Given that:

$\frac{x}{a} + \frac{y}{b} = 0$ and the second equation:

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

We will solve this system of equations step by step.

Step 1: Solve the first equation for $x$ or $y$

Start with: $\frac{x}{a} + \frac{y}{b} = 0$ Multiply through by $ab$ to eliminate the denominators: $b x + a y = 0$ So, we have: $bx = -ay \quad \text{(Equation 1)}$

Step 2: Solve for $x$ in terms of $y$

From $bx = -ay$, solve for $x$: $x = -\frac{a}{b} y \quad \text{(Equation 2)}$

Step 3: Substitute into the second equation

Now substitute $x = -\frac{a}{b} y$ into the second equation $ax + by = a^2 + b^2$:

$a\left(-\frac{a}{b} y\right) + by = a^2 + b^2$

Simplify the left-hand side: $-\frac{a^2}{b} y + by = a^2 + b^2$

Multiply through by $b$ to eliminate the denominator: $-a^2 y + b^2 y = b(a^2 + b^2)$

Step 4: Solve for $y$

Factor out $y$ on the left-hand side: $(b^2 - a^2) y = b(a^2 + b^2)$

Now solve for $y$: $y = \frac{b(a^2 + b^2)}{b^2 - a^2}$

Step 5: Solve for $x$

Now that we have $y$, substitute it into Equation 2 $x = -\frac{a}{b} y$:

$x = -\frac{a}{b} \cdot \frac{b(a^2 + b^2)}{b^2 - a^2}$

Simplify: $x = -\frac{a(a^2 + b^2)}{b^2 - a^2}$

Final Answers:

$x = -\frac{a(a^2 + b^2)}{b^2 - a^2}, \quad y = \frac{b(a^2 + b^2)}{b^2 - a^2}$

Let me know if you'd like more explanation on any of the steps!

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Relative Questions:

1. Can these equations be solved using matrices or Cramer's Rule?
2. What happens when $a = b$?
3. How does the solution change if the right-hand side of the second equation is different?
4. Can you apply this method to non-linear systems?
5. How do the values of $a$ and $b$ affect the solution behavior?

Tip: Always check for special cases like $a = b$ to simplify calculations!