We need to solve the equation $\frac{3-5x}{7} - \frac{x}{4} = \frac{1}{2} + \frac{5-4x}{8}$ for $x$. We also need to solve the equation $\frac{3(x-1)}{4} - 2 = 5x - \frac{x-2}{2}$ for $x$.

1. Problem Description

We need to solve the equation $\frac{3-5x}{7} - \frac{x}{4} = \frac{1}{2} + \frac{5-4x}{8}$ for $x$.

We also need to solve the equation $\frac{3(x-1)}{4} - 2 = 5x - \frac{x-2}{2}$ for $x$.

2. Solution Steps

Let's first solve the equation $\frac{3-5x}{7} - \frac{x}{4} = \frac{1}{2} + \frac{5-4x}{8}$.

First, we find the least common multiple (LCM) of the denominators 7, 4, 2, and 8, which is

5

6. Multiply both sides of the equation by 56:

$56 \cdot (\frac{3-5x}{7} - \frac{x}{4}) = 56 \cdot (\frac{1}{2} + \frac{5-4x}{8})$

$8(3-5x) - 14x = 28 + 7(5-4x)$

$24 - 40x - 14x = 28 + 35 - 28x$

$24 - 54x = 63 - 28x$

$-54x + 28x = 63 - 24$

$-26x = 39$

$x = \frac{39}{-26}$

$x = -\frac{3}{2}$

Now, let's solve the equation $\frac{3(x-1)}{4} - 2 = 5x - \frac{x-2}{2}$.

First, we find the LCM of the denominators 4 and 2, which is

4. Multiply both sides of the equation by 4:

$4 \cdot (\frac{3(x-1)}{4} - 2) = 4 \cdot (5x - \frac{x-2}{2})$

$3(x-1) - 8 = 20x - 2(x-2)$

$3x - 3 - 8 = 20x - 2x + 4$

$3x - 11 = 18x + 4$

$3x - 18x = 4 + 11$

$-15x = 15$

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

$x = -1$

3. Final Answer

The solution to the first equation is $x = -\frac{3}{2}$.

The solution to the second equation is $x = -1$.