Worksheet #6 Section 2.2 — Fractional Linear Equations - Free Printable

Problem: Solve each equation, leaving your final answer as a whole number or a fraction in reduced form.

The worksheet contains 30 equations divided into two columns. Below, I will solve the equations step by step for clarity.

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#### Column 1

1. Equation (1): $\frac{2}{3}x = 3$

- Multiply both sides by $\frac{3}{2}$ to isolate $x$:
\[
x = 3 \cdot \frac{3}{2} = \frac{9}{2}
\]
- Final answer: $\boxed{\frac{9}{2}}$

2. Equation (2): $5\frac{x}{7} = 10$

- Rewrite $5\frac{x}{7}$ as $\frac{5x}{7}$:
\[
\frac{5x}{7} = 10
\]
- Multiply both sides by 7:
\[
5x = 70
\]
- Divide both sides by 5:
\[
x = 14
\]
- Final answer: $\boxed{14}$

3. Equation (3): $2 + \frac{3}{4} = 4$

- Subtract 2 from both sides:
\[
\frac{3}{4} = 2
\]
- This equation is not solvable for $x$, as it is a statement rather than an equation with a variable. It appears there might be a typo. Assuming the equation should involve $x$, let's move on.

4. Equation (4): $\frac{s + 2}{3} = 8$

- Multiply both sides by 3:
\[
s + 2 = 24
\]
- Subtract 2 from both sides:
\[
s = 22
\]
- Final answer: $\boxed{22}$

5. Equation (5): $\frac{3x - 1}{4} = 2x + \frac{1}{3}$

- Eliminate the fractions by finding a common denominator (12):
\[
12 \cdot \frac{3x - 1}{4} = 12 \cdot \left(2x + \frac{1}{3}\right)
\]
\[
3(3x - 1) = 12 \cdot 2x + 12 \cdot \frac{1}{3}
\]
\[
9x - 3 = 24x + 4
\]
- Subtract $9x$ from both sides:
\[
-3 = 15x + 4
\]
- Subtract 4 from both sides:
\[
-7 = 15x
\]
- Divide by 15:
\[
x = -\frac{7}{15}
\]
- Final answer: $\boxed{-\frac{7}{15}}$

6. Equation (6): $\frac{2z}{3} - \frac{3}{4} = \frac{5}{12}$

- Find a common denominator (12) for all terms:
\[
\frac{8z}{12} - \frac{9}{12} = \frac{5}{12}
\]
- Combine the fractions on the left side:
\[
\frac{8z - 9}{12} = \frac{5}{12}
\]
- Since the denominators are the same, equate the numerators:
\[
8z - 9 = 5
\]
- Add 9 to both sides:
\[
8z = 14
\]
- Divide by 8:
\[
z = \frac{14}{8} = \frac{7}{4}
\]
- Final answer: $\boxed{\frac{7}{4}}$

7. Equation (7): $\frac{y + 1}{3} = \frac{3y - 1}{4} - 2$

- Eliminate the fractions by finding a common denominator (12):
\[
12 \cdot \frac{y + 1}{3} = 12 \cdot \left(\frac{3y - 1}{4} - 2\right)
\]
\[
4(y + 1) = 3(3y - 1) - 24
\]
\[
4y + 4 = 9y - 3 - 24
\]
\[
4y + 4 = 9y - 27
\]
- Subtract $4y$ from both sides:
\[
4 = 5y - 27
\]
- Add 27 to both sides:
\[
31 = 5y
\]
- Divide by 5:
\[
y = \frac{31}{5}
\]
- Final answer: $\boxed{\frac{31}{5}}$

8. Equation (8): $\frac{3}{4}(4x - 2) = 4$

- Distribute $\frac{3}{4}$:
\[
\frac{3}{4} \cdot 4x - \frac{3}{4} \cdot 2 = 4
\]
\[
3x - \frac{3}{2} = 4
\]
- Add $\frac{3}{2}$ to both sides:
\[
3x = 4 + \frac{3}{2}
\]
\[
3x = \frac{8}{2} + \frac{3}{2} = \frac{11}{2}
\]
- Divide by 3:
\[
x = \frac{11}{2} \cdot \frac{1}{3} = \frac{11}{6}
\]
- Final answer: $\boxed{\frac{11}{6}}$

9. Equation (9): $\frac{5t + 2}{3} = \frac{2t - 1}{4}$

- Eliminate the fractions by finding a common denominator (12):
\[
12 \cdot \frac{5t + 2}{3} = 12 \cdot \frac{2t - 1}{4}
\]
\[
4(5t + 2) = 3(2t - 1)
\]
\[
20t + 8 = 6t - 3
\]
- Subtract $6t$ from both sides:
\[
14t + 8 = -3
\]
- Subtract 8 from both sides:
\[
14t = -11
\]
- Divide by 14:
\[
t = -\frac{11}{14}
\]
- Final answer: $\boxed{-\frac{11}{14}}$

10. Equation (10): $\frac{4x + 3}{2} = 5$

- Multiply both sides by 2:
\[
4x + 3 = 10
\]
- Subtract 3 from both sides:
\[
4x = 7
\]
- Divide by 4:
\[
x = \frac{7}{4}
\]
- Final answer: $\boxed{\frac{7}{4}}$

---

#### Column 2

1. Equation (1): $2 + \frac{x}{2} = \frac{7}{4}$

- Subtract 2 from both sides:
\[
\frac{x}{2} = \frac{7}{4} - 2
\]
\[
\frac{x}{2} = \frac{7}{4} - \frac{8}{4} = -\frac{1}{4}
\]
- Multiply both sides by 2:
\[
x = -\frac{1}{4} \cdot 2 = -\frac{1}{2}
\]
- Final answer: $\boxed{-\frac{1}{2}}$

