1. Stating the problems: Solve the systems of linear equations: (1) $10x - 5y = 20$ and $-4x + 4y = -4$ (2) $5x - 7y = 3$ and $2x + 2y = 3$ 2. Formula and method: We will use the elimination method to solve each system. This involves multiplying equations to align coefficients and then adding or subtracting to eliminate one variable. --- **Problem 1:** 3. Multiply the second equation by 5 to align the coefficients of $y$: $5 \times (-4x + 4y) = 5 \times (-4)$ $-20x + 20y = -20$ 4. Multiply the first equation by 4 to align the coefficients of $x$: $4 \times (10x - 5y) = 4 \times 20$ $40x - 20y = 80$ 5. Add the two new equations: $40x - 20y + (-20x + 20y) = 80 + (-20)$ Simplify: $$40x - 20y - 20x + 20y = 60$$ $$20x = 60$$ 6. Solve for $x$: $$x = \frac{60}{20} = 3$$ 7. Substitute $x=3$ into the first original equation: $$10(3) - 5y = 20$$ $$30 - 5y = 20$$ 8. Solve for $y$: $$-5y = 20 - 30 = -10$$ $$y = \frac{-10}{-5} = 2$$ --- **Problem 2:** 9. Multiply the second equation by 7 to align the coefficients of $y$: $$7 \times (2x + 2y) = 7 \times 3$$ $$14x + 14y = 21$$ 10. Multiply the first equation by 2 to align the coefficients of $x$: $$2 \times (5x - 7y) = 2 \times 3$$ $$10x - 14y = 6$$ 11. Subtract the first new equation from the second new equation: $$(14x + 14y) - (10x - 14y) = 21 - 6$$ Simplify: $$14x + 14y - 10x + 14y = 15$$ $$4x + 28y = 15$$ 12. This did not eliminate a variable as intended, so instead add the two equations: $$(10x - 14y) + (14x + 14y) = 6 + 21$$ Simplify: $$24x = 27$$ 13. Solve for $x$: $$x = \frac{27}{24} = \frac{9}{8}$$ 14. Substitute $x=\frac{9}{8}$ into the second original equation: $$2 \times \frac{9}{8} + 2y = 3$$ $$\frac{18}{8} + 2y = 3$$ 15. Simplify and solve for $y$: $$2y = 3 - \frac{18}{8} = \frac{24}{8} - \frac{18}{8} = \frac{6}{8} = \frac{3}{4}$$ $$y = \frac{3}{8}$$ --- **Final answers:** (1) $x=3, y=2$ (2) $x=\frac{9}{8}, y=\frac{3}{8}$