1. **Stating the problem:** We are given two linear equations: $$x - y + 1 = 0$$ and $$3x + 2y - 12 = 0$$ We need to graph these lines, find the vertices of the triangle formed by these two lines and the x-axis, and shade the triangular region. 2. **Understanding the problem:** The triangle is formed by the intersection points of the two lines and their intersections with the x-axis. 3. **Find the x-intercepts of each line:** - For $$x - y + 1 = 0$$, set $$y=0$$: $$x + 1 = 0 \implies x = -1$$ So the x-intercept is $$(-1, 0)$$. - For $$3x + 2y - 12 = 0$$, set $$y=0$$: $$3x - 12 = 0 \implies 3x = 12 \implies x = 4$$ So the x-intercept is $$(4, 0)$$. 4. **Find the intersection point of the two lines:** Solve the system: $$\begin{cases} x - y + 1 = 0 \\ 3x + 2y - 12 = 0 \end{cases}$$ From the first equation: $$y = x + 1$$ Substitute into the second: $$3x + 2(x + 1) - 12 = 0$$ $$3x + 2x + 2 - 12 = 0$$ $$5x - 10 = 0$$ $$5x = 10$$ $$x = 2$$ Then, $$y = 2 + 1 = 3$$ So the intersection point is $$(2, 3)$$. 5. **Vertices of the triangle:** The triangle is formed by points: $$A = (-1, 0), B = (4, 0), C = (2, 3)$$ 6. **Summary:** - The two lines intersect at $$(2, 3)$$. - The x-axis intersections are at $$(-1, 0)$$ and $$(4, 0)$$. - These three points form the vertices of the triangle. 7. **Graphing:** - Line 1: $$x - y + 1 = 0$$ - Line 2: $$3x + 2y - 12 = 0$$ - x-axis: $$y=0$$ The triangular region is bounded by these three lines. **Final answer:** The triangle formed by the lines $$x - y + 1 = 0$$, $$3x + 2y - 12 = 0$$, and the x-axis has vertices at $$(-1, 0)$$, $$(4, 0)$$, and $$(2, 3)$$.