1. **State the problem:** Solve the system of equations by substitution, clearing $x$: $$4x + 5y = 4$$ $$-16x^2 + 15y^2 = -16$$ 2. **Isolate $x$ from the linear equation:** $$4x = 4 - 5y$$ $$x = \frac{4 - 5y}{4}$$ 3. **Substitute $x$ into the hyperbola equation:** $$-16\left(\frac{4 - 5y}{4}\right)^2 + 15y^2 = -16$$ 4. **Simplify the squared term:** $$-16 \cdot \frac{(4 - 5y)^2}{16} + 15y^2 = -16$$ 5. **Cancel the 16 in numerator and denominator:** $$\cancel{-16} \cdot \frac{(4 - 5y)^2}{\cancel{16}} + 15y^2 = -16$$ $$-(4 - 5y)^2 + 15y^2 = -16$$ 6. **Expand $(4 - 5y)^2$:** $$-(16 - 40y + 25y^2) + 15y^2 = -16$$ 7. **Distribute the negative sign:** $$-16 + 40y - 25y^2 + 15y^2 = -16$$ 8. **Combine like terms:** $$-16 + 40y - 10y^2 = -16$$ 9. **Add 16 to both sides:** $$-16 + 40y - 10y^2 + 16 = -16 + 16$$ $$40y - 10y^2 = 0$$ 10. **Factor out $y$:** $$y(40 - 10y) = 0$$ 11. **Set each factor equal to zero:** $$y = 0$$ $$40 - 10y = 0$$ 12. **Solve for $y$ in second equation:** $$40 = 10y$$ $$y = \frac{40}{10} = 4$$ 13. **Find corresponding $x$ values using $x = \frac{4 - 5y}{4}$:** - For $y=0$: $$x = \frac{4 - 5 \cdot 0}{4} = \frac{4}{4} = 1$$ - For $y=4$: $$x = \frac{4 - 5 \cdot 4}{4} = \frac{4 - 20}{4} = \frac{-16}{4} = -4$$ 14. **Final solutions:** $$(x,y) = (1,0) \text{ and } (-4,4)$$