We are tasked with finding integer solutions for the equation:

$\frac{a}{b} - \frac{b}{a} = \frac{a + b}{ab}$ where $-9 \leq a \leq 9$ and $-9 \leq b \leq 9$.

Step-by-Step Solution:

1. Rewrite the equation:

   Let's simplify both sides of the equation. $\frac{a}{b} - \frac{b}{a} = \frac{a + b}{ab}$ The left-hand side (LHS) can be combined by finding a common denominator: $\frac{a}{b} - \frac{b}{a} = \frac{a^2 - b^2}{ab}$

2. Substitute back into the equation:

   Now, substitute this into the original equation: $\frac{a^2 - b^2}{ab} = \frac{a + b}{ab}$

3. Simplify the equation:

   Both sides have the same denominator, so we can cancel the denominator: $a^2 - b^2 = a + b$

4. Rearrange the equation:

   $a^2 - b^2 - a - b = 0$ Factor this expression: $(a - b)(a + b) - (a + b) = 0$ Factor out $(a + b)$: $(a + b)(a - b - 1) = 0$

5. Solve the factored equation:

   For this product to be zero, either $a + b = 0$ or $a - b - 1 = 0$.

   • If $a + b = 0$, then $a = -b$.
   • If $a - b - 1 = 0$, then $a = b + 1$.

Solution 1: $a = -b$

Substitute $a = -b$ into the range $-9 \leq a \leq 9$ and $-9 \leq b \leq 9$. The possible pairs are: $(a, b) = (-9, 9), (-8, 8), (-7, 7), \ldots, (9, -9)$

Solution 2: $a = b + 1$

Substitute $a = b + 1$ into the range $-9 \leq a \leq 9$ and $-9 \leq b \leq 9$. The possible pairs are: $(a, b) = (-8, -9), (-7, -8), (-6, -7), \ldots, (9, 8)$

Final Answer:

The integer solutions for $(a, b)$ are:

1. From $a = -b$: $(a, b) = (-9, 9), (-8, 8), (-7, 7), \ldots, (9, -9)$

2. From $a = b + 1$: $(a, b) = (-8, -9), (-7, -8), (-6, -7), \ldots, (9, 8)$

Do you have any questions or would you like further clarification?

Related Questions:

1. What happens if $a$ and $b$ are restricted to non-negative values?
2. Can the equation $\frac{a}{b} - \frac{b}{a} = \frac{a + b}{ab}$ be solved for real numbers?
3. What would the solutions be if $a$ and $b$ were restricted to positive values only?
4. How does the solution change if $-b = a + 1$?
5. Can this equation be solved graphically?

Tip: Factoring is often the most effective method when solving polynomial equations. Always check for common factors before expanding terms!