \frac{-15}{-3}
\]
\[
x = 5
\]

Solution: \( x = 5 \)

---

Equation 9: \( 9x - 6(x + 2) - 6 = 12 \)

1. Distribute the -6:
\[
-6(x + 2) = -6x - 12
\]
So the equation becomes:
\[
9x - 6x - 12 - 6 = 12
\]

2. Combine like terms:
\[
(9x - 6x) + (-12 - 6) = 12
\]
\[
3x - 18 = 12
\]

3. Add 18 to both sides:
\[
3x - 18 + 18 = 12 + 18
\]
\[
3x = 30
\]

4. Divide both sides by 3:
\[
\frac{3x}{3} = \frac{30}{3}
\]
\[
x = 10
\]

Solution: \( x = 10 \)

---

Equation 10: \( 10 + 3x + 2(x - 3) = 29 \)

1. Distribute the 2:
\[
2(x - 3) = 2x - 6
\]
So the equation becomes:
\[
10 + 3x + 2x - 6 = 29
\]

2. Combine like terms:
\[
(3x + 2x) + (10 - 6) = 29
\]
\[
5x + 4 = 29
\]

3. Subtract 4 from both sides:
\[
5x + 4 - 4 = 29 - 4
\]
\[
5x = 25
\]

4. Divide both sides by 5:
\[
\frac{5x}{5} = \frac{25}{5}
\]
\[
x = 5
\]

Solution: \( x = 5 \)

---

Part 2: Solving for \( x \) where \( y \) is given

#### Equation 1: \( y = 4, x = ? \)
Given: \( 7x - 3 + 4y = 27 \)

1. Substitute \( y = 4 \):
\[
7x - 3 + 4(4) = 27
\]
\[
7x - 3 + 16 = 27
\]

2. Simplify:
\[
7x + 13 = 27
\]

3. Subtract 13 from both sides:
\[
7x + 13 - 13 = 27 - 13
\]
\[
7x = 14
\]

4. Divide both sides by 7:
\[
\frac{7x}{7} = \frac{14}{7}
\]
\[
x = 2
\]

Solution: \( x = 2 \)

---

#### Equation 2: \( y = 2, x = ? \)
Given: \( 2y + 8 + 5x = 27 \)

1. Substitute \( y = 2 \):
\[
2(2) + 8 + 5x = 27
\]
\[
4 + 8 + 5x = 27
\]

2. Simplify:
\[
12 + 5x = 27
\]

3. Subtract 12 from both sides:
\[
12 + 5x - 12 = 27 - 12
\]
\[
5x = 15
\]

4. Divide both sides by 5:
\[
\frac{5x}{5} = \frac{15}{5}
\]
\[
x = 3
\]

Solution: \( x = 3 \)

---

#### Equation 3: \( y = 5, x = ? \)
Given: \( 4(x - 3) + y = 21 \)

1. Substitute \( y = 5 \):
\[
4(x - 3) + 5 = 21
\]

2. Subtract 5 from both sides:
\[
4(x - 3) + 5 - 5 = 21 - 5
\]
\[
4(x - 3) = 16
\]

3. Divide both sides by 4:
\[
\frac{4(x - 3)}{4} = \frac{16}{4}
\]
\[
x - 3 = 4
\]

4. Add 3 to both sides:
\[
x - 3 + 3 = 4 + 3
\]
\[
x = 7
\]

Solution: \( x = 7 \)

---

#### Equation 4: \( y = 3, x = ? \)
Given: \( 4y + 6(5 - 2x) = 6 \)

1. Substitute \( y = 3 \):
\[
4(3) + 6(5 - 2x) = 6
\]
\[
12 + 6(5 - 2x) = 6
\]

2. Distribute the 6:
\[
6(5 - 2x) = 30 - 12x
\]
So the equation becomes:
\[
12 + 30 - 12x = 6
\]

3. Simplify:
\[
42 - 12x = 6
\]

4. Subtract 42 from both sides:
\[
42 - 12x - 42 = 6 - 42
\]
\[
-12x = -36
\]

5. Divide both sides by -12:
\[
\frac{-12x}{-12} = \frac{-36}{-12}
\]
\[
x = 3
\]

Solution: \( x = 3 \)

---

Final Answers:

1. \( x = 2 \)
2. \( x = 7 \)
3. \( x = 23 \)
4. \( x = 4 \)
5. \( x = 1 \)
6. \( x = 4 \)
7. \( x = 4 \)
8. \( x = 5 \)
9. \( x = 10 \)
10. \( x = 5 \)

For the second part:
1. \( x = 2 \)
2. \( x = 3 \)
3. \( x = 7 \)
4. \( x = 3 \)

\[
\boxed{x = 2, 7, 23, 4, 1, 4, 4, 5, 10, 5, 2, 3, 7, 3}
\]