The problem consists of two parts. Part g) asks us to solve the equation $\frac{x}{5-y} = m-1$ for $y$. Part h) asks us to solve the equation $k-y = \frac{e-b}{x+a}$ for $x$.

1. Problem Description

The problem consists of two parts.

Part g) asks us to solve the equation $\frac{x}{5-y} = m-1$ for $y$.

Part h) asks us to solve the equation $k-y = \frac{e-b}{x+a}$ for $x$.

2. Solution Steps

Part g) Solving for $y$:

We are given the equation $\frac{x}{5-y} = m-1$.

First, multiply both sides of the equation by $5-y$:

$x = (m-1)(5-y)$

Next, divide both sides by $m-1$:

$\frac{x}{m-1} = 5-y$

Then, add $y$ to both sides and subtract $\frac{x}{m-1}$ from both sides:

$y = 5 - \frac{x}{m-1}$

Finally, we can write it as:

$y = \frac{5(m-1)-x}{m-1}$

$y = \frac{5m - 5 - x}{m-1}$

Part h) Solving for $x$:

We are given the equation $k-y = \frac{e-b}{x+a}$.

First, multiply both sides of the equation by $x+a$:

$(k-y)(x+a) = e-b$

Next, divide both sides by $k-y$:

$x+a = \frac{e-b}{k-y}$

Then, subtract $a$ from both sides:

$x = \frac{e-b}{k-y} - a$

Finally, we can write it as:

$x = \frac{e-b - a(k-y)}{k-y}$

$x = \frac{e-b - ak + ay}{k-y}$

$x = \frac{e-b-ak+ay}{k-y}$

3. Final Answer

For part g): $y = \frac{5m-5-x}{m-1}$

For part h): $x = \frac{e-b-ak+ay}{k-y}$