1. **State the problem:** We need to find the coordinates of the vertices of the triangle formed by the intersection of the three lines: $$y = x + 1$$ $$y = 4 - 2x$$ $$y = -x - 5$$ 2. **Method:** The vertices of the triangle are the points where each pair of lines intersect. We will find these points algebraically by solving the systems of equations pairwise. 3. **Find intersection of** $y = x + 1$ **and** $y = 4 - 2x$: Set the right sides equal: $$x + 1 = 4 - 2x$$ Add $2x$ to both sides: $$x + 2x + 1 = 4$$ $$3x + 1 = 4$$ Subtract 1 from both sides: $$3x = 3$$ Divide both sides by 3: $$\cancel{3}x = \cancel{3}1$$ $$x = 1$$ Substitute $x=1$ into $y = x + 1$: $$y = 1 + 1 = 2$$ So, the first vertex is $(1, 2)$. 4. **Find intersection of** $y = 4 - 2x$ **and** $y = -x - 5$: Set equal: $$4 - 2x = -x - 5$$ Add $2x$ to both sides: $$4 = -x + 2x - 5$$ $$4 = x - 5$$ Add 5 to both sides: $$4 + 5 = x$$ $$9 = x$$ Substitute $x=9$ into $y = 4 - 2x$: $$y = 4 - 2(9) = 4 - 18 = -14$$ So, the second vertex is $(9, -14)$. 5. **Find intersection of** $y = x + 1$ **and** $y = -x - 5$: Set equal: $$x + 1 = -x - 5$$ Add $x$ to both sides: $$x + x + 1 = -5$$ $$2x + 1 = -5$$ Subtract 1 from both sides: $$2x = -6$$ Divide both sides by 2: $$\cancel{2}x = \cancel{2}(-3)$$ $$x = -3$$ Substitute $x = -3$ into $y = x + 1$: $$y = -3 + 1 = -2$$ So, the third vertex is $(-3, -2)$. 6. **Final answer:** The vertices of the triangle are: $$(1, 2), (9, -14), (-3, -2)$$ 7. **Graphical solution:** Plotting these three lines on a graph will show that they intersect at these points, forming a triangle with the vertices found algebraically. This completes the solution.