1. **State the problem:** We need to find the solution to the system of linear equations: $$-\frac{1}{2} y = \frac{1}{2} x + 5$$ and $$y = 2x + 2$$ This means finding the point $(x,y)$ where both lines intersect. 2. **Rewrite the first equation to express $y$ in terms of $x$: ** Multiply both sides by $-2$ to isolate $y$: $$-2 \times \left(-\frac{1}{2} y\right) = -2 \times \left(\frac{1}{2} x + 5\right)$$ $$\cancel{-2} \times \cancel{-\frac{1}{2}} y = \cancel{-2} \times \left(\frac{1}{2} x + 5\right)$$ Simplifying the left side: $$y = -2 \times \frac{1}{2} x - 2 \times 5$$ $$y = -x - 10$$ 3. **Set the two expressions for $y$ equal to find $x$: ** Since $y = -x - 10$ and $y = 2x + 2$, set them equal: $$-x - 10 = 2x + 2$$ 4. **Solve for $x$: ** Add $x$ to both sides: $$-x - 10 + x = 2x + 2 + x$$ $$-10 = 3x + 2$$ Subtract 2 from both sides: $$-10 - 2 = 3x + 2 - 2$$ $$-12 = 3x$$ Divide both sides by 3: $$\frac{-12}{\cancel{3}} = \frac{3x}{\cancel{3}}$$ $$-4 = x$$ 5. **Find $y$ by substituting $x = -4$ into one of the equations: ** Using $y = 2x + 2$: $$y = 2(-4) + 2 = -8 + 2 = -6$$ 6. **Final answer:** The solution to the system is $$(x,y) = (-4, -6)$$ This means the two lines intersect at the point $(-4, -6)$ on the coordinate plane.