Worksheet for solving multi-step equations with fractions.

Problem: Solve each equation involving fractions.

We will solve each equation step by step.

---

#### 1) \( 5\left(2b + \frac{1}{3}\right) = 7b + 1 \)

Step 1: Distribute the 5 on the left-hand side.
\[
5 \cdot 2b + 5 \cdot \frac{1}{3} = 7b + 1
\]
\[
10b + \frac{5}{3} = 7b + 1
\]

Step 2: Eliminate the fraction by multiplying every term by 3 (the least common denominator).
\[
3 \cdot 10b + 3 \cdot \frac{5}{3} = 3 \cdot 7b + 3 \cdot 1
\]
\[
30b + 5 = 21b + 3
\]

Step 3: Isolate the variable \( b \). Subtract \( 21b \) from both sides.
\[
30b - 21b + 5 = 21b - 21b + 3
\]
\[
9b + 5 = 3
\]

Step 4: Subtract 5 from both sides.
\[
9b + 5 - 5 = 3 - 5
\]
\[
9b = -2
\]

Step 5: Divide both sides by 9.
\[
b = \frac{-2}{9}
\]

Solution:
\[
\boxed{b = -\frac{2}{9}}
\]

---

#### 2) \( \frac{9}{7}(1 + x) - \frac{6}{7}x + 1 = 1 \)

Step 1: Simplify the equation. Start by distributing \( \frac{9}{7} \) on the left-hand side.
\[
\frac{9}{7} \cdot 1 + \frac{9}{7} \cdot x - \frac{6}{7}x + 1 = 1
\]
\[
\frac{9}{7} + \frac{9}{7}x - \frac{6}{7}x + 1 = 1
\]

Step 2: Combine like terms involving \( x \).
\[
\frac{9}{7}x - \frac{6}{7}x = \frac{3}{7}x
\]
So the equation becomes:
\[
\frac{9}{7} + \frac{3}{7}x + 1 = 1
\]

Step 3: Combine the constant terms on the left-hand side.
\[
\frac{9}{7} + 1 = \frac{9}{7} + \frac{7}{7} = \frac{16}{7}
\]
So the equation is:
\[
\frac{16}{7} + \frac{3}{7}x = 1
\]

Step 4: Subtract \( \frac{16}{7} \) from both sides.
\[
\frac{16}{7} + \frac{3}{7}x - \frac{16}{7} = 1 - \frac{16}{7}
\]
\[
\frac{3}{7}x = \frac{7}{7} - \frac{16}{7}
\]
\[
\frac{3}{7}x = \frac{-9}{7}
\]

Step 5: Divide both sides by \( \frac{3}{7} \). This is equivalent to multiplying by \( \frac{7}{3} \).
\[
x = \frac{-9}{7} \cdot \frac{7}{3}
\]
\[
x = \frac{-9 \cdot 7}{7 \cdot 3}
\]
\[
x = \frac{-9}{3}
\]
\[
x = -3
\]

Solution:
\[
\boxed{x = -3}
\]

---

#### 3) \( \frac{7}{8}t - \frac{4}{7} = \frac{5}{6}t - \frac{3}{4} \)

Step 1: Eliminate the fractions by finding the least common denominator (LCD). The denominators are 8, 7, 6, and 4. The LCD is 168.

Multiply every term by 168:
\[
168 \cdot \frac{7}{8}t - 168 \cdot \frac{4}{7} = 168 \cdot \frac{5}{6}t - 168 \cdot \frac{3}{4}
\]

Simplify each term:
\[
168 \cdot \frac{7}{8}t = 21 \cdot 7t = 147t
\]
\[
168 \cdot \frac{4}{7} = 24 \cdot 4 = 96
\]
\[
168 \cdot \frac{5}{6}t = 28 \cdot 5t = 140t
\]
\[
168 \cdot \frac{3}{4} = 42 \cdot 3 = 126
\]

So the equation becomes:
\[
147t - 96 = 140t - 126
\]

Step 2: Isolate the variable \( t \). Subtract \( 140t \) from both sides.
\[
147t - 140t - 96 = 140t - 140t - 126
\]
\[
7t - 96 = -126
\]

Step 3: Add 96 to both sides.
\[
7t - 96 + 96 = -126 + 96
\]
\[
7t = -30
\]

Step 4: Divide both sides by 7.
\[
t = \frac{-30}{7}
\]

Solution:
\[
\boxed{t = -\frac{30}{7}}
\]

---

#### 4) \( \frac{8}{5} + 3z = 10z - 4 \)

Step 1: Isolate the variable \( z \). Subtract \( 3z \) from both sides.
\[
\frac{8}{5} + 3z - 3z = 10z - 3z - 4
\]
\[
\frac{8}{5} = 7z - 4
\]

Step 2: Add 4 to both sides.
\[
\frac{8}{5} + 4 = 7z - 4 + 4
\]
\[
\frac{8}{5} + 4 = 7z
\]

