1. Problem a: Solve the system using substitution: $$\frac{x}{3} + \frac{y}{5} = 2$$ $$2x - y = 1$$ 2. From the second equation, express $y$ in terms of $x$: $$y = 2x - 1$$ 3. Substitute $y = 2x - 1$ into the first equation: $$\frac{x}{3} + \frac{2x - 1}{5} = 2$$ 4. Multiply both sides by 15 (LCM of 3 and 5) to clear denominators: $$5x + 3(2x - 1) = 30$$ 5. Simplify: $$5x + 6x - 3 = 30$$ $$11x = 33$$ $$x = 3$$ 6. Substitute $x=3$ back into $y = 2x - 1$: $$y = 2(3) - 1 = 6 - 1 = 5$$ 7. Solution for a: $x=3$, $y=5$ --- 1. Problem b: Solve the system: $$\frac{x}{5} + \frac{y}{3} = -2$$ $$\frac{x}{3} - y = 10$$ 2. From the second equation, express $y$: $$y = \frac{x}{3} - 10$$ 3. Substitute into the first equation: $$\frac{x}{5} + \frac{\frac{x}{3} - 10}{3} = -2$$ 4. Simplify the second term: $$\frac{x}{5} + \frac{x}{9} - \frac{10}{3} = -2$$ 5. Multiply both sides by 45 (LCM of 5,9,3): $$9x + 5x - 150 = -90$$ 6. Simplify: $$14x - 150 = -90$$ $$14x = 60$$ $$x = \frac{60}{14} = \frac{30}{7}$$ 7. Substitute $x$ back into $y = \frac{x}{3} - 10$: $$y = \frac{30/7}{3} - 10 = \frac{30}{21} - 10 = \frac{10}{7} - 10 = \frac{10 - 70}{7} = -\frac{60}{7}$$ 8. Solution for b: $x=\frac{30}{7}$, $y=-\frac{60}{7}$ --- 1. Problem c: Solve the system: $$\frac{2x}{5} + \frac{y}{4} = 1$$ $$\frac{x}{5} + 1 = y$$ 2. Express $y$ from second equation: $$y = \frac{x}{5} + 1$$ 3. Substitute into first equation: $$\frac{2x}{5} + \frac{\frac{x}{5} + 1}{4} = 1$$ 4. Multiply both sides by 20 (LCM of 5 and 4): $$4(2x) + 5\left(\frac{x}{5} + 1\right) = 20$$ 5. Simplify: $$8x + x + 5 = 20$$ $$9x + 5 = 20$$ $$9x = 15$$ $$x = \frac{15}{9} = \frac{5}{3}$$ 6. Substitute $x$ back into $y = \frac{x}{5} + 1$: $$y = \frac{5/3}{5} + 1 = \frac{5}{15} + 1 = \frac{1}{3} + 1 = \frac{4}{3}$$ 7. Solution for c: $x=\frac{5}{3}$, $y=\frac{4}{3}$ --- 1. Problem d: Solve the system: $$3x + \frac{y}{2} = 4$$ $$\frac{2x}{3} + \frac{y}{6} = 1$$ 2. Multiply second equation by 6 to clear denominators: $$4x + y = 6$$ 3. From first equation, multiply both sides by 2: $$6x + y = 8$$ 4. Subtract second equation from this: $$(6x + y) - (4x + y) = 8 - 6$$ $$2x = 2$$ $$x = 1$$ 5. Substitute $x=1$ into $4x + y = 6$: $$4(1) + y = 6$$ $$4 + y = 6$$ $$y = 2$$ 6. Solution for d: $x=1$, $y=2$