1. **State the problem:** Solve the system of equations algebraically using substitution or elimination. Given: $$y = -\frac{1}{2}x + 3$$ $$y = 3x - 4$$ 2. **Use substitution method:** Since both expressions equal $y$, set them equal to each other: $$-\frac{1}{2}x + 3 = 3x - 4$$ 3. **Solve for $x$:** Multiply both sides by 2 to clear the fraction: $$2\left(-\frac{1}{2}x + 3\right) = 2(3x - 4)$$ $$\cancel{2} \times -\frac{1}{\cancel{2}}x + 6 = 6x - 8$$ Simplifies to: $$-x + 6 = 6x - 8$$ 4. **Add $x$ to both sides:** $$-x + x + 6 = 6x + x - 8$$ $$6 = 7x - 8$$ 5. **Add 8 to both sides:** $$6 + 8 = 7x - 8 + 8$$ $$14 = 7x$$ 6. **Divide both sides by 7:** $$\frac{14}{\cancel{7}} = \frac{7x}{\cancel{7}}$$ $$2 = x$$ 7. **Substitute $x=2$ into one of the original equations to find $y$:** Using $y = 3x - 4$: $$y = 3(2) - 4 = 6 - 4 = 2$$ 8. **Final solution:** $$(x, y) = (2, 2)$$