1. **State the problem:** Find the vertices of the triangle formed by the lines $x - 2y = -4$, $x + y = 2$, and the x-axis ($y=0$). 2. **Find the intersection of the two lines:** From $x + y = 2$, express $x = 2 - y$. Substitute into $x - 2y = -4$: $$ (2 - y) - 2y = -4 $$ $$ 2 - 3y = -4 $$ $$ -3y = -6 $$ $$ y = 2 $$ Then, $x = 2 - 2 = 0$. So, the lines intersect at point $(0, 2)$. 3. **Find the intersection of $x - 2y = -4$ with the x-axis ($y=0$):** Substitute $y=0$: $$ x - 2(0) = -4 ightarrow x = -4 $$ So, the intersection is $(-4, 0)$. 4. **Find the intersection of $x + y = 2$ with the x-axis ($y=0$):** Substitute $y=0$: $$ x + 0 = 2 ightarrow x = 2 $$ So, the intersection is $(2, 0)$. 5. **Vertices of the triangle:** The triangle is formed by points $(-4, 0)$, $(2, 0)$, and $(0, 2)$. 6. **Summary:** The triangle is bounded by the lines $x - 2y = -4$, $x + y = 2$, and the x-axis, with vertices at $(-4, 0)$, $(2, 0)$, and $(0, 2)$.