Solve the equation $\frac{1}{4}x + \frac{1}{5} = \frac{9}{8}(3x - 3)$ for $x$.

1. Problem Description

Solve the equation $\frac{1}{4}x + \frac{1}{5} = \frac{9}{8}(3x - 3)$ for $x$.

2. Solution Steps

First, we expand the right side of the equation:

$\frac{1}{4}x + \frac{1}{5} = \frac{9}{8}(3x) - \frac{9}{8}(3)$

$\frac{1}{4}x + \frac{1}{5} = \frac{27}{8}x - \frac{27}{8}$

Now, we want to eliminate the fractions. The least common multiple of 4, 5, and 8 is

4

0. We multiply both sides of the equation by 40:

$40(\frac{1}{4}x + \frac{1}{5}) = 40(\frac{27}{8}x - \frac{27}{8})$

$40(\frac{1}{4}x) + 40(\frac{1}{5}) = 40(\frac{27}{8}x) - 40(\frac{27}{8})$

$10x + 8 = 5(27x) - 5(27)$

$10x + 8 = 135x - 135$

Next, we isolate $x$ terms on one side and constants on the other side. Subtract $10x$ from both sides:

$10x + 8 - 10x = 135x - 135 - 10x$

$8 = 125x - 135$

Add 135 to both sides:

$8 + 135 = 125x - 135 + 135$

$143 = 125x$

Finally, divide both sides by 125:

$\frac{143}{125} = \frac{125x}{125}$

$x = \frac{143}{125}$

3. Final Answer

$\frac{143}{125}$