We are given the following system of equations: $ax + by = c^2$ $\frac{a+x}{b} = \frac{b+y}{a}$ $x - y + z = 5$ where $a \neq 0$ and $b \neq 0$.

1. Problem Description

We are given the following system of equations:

$ax + by = c^2$

$\frac{a+x}{b} = \frac{b+y}{a}$

$x - y + z = 5$

where $a \neq 0$ and $b \neq 0$.

2. Solution Steps

From the second equation, we have:

$a(a+x) = b(b+y)$

$a^2 + ax = b^2 + by$

$ax - by = b^2 - a^2$

We are given $ax + by = c^2$.

Adding the two equations, we get:

$ax - by + ax + by = b^2 - a^2 + c^2$

$2ax = b^2 - a^2 + c^2$

$x = \frac{b^2 - a^2 + c^2}{2a}$

Subtracting the two equations, we get:

$ax + by - (ax - by) = c^2 - (b^2 - a^2)$

$2by = c^2 - b^2 + a^2$

$y = \frac{c^2 - b^2 + a^2}{2b}$

Now, we substitute $x$ and $y$ into the third equation $x - y + z = 5$:

$\frac{b^2 - a^2 + c^2}{2a} - \frac{c^2 - b^2 + a^2}{2b} + z = 5$

$z = 5 - \frac{b^2 - a^2 + c^2}{2a} + \frac{c^2 - b^2 + a^2}{2b}$

$z = 5 + \frac{c^2 - b^2 + a^2}{2b} - \frac{b^2 - a^2 + c^2}{2a}$

$z = 5 + \frac{a(c^2 - b^2 + a^2) - b(b^2 - a^2 + c^2)}{2ab}$

$z = 5 + \frac{ac^2 - ab^2 + a^3 - b^3 + ba^2 - bc^2}{2ab}$

$z = 5 + \frac{ac^2 - bc^2 - ab^2 + a^2b + a^3 - b^3}{2ab}$

$z = 5 + \frac{c^2(a-b) + ab(a-b) + (a-b)(a^2 + ab + b^2)}{2ab}$

$z = 5 + \frac{(a-b)(c^2 + ab + a^2 + ab + b^2)}{2ab}$

$z = 5 + \frac{(a-b)(c^2 + a^2 + 2ab + b^2)}{2ab}$

$z = 5 + \frac{(a-b)(c^2 + (a+b)^2)}{2ab}$

3. Final Answer

$x = \frac{b^2 - a^2 + c^2}{2a}$

$y = \frac{a^2 - b^2 + c^2}{2b}$

$z = 5 + \frac{(a-b)(c^2 + (a+b)^2)}{2ab}$