This worksheet offers eight practice problems for solving multi-step linear equations where the variable appears on both sides of the equation.

Problem Analysis:

The worksheet contains multi-step linear equations and a geometry problem. Let's solve each problem step by step.

---

Problem 1: Solve the linear equation for \( b \).

\[ 3(b - 8) = -5b - 48 \]

#### Solution:
1. Distribute the 3 on the left side:
\[
3b - 24 = -5b - 48
\]

2. Add \( 5b \) to both sides to get all \( b \)-terms on one side:
\[
3b + 5b - 24 = -48
\]
\[
8b - 24 = -48
\]

3. Add 24 to both sides to isolate the term with \( b \):
\[
8b = -24
\]

4. Divide both sides by 8:
\[
b = -3
\]

#### Final Answer:
\[
\boxed{-3}
\]

---

Problem 2: Solve the linear equation for \( z \).

\[ -3(2z - 8) = 3(2z + 4) + 24 \]

#### Solution:
1. Distribute the constants on both sides:
\[
-6z + 24 = 6z + 12 + 24
\]
\[
-6z + 24 = 6z + 36
\]

2. Subtract \( 6z \) from both sides to get all \( z \)-terms on one side:
\[
-6z - 6z + 24 = 36
\]
\[
-12z + 24 = 36
\]

3. Subtract 24 from both sides to isolate the term with \( z \):
\[
-12z = 12
\]

4. Divide both sides by -12:
\[
z = -1
\]

#### Final Answer:
\[
\boxed{-1}
\]

---

Problem 3: Solve the linear equation for \( x \).

\[ 2(2x + 9) - 52 = -(5x - 11) \]

#### Solution:
1. Distribute the constants on both sides:
\[
4x + 18 - 52 = -5x + 11
\]
\[
4x - 34 = -5x + 11
\]

2. Add \( 5x \) to both sides to get all \( x \)-terms on one side:
\[
4x + 5x - 34 = 11
\]
\[
9x - 34 = 11
\]

3. Add 34 to both sides to isolate the term with \( x \):
\[
9x = 45
\]

4. Divide both sides by 9:
\[
x = 5
\]

#### Final Answer:
\[
\boxed{5}
\]

---

Problem 4: Solve the linear equation for \( a \).

\[ -3 + \frac{3}{4}a + 13 = 1 - \frac{3}{2}a \]

#### Solution:
1. Simplify the left side:
\[
10 + \frac{3}{4}a = 1 - \frac{3}{2}a
\]

2. Add \( \frac{3}{2}a \) to both sides to get all \( a \)-terms on one side:
\[
10 + \frac{3}{4}a + \frac{3}{2}a = 1
\]

3. Combine the \( a \)-terms. Note that \( \frac{3}{2} = \frac{6}{4} \):
\[
10 + \left( \frac{3}{4} + \frac{6}{4} \right)a = 1
\]
\[
10 + \frac{9}{4}a = 1
\]

4. Subtract 10 from both sides:
\[
\frac{9}{4}a = 1 - 10
\]
\[
\frac{9}{4}a = -9
\]

5. Multiply both sides by \( \frac{4}{9} \) to solve for \( a \):
\[
a = -9 \cdot \frac{4}{9}
\]
\[
a = -4
\]

#### Final Answer:
\[
\boxed{-4}
\]

---

Problem 5: Solve the linear equation for \( k \).

\[ 5k - \frac{282}{5} = 3\left(-\frac{3}{5}k - 12\right) \]

#### Solution:
1. Distribute the 3 on the right side:
\[
5k - \frac{282}{5} = -\frac{9}{5}k - 36
\]

2. Eliminate the fractions by multiplying every term by 5:
\[
5 \cdot 5k - 5 \cdot \frac{282}{5} = 5 \cdot \left( -\frac{9}{5}k \right) - 5 \cdot 36
\]
\[
25k - 282 = -9k - 180
\]

3. Add \( 9k \) to both sides to get all \( k \)-terms on one side:
\[
25k + 9k - 282 = -180
\]
\[
34k - 282 = -180
\]

4. Add 282 to both sides to isolate the term with \( k \):
\[
34k = 102
\]

5. Divide both sides by 34:
\[
k = 3
\]

#### Final Answer:
\[
\boxed{3}
\]

---

Problem 6: Solve the linear equation for \( n \).

\[ 3\left(\frac{7}{4}n - 9\right) = \frac{1}{6}n - \frac{629}{12} \]

#### Solution:
1. Distribute the 3 on the left side:
\[
\frac{21}{4}n - 27 = \frac{1}{6}n - \frac{629}{12}
\]

2. Eliminate the fractions by finding a common denominator (12):
\[
12 \cdot \frac{21}{4}n - 12 \cdot 27 = 12 \cdot \frac{1}{6}n - 12 \cdot \frac{629}{12}
\]
\[
3 \cdot 21n - 324 = 2n - 629
\]
\[
63n - 324 = 2n - 629
\]

3. Subtract \( 2n \) from both sides to get all \( n \)-terms on one side:
\[
63n - 2n - 324 = -629
\]
\[
61n - 324 = -629
\]

4. Add 324 to both sides to isolate the term with \( n \):
\[
61n = -305
\]

5. Divide both sides by 61:
\[
n = -5
\]

#### Final Answer:
\[
\boxed{-5}
\]

---

Problem 7: Geometry Problem

Line segment \( AB \) is on square \( ABCD \). Segment \( EF \) on equilateral triangle \( EFG \) is 2 units longer than \( AB \). Square \( ABCD \) and triangle \( EFG \) have equal perimeters. What is the length of \( AB \)?

#### Solution:
1. Let the length of \( AB \) be \( x \). Since \( ABCD \) is a square, its perimeter is:
\[
4x
\]

2. The length of \( EF \) is \( x + 2 \). Since \( EFG \) is an equilateral triangle, its perimeter is:
\[
3(x + 2)
\]

3. The perimeters of the square and the triangle are equal:
\[
4x = 3(x + 2)
\]

4. Distribute the 3 on the right side:
\[
4x = 3x + 6
\]

5. Subtract \( 3x \) from both sides:
\[
x = 6
\]

#### Final Answer:
\[
\boxed{6}
\]

---

Problem 8: Geometry Problem

Line segment \( AB \) is on square \( ABCD \). Segment \( EF \) on equilateral triangle \( EFG \) is 14 units longer than \( AB \). Square \( ABCD \) and triangle \( EFG \) have equal perimeters. What is the length of \( AB \)?

#### Solution:
1. Let the length of \( AB \) be \( x \). Since \( ABCD \) is a square, its perimeter is:
\[
4x
\]

2. The length of \( EF \) is \( x + 14 \). Since \( EFG \) is an equilateral triangle, its perimeter is:
\[
3(x + 14)
\]

3. The perimeters of the square and the triangle are equal:
\[
4x = 3(x + 14)
\]

4. Distribute the 3 on the right side:
\[
4x = 3x + 42
\]