1. **Problem statement:** Solve the linear systems using the substitution method for g) and h). 2. **Recall substitution method:** Solve one equation for one variable, then substitute into the other. --- g) System: (I) $x - 3y = 1$ (II) $2x - 5y = 3$ Step 1: Solve (I) for $x$: $$x = 1 + 3y$$ Step 2: Substitute into (II): $$2(1 + 3y) - 5y = 3$$ $$2 + 6y - 5y = 3$$ $$2 + y = 3$$ Step 3: Solve for $y$: $$y = 3 - 2 = 1$$ Step 4: Substitute $y=1$ back into $x = 1 + 3y$: $$x = 1 + 3(1) = 4$$ Step 5: Check by substitution: (I) $4 - 3(1) = 4 - 3 = 1$ ✓ (II) $2(4) - 5(1) = 8 - 5 = 3$ ✓ Solution for g): $\boxed{(4, 1)}$ --- h) System: (I) $-x + 3y = 1$ (II) $-3x + 8y = 1$ Step 1: Solve (I) for $x$: $$-x + 3y = 1 \implies -x = 1 - 3y \implies x = 3y - 1$$ Step 2: Substitute into (II): $$-3(3y - 1) + 8y = 1$$ $$-9y + 3 + 8y = 1$$ $$-y + 3 = 1$$ Step 3: Solve for $y$: $$-y = 1 - 3 = -2 \implies y = 2$$ Step 4: Substitute $y=2$ back into $x = 3y - 1$: $$x = 3(2) - 1 = 6 - 1 = 5$$ Step 5: Check by substitution: (I) $-5 + 3(2) = -5 + 6 = 1$ ✓ (II) $-3(5) + 8(2) = -15 + 16 = 1$ ✓ Solution for h): $\boxed{(5, 2)}$