For real numbers $a$ and $b$, where $|a| \neq|b|$ and $a \neq 0$, the following holds:

$$
\frac{a-b}{a^{2}+a b}+\frac{a+b}{a^{2}-a b}=\frac{3 a-b}{a^{2}-b^{2}}
$$

Determine the value of the expression $\frac{b}{a}$.
$\mathbf{I} / 1$. Multiply the equation by $a(a+b)(a-b)$ and we get

$$
(a-b)^{2}+(a+b)^{2}=a(3 a-b)
$$

When we remove the parentheses and move all terms to the right side, we get

$$
0=a^{2}-a b-2 b^{2}=(a-2 b)(a+b)
$$

Since $a \neq -b$, it must be that $a-2 b=0$ or $a=2 b$. Since $a$ is not equal to 0, then $\frac{b}{a}=\frac{1}{2}$.

Elimination of fractions 1 point

Simplified equation in $0=a^{2}-a b-2 b^{2}$ 2 points

Factored expression to $(a-2 b)(a+b)$ 2 points

![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-06.jpg?height=49&width=1625&top_left_y=1112&top_left_x=221)

Justification that we can divide by $a$ and the calculated value of the expression $\frac{b}{a}=\frac{1}{2} \ldots \ldots . .1$ point