2. Equation (2): $\frac{2x + 3}{4} = 5$

- Multiply both sides by 4:
\[
2x + 3 = 20
\]
- Subtract 3 from both sides:
\[
2x = 17
\]
- Divide by 2:
\[
x = \frac{17}{2}
\]
- Final answer: $\boxed{\frac{17}{2}}$

3. Equation (3): $2z + \frac{3z}{5} = \frac{4}{5}$

- Combine like terms on the left side:
\[
2z + \frac{3z}{5} = \frac{10z}{5} + \frac{3z}{5} = \frac{13z}{5}
\]
- Set the equation:
\[
\frac{13z}{5} = \frac{4}{5}
\]
- Since the denominators are the same, equate the numerators:
\[
13z = 4
\]
- Divide by 13:
\[
z = \frac{4}{13}
\]
- Final answer: $\boxed{\frac{4}{13}}$

4. Equation (4): $\frac{z^2 - 1}{z - 1} = 5$

- Factor the numerator:
\[
\frac{(z - 1)(z + 1)}{z - 1} = 5
\]
- Cancel $z - 1$ (assuming $z \neq 1$):
\[
z + 1 = 5
\]
- Subtract 1 from both sides:
\[
z = 4
\]
- Final answer: $\boxed{4}$

5. Equation (5): $\frac{x - 2}{2x + 3} = \frac{3}{5}$

- Cross-multiply:
\[
5(x - 2) = 3(2x + 3)
\]
\[
5x - 10 = 6x + 9
\]
- Subtract $5x$ from both sides:
\[
-10 = x + 9
\]
- Subtract 9 from both sides:
\[
x = -19
\]
- Final answer: $\boxed{-19}$

6. Equation (6): $\frac{3}{4} \cdot \frac{2x - 6}{6} - \frac{4}{3}$

- Simplify $\frac{2x - 6}{6}$:
\[
\frac{2x - 6}{6} = \frac{2(x - 3)}{6} = \frac{x - 3}{3}
\]
- Substitute back:
\[
\frac{3}{4} \cdot \frac{x - 3}{3} - \frac{4}{3}
\]
\[
\frac{x - 3}{4} - \frac{4}{3}
\]
- Find a common denominator (12):
\[
\frac{3(x - 3)}{12} - \frac{16}{12} = \frac{3x - 9 - 16}{12} = \frac{3x - 25}{12}
\]
- Final answer: $\boxed{\frac{3x - 25}{12}}$

7. Equation (7): $\frac{y + 1}{3} = \frac{3y - 1}{4} - 2$

- This equation is identical to Column 1, Equation (7). The solution is:
\[
y = \frac{31}{5}
\]
- Final answer: $\boxed{\frac{31}{5}}$

8. Equation (8): $\frac{3}{4} \cdot \frac{2x - 6}{6} - \frac{4}{3}$

- This equation is identical to Column 1, Equation (6). The solution is:
\[
x = \frac{7}{4}
\]
- Final answer: $\boxed{\frac{7}{4}}$

9. Equation (9): $\frac{t - 2}{7} - \frac{2t}{14} = \frac{3}{7}$

- Simplify $\frac{2t}{14}$:
\[
\frac{2t}{14} = \frac{t}{7}
\]
- Substitute back:
\[
\frac{t - 2}{7} - \frac{t}{7} = \frac{3}{7}
\]
- Combine the fractions on the left side:
\[
\frac{t - 2 - t}{7} = \frac{3}{7}
\]
\[
\frac{-2}{7} = \frac{3}{7}
\]
- This equation is not solvable for $t$, as it leads to a contradiction. Assuming there is a typo, we move on.

10. Equation (10): $\frac{3}{4}(x + 2) - (x + 4) = 3x + 2$

- Distribute $\frac{3}{4}$:
\[
\frac{3}{4}x + \frac{3}{2} - x - 4 = 3x + 2
\]
- Combine like terms:
\[
\left(\frac{3}{4}x - x\right) + \left(\frac{3}{2} - 4\right) = 3x + 2
\]
\[
\left(-\frac{1}{4}x\right) + \left(\frac{3}{2} - \frac{8}{2}\right) = 3x + 2
\]
\[
-\frac{1}{4}x - \frac{5}{2} = 3x + 2
\]
- Eliminate the fractions by multiplying through by 4:
\[
4 \left(-\frac{1}{4}x\right) + 4 \left(-\frac{5}{2}\right) = 4(3x) + 4(2)
\]
\[
-x - 10 = 12x + 8
\]
- Add $x$ to both sides:
\[
-10 = 13x + 8
\]
- Subtract 8 from both sides:
\[
-18 = 13x
\]
- Divide by 13:
\[
x = -\frac{18}{13}
\]
- Final answer: $\boxed{-\frac{18}{13}}$

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{Column 1:} & \frac{9}{2}, 14, 22, -\frac{7}{15}, \frac{7}{4}, \frac{31}{5}, \frac{11}{6}, -\frac{11}{14}, \frac{7}{4} \\
\text{Column 2:} & -\frac{1}{2}, \frac{17}{2}, \frac{4}{13}, 4, -19, \frac{3x - 25}{12}, \frac{31}{5}, \frac{7}{4}, \text{No solution}, -\frac{18}{13}
\end{array}
}
\]