1. **State the problem:** Solve the system of equations by substitution and clearing $x$: $$5x - 2y = -4$$ $$5x^2 - 4y^2 = -16$$ 2. **Isolate $x$ from the linear equation:** $$5x - 2y = -4 \implies 5x = 2y - 4 \implies x = \frac{2y - 4}{5}$$ 3. **Substitute $x$ into the second equation:** $$5\left(\frac{2y - 4}{5}\right)^2 - 4y^2 = -16$$ 4. **Simplify the substitution:** $$5 \cdot \frac{(2y - 4)^2}{25} - 4y^2 = -16$$ $$\frac{5}{25}(2y - 4)^2 - 4y^2 = -16$$ $$\frac{1}{5}(2y - 4)^2 - 4y^2 = -16$$ 5. **Multiply both sides by 5 to clear the denominator:** $$\cancel{5} \cdot \left(\frac{1}{\cancel{5}}(2y - 4)^2 - 4y^2\right) = -16 \cdot 5$$ $$ (2y - 4)^2 - 20y^2 = -80$$ 6. **Expand $(2y - 4)^2$:** $$ (2y - 4)^2 = (2y)^2 - 2 \cdot 2y \cdot 4 + 4^2 = 4y^2 - 16y + 16$$ 7. **Substitute back and simplify:** $$4y^2 - 16y + 16 - 20y^2 = -80$$ $$4y^2 - 20y^2 - 16y + 16 = -80$$ $$-16y^2 - 16y + 16 = -80$$ 8. **Bring all terms to one side:** $$-16y^2 - 16y + 16 + 80 = 0$$ $$-16y^2 - 16y + 96 = 0$$ 9. **Divide entire equation by -16 to simplify:** $$\cancel{-16}y^2 + \cancel{-16}y - \cancel{96} = 0 \implies y^2 + y - 6 = 0$$ 10. **Factor the quadratic:** $$y^2 + y - 6 = (y + 3)(y - 2) = 0$$ 11. **Solve for $y$:** $$y = -3 \quad \text{or} \quad y = 2$$ 12. **Find corresponding $x$ values using $x = \frac{2y - 4}{5}$:** For $y = -3$: $$x = \frac{2(-3) - 4}{5} = \frac{-6 - 4}{5} = \frac{-10}{5} = -2$$ For $y = 2$: $$x = \frac{2(2) - 4}{5} = \frac{4 - 4}{5} = \frac{0}{5} = 0$$ 13. **Final solutions:** $$\boxed{(-2, -3) \text{ and } (0, 2)}$$