1. **State the problem:** We have the simultaneous equations: $$ax + by = a + b$$ $$bx + ay = \frac{b^2}{a} + \frac{a^2}{b}$$ where $a$ and $b$ are different positive numbers. We want to: a) Use elimination to show that $y = \frac{a}{b}$. b) Find an expression for $x$ in terms of $a$ and $b$, fully simplified. 2. **Use elimination to find $y$:** Multiply the first equation by $a$ and the second by $b$ to align coefficients of $x$: $$a(ax + by) = a(a + b) \Rightarrow a^2x + aby = a^2 + ab$$ $$b(bx + ay) = b\left(\frac{b^2}{a} + \frac{a^2}{b}\right) \Rightarrow b^2x + aby = b\left(\frac{b^2}{a} + \frac{a^2}{b}\right)$$ 3. **Subtract the first new equation from the second:** $$\left(b^2x + aby\right) - \left(a^2x + aby\right) = b\left(\frac{b^2}{a} + \frac{a^2}{b}\right) - (a^2 + ab)$$ Simplify left side: $$b^2x - a^2x + \cancel{aby} - \cancel{aby} = (b^2 - a^2)x$$ Simplify right side: $$b\left(\frac{b^2}{a} + \frac{a^2}{b}\right) - (a^2 + ab) = \frac{b^3}{a} + a^2 - a^2 - ab = \frac{b^3}{a} - ab$$ 4. **Set up the equation:** $$ (b^2 - a^2)x = \frac{b^3}{a} - ab $$ 5. **Factor the right side:** $$ \frac{b^3}{a} - ab = b \left( \frac{b^2}{a} - a \right) = b \left( \frac{b^2 - a^2}{a} \right) $$ 6. **Rewrite the equation:** $$ (b^2 - a^2)x = b \frac{b^2 - a^2}{a} $$ 7. **Divide both sides by $b^2 - a^2$ (nonzero since $a \neq b$):** $$ x = \frac{b}{a} $$ 8. **Now substitute $x = \frac{b}{a}$ into the first original equation to find $y$:** $$ a \left( \frac{b}{a} \right) + by = a + b $$ Simplify: $$ b + by = a + b $$ Subtract $b$ from both sides: $$ by = a $$ Divide both sides by $b$: $$ y = \frac{a}{b} $$ **Answer:** a) $y = \frac{a}{b}$ b) $x = \frac{b}{a}$