Convert 4 to a fraction with a denominator of 5:
\[
4 = \frac{20}{5}
\]
So:
\[
\frac{8}{5} + \frac{20}{5} = 7z
\]
\[
\frac{28}{5} = 7z
\]

Step 3: Divide both sides by 7.
\[
z = \frac{\frac{28}{5}}{7}
\]
\[
z = \frac{28}{5} \cdot \frac{1}{7}
\]
\[
z = \frac{28}{35}
\]
\[
z = \frac{4}{5}
\]

Solution:
\[
\boxed{z = \frac{4}{5}}
\]

---

#### 5) \( \frac{3}{2}h - 10 = 4h \)

Step 1: Eliminate the fraction by multiplying every term by 2 (the denominator of \( \frac{3}{2} \)).
\[
2 \cdot \frac{3}{2}h - 2 \cdot 10 = 2 \cdot 4h
\]
\[
3h - 20 = 8h
\]

Step 2: Isolate the variable \( h \). Subtract \( 3h \) from both sides.
\[
3h - 3h - 20 = 8h - 3h
\]
\[
-20 = 5h
\]

Step 3: Divide both sides by 5.
\[
h = \frac{-20}{5}
\]
\[
h = -4
\]

Solution:
\[
\boxed{h = -4}
\]

---

#### 6) \( \frac{3 + 4n}{4} = \frac{8}{4}n \)

Step 1: Simplify the right-hand side.
\[
\frac{3 + 4n}{4} = 2n
\]

Step 2: Eliminate the fraction by multiplying every term by 4.
\[
4 \cdot \frac{3 + 4n}{4} = 4 \cdot 2n
\]
\[
3 + 4n = 8n
\]

Step 3: Isolate the variable \( n \). Subtract \( 4n \) from both sides.
\[
3 + 4n - 4n = 8n - 4n
\]
\[
3 = 4n
\]

Step 4: Divide both sides by 4.
\[
n = \frac{3}{4}
\]

Solution:
\[
\boxed{n = \frac{3}{4}}
\]

---

#### 7) \( \frac{u - 5}{8} = 2u - 1 \)

Step 1: Eliminate the fraction by multiplying every term by 8.
\[
8 \cdot \frac{u - 5}{8} = 8 \cdot (2u - 1)
\]
\[
u - 5 = 16u - 8
\]

Step 2: Isolate the variable \( u \). Subtract \( u \) from both sides.
\[
u - u - 5 = 16u - u - 8
\]
\[
-5 = 15u - 8
\]

Step 3: Add 8 to both sides.
\[
-5 + 8 = 15u - 8 + 8
\]
\[
3 = 15u
\]

Step 4: Divide both sides by 15.
\[
u = \frac{3}{15}
\]
\[
u = \frac{1}{5}
\]

Solution:
\[
\boxed{u = \frac{1}{5}}
\]

---

#### 8) \( 6v + \frac{4}{9} = 4\left(\frac{1}{2} + v\right) \)

Step 1: Distribute the 4 on the right-hand side.
\[
6v + \frac{4}{9} = 4 \cdot \frac{1}{2} + 4 \cdot v
\]
\[
6v + \frac{4}{9} = 2 + 4v
\]

Step 2: Isolate the variable \( v \). Subtract \( 4v \) from both sides.
\[
6v - 4v + \frac{4}{9} = 2 + 4v - 4v
\]
\[
2v + \frac{4}{9} = 2
\]

Step 3: Subtract \( \frac{4}{9} \) from both sides.
\[
2v + \frac{4}{9} - \frac{4}{9} = 2 - \frac{4}{9}
\]
\[
2v = \frac{18}{9} - \frac{4}{9}
\]
\[
2v = \frac{14}{9}
\]

Step 4: Divide both sides by 2.
\[
v = \frac{\frac{14}{9}}{2}
\]
\[
v = \frac{14}{9} \cdot \frac{1}{2}
\]
\[
v = \frac{14}{18}
\]
\[
v = \frac{7}{9}
\]

Solution:
\[
\boxed{v = \frac{7}{9}}
\]

---

Final Answers:

1. \( b = -\frac{2}{9} \)
2. \( x = -3 \)
3. \( t = -\frac{30}{7} \)
4. \( z = \frac{4}{5} \)
5. \( h = -4 \)
6. \( n = \frac{3}{4} \)
7. \( u = \frac{1}{5} \)
8. \( v = \frac{7}{9} \)

\[
\boxed{
\begin{aligned}
1) & \quad b = -\frac{2}{9} \\
2) & \quad x = -3 \\
3) & \quad t = -\frac{30}{7} \\
4) & \quad z = \frac{4}{5} \\
5) & \quad h = -4 \\
6) & \quad n = \frac{3}{4} \\
7) & \quad u = \frac{1}{5} \\
8) & \quad v = \frac{7}{9}
\end{aligned}
}
\]