1. **State the problem:** Solve the system of equations: $$y^2 + xy = 5$$ $$x^2 + xy = 20$$ 2. **Express $y$ from the second equation:** From $$x^2 + xy = 20$$, isolate $y$: $$xy = 20 - x^2$$ $$y = \frac{20 - x^2}{x}$$ 3. **Substitute $y$ into the first equation:** Replace $y$ in $$y^2 + xy = 5$$ with $$\frac{20 - x^2}{x}$$: $$\left(\frac{20 - x^2}{x}\right)^2 + x \cdot \frac{20 - x^2}{x} = 5$$ 4. **Simplify the equation:** $$\frac{(20 - x^2)^2}{x^2} + (20 - x^2) = 5$$ Multiply both sides by $x^2$ to clear the denominator: $$\cancel{x^2} \cdot \frac{(20 - x^2)^2}{\cancel{x^2}} + (20 - x^2) x^2 = 5 x^2$$ which simplifies to: $$(20 - x^2)^2 + (20 - x^2) x^2 = 5 x^2$$ 5. **Expand and simplify:** Expand $$(20 - x^2)^2 = 400 - 40 x^2 + x^4$$ So the equation becomes: $$400 - 40 x^2 + x^4 + 20 x^2 - x^4 = 5 x^2$$ Simplify terms: $$400 - 20 x^2 = 5 x^2$$ 6. **Solve for $x^2$:** Bring all terms to one side: $$400 = 25 x^2$$ Divide both sides by 25: $$\cancel{25} x^2 = \frac{400}{\cancel{25}}$$ $$x^2 = 16$$ 7. **Find $x$ values:** $$x = \pm 4$$ 8. **Find corresponding $y$ values:** For $x=4$: $$y = \frac{20 - 4^2}{4} = \frac{20 - 16}{4} = \frac{4}{4} = 1$$ For $x=-4$: $$y = \frac{20 - (-4)^2}{-4} = \frac{20 - 16}{-4} = \frac{4}{-4} = -1$$ 9. **Final solutions:** $$(x,y) = (4,1) \text{ and } (-4,-1)$$