We are asked to solve the following equations: 28. $\frac{4}{x+1} = \frac{7}{3x-2}$ 29. $\frac{x+1}{2} + \frac{x-1}{3} = \frac{1}{6}$ 31. $\frac{1}{2}(x-1) - \frac{1}{6}(x+1) = 0$ 32. $\frac{1}{4}(x+5) - \frac{2x}{3} = 0$

1. Problem Description

We are asked to solve the following equations:

2

8. $\frac{4}{x+1} = \frac{7}{3x-2}$

2

9. $\frac{x+1}{2} + \frac{x-1}{3} = \frac{1}{6}$

3

1. $\frac{1}{2}(x-1) - \frac{1}{6}(x+1) = 0$

3

2. $\frac{1}{4}(x+5) - \frac{2x}{3} = 0$

2. Solution Steps

2

8. $\frac{4}{x+1} = \frac{7}{3x-2}$

Cross-multiply to get:

$4(3x-2) = 7(x+1)$

$12x - 8 = 7x + 7$

$12x - 7x = 7 + 8$

$5x = 15$

$x = \frac{15}{5}$

$x = 3$

2

9. $\frac{x+1}{2} + \frac{x-1}{3} = \frac{1}{6}$

Multiply both sides by 6:

$6(\frac{x+1}{2} + \frac{x-1}{3}) = 6(\frac{1}{6})$

$3(x+1) + 2(x-1) = 1$

$3x + 3 + 2x - 2 = 1$

$5x + 1 = 1$

$5x = 0$

$x = 0$

3

1. $\frac{1}{2}(x-1) - \frac{1}{6}(x+1) = 0$

Multiply both sides by 6:

$6(\frac{1}{2}(x-1) - \frac{1}{6}(x+1)) = 6(0)$

$3(x-1) - (x+1) = 0$

$3x - 3 - x - 1 = 0$

$2x - 4 = 0$

$2x = 4$

$x = 2$

3

2. $\frac{1}{4}(x+5) - \frac{2x}{3} = 0$

Multiply both sides by 12:

$12(\frac{1}{4}(x+5) - \frac{2x}{3}) = 12(0)$

$3(x+5) - 4(2x) = 0$

$3x + 15 - 8x = 0$

$-5x + 15 = 0$

$-5x = -15$

$x = \frac{-15}{-5}$

$x = 3$

3. Final Answer

2

8. $x = 3$

2

9. $x = 0$

3

1. $x = 2$

3

2. $x = 3$