1. **State the problem:** We are given the system of equations: $$5x - 2y = -4$$ $$5x^2 - 4y^2 = -16$$ We want to analyze and solve this system. 2. **Rewrite the first equation to express $y$ in terms of $x$: ** $$5x - 2y = -4 \implies -2y = -4 - 5x \implies y = \frac{4 + 5x}{2}$$ 3. **Substitute $y$ into the second equation:** $$5x^2 - 4\left(\frac{4 + 5x}{2}\right)^2 = -16$$ 4. **Simplify the substitution:** $$5x^2 - 4 \cdot \frac{(4 + 5x)^2}{4} = -16$$ $$5x^2 - (4 + 5x)^2 = -16$$ 5. **Expand the square:** $$(4 + 5x)^2 = 16 + 40x + 25x^2$$ 6. **Rewrite the equation:** $$5x^2 - (16 + 40x + 25x^2) = -16$$ 7. **Distribute the negative sign:** $$5x^2 - 16 - 40x - 25x^2 = -16$$ 8. **Combine like terms:** $$5x^2 - 25x^2 - 40x - 16 = -16$$ $$-20x^2 - 40x - 16 = -16$$ 9. **Add 16 to both sides:** $$-20x^2 - 40x - 16 + 16 = -16 + 16$$ $$-20x^2 - 40x = 0$$ 10. **Factor out common terms:** $$-20x^2 - 40x = -20x(x + 2) = 0$$ 11. **Set each factor equal to zero:** $$-20x = 0 \implies x = 0$$ $$x + 2 = 0 \implies x = -2$$ 12. **Find corresponding $y$ values using $y = \frac{4 + 5x}{2}$:** - For $x=0$: $$y = \frac{4 + 5 \cdot 0}{2} = \frac{4}{2} = 2$$ - For $x=-2$: $$y = \frac{4 + 5 \cdot (-2)}{2} = \frac{4 - 10}{2} = \frac{-6}{2} = -3$$ 13. **Final solutions:** $$(x,y) = (0, 2) \text{ and } (-2, -3)$$ These are the points where the two equations intersect.