Linear Substitution C17D7A

1. Solve the system: $$-x = 4 - y$$ $$8x + 2y = 10$$ 2. Solve the system: $$x - 1 = 5y$$ $$y + 4x = 19$$ --- ### Problem 1 1. Start with the first equation: $$-x = 4 - y$$ Rearrange to express $x$ in terms of $y$: $$-x = 4 - y \implies x = y - 4$$ 2. Substitute $x = y - 4$ into the second equation: $$8x + 2y = 10$$ $$8(y - 4) + 2y = 10$$ 3. Expand and simplify: $$8y - 32 + 2y = 10$$ $$10y - 32 = 10$$ 4. Add 32 to both sides: $$10y - 32 + 32 = 10 + 32$$ $$10y = 42$$ 5. Divide both sides by 10: $$\frac{\cancel{10}y}{\cancel{10}} = \frac{42}{10}$$ $$y = \frac{42}{10} = 4.2$$ 6. Substitute $y = 4.2$ back into $x = y - 4$: $$x = 4.2 - 4 = 0.2$$ **Solution for system 1:** $$x = 0.2, \quad y = 4.2$$ --- ### Problem 2 1. Start with the first equation: $$x - 1 = 5y$$ Rearrange to express $x$ in terms of $y$: $$x = 5y + 1$$ 2. Substitute $x = 5y + 1$ into the second equation: $$y + 4x = 19$$ $$y + 4(5y + 1) = 19$$ 3. Expand and simplify: $$y + 20y + 4 = 19$$ $$21y + 4 = 19$$ 4. Subtract 4 from both sides: $$21y + 4 - 4 = 19 - 4$$ $$21y = 15$$ 5. Divide both sides by 21: $$\frac{\cancel{21}y}{\cancel{21}} = \frac{15}{21}$$ $$y = \frac{15}{21} = \frac{5}{7} \approx 0.714$$ 6. Substitute $y = \frac{5}{7}$ back into $x = 5y + 1$: $$x = 5 \times \frac{5}{7} + 1 = \frac{25}{7} + 1 = \frac{25}{7} + \frac{7}{7} = \frac{32}{7} \approx 4.571$$ **Solution for system 2:** $$x = \frac{32}{7}, \quad y = \frac{5}{7}$$