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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.

This guide demonstrates how to solve a rational equation involving fractions with detailed steps, including fraction simplification, equation manipulation, and isolating variables. Ideal for high school students in Grades 10-12.

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