The given equation is:

$\frac{\frac{1}{1+r} - a}{\frac{1}{1+r} + b} = c$

We will solve this equation for $r$ step by step.

Step 1: Eliminate the fraction

Multiply both sides of the equation by the denominator of the left-hand side:

$\left( \frac{1}{1+r} + b \right) \times \frac{\frac{1}{1+r} - a}{\frac{1}{1+r} + b} = c \times \left( \frac{1}{1+r} + b \right)$

This simplifies to:

$\frac{1}{1+r} - a = c \times \left( \frac{1}{1+r} + b \right)$

Step 2: Expand the right-hand side

Expand the right side of the equation:

$\frac{1}{1+r} - a = c \times \frac{1}{1+r} + c \times b$

Step 3: Isolate the term with $\frac{1}{1+r}$

Move all terms with $\frac{1}{1+r}$ to one side of the equation:

$\frac{1}{1+r} - c \times \frac{1}{1+r} = c \times b + a$

Factor out $\frac{1}{1+r}$ from the left side:

$\frac{1}{1+r} (1 - c) = c \times b + a$

Step 4: Solve for $\frac{1}{1+r}$

Now, divide both sides by $1 - c$:

$\frac{1}{1+r} = \frac{c \times b + a}{1 - c}$

Step 5: Solve for $r$

Take the reciprocal of both sides:

$1 + r = \frac{1 - c}{c \times b + a}$

Finally, solve for $r$:

$r = \frac{1 - c}{c \times b + a} - 1$

This is the solution for $r$.

Would you like any clarifications or further details on these steps?

Here are 5 related questions for further practice:

1. How would the solution change if $a = 0$?
2. What happens if $b = 0$ in the equation?
3. Can you generalize the solution for similar equations with different exponents in the denominator?
4. How would the equation change if $c$ were a function of $r$?
5. What would the solution be if both $a$ and $b$ were complex numbers?

Tip: Always simplify the equation as much as possible before solving for the variable, as it makes the process easier and avoids potential errors.