What is the area, in square units, of the triangle bounded by $y = 0$, $y = x + 4$ and $x + 3y = 12$?
I want to find the area of the triangle, which is half the product of the base and the height. To do that, I need to find the vertices of the triangle, where the three lines intersect. I can use the equations of the lines to solve for $x$ and $y$. For example, where $y = 0$ and $y = x + 4$ intersect, I can substitute $0$ for $y$ in the second equation and get $0 = x + 4$, which implies $x = -4$. So one vertex is $(-4, 0)$. Similarly, where $y = 0$ and $x + 3y = 12$ intersect, I can substitute $0$ for $y$ in the third equation and get $x = 12$, which implies $x = 12$. So another vertex is $(12, 0)$. To find the third vertex, where $y = x + 4$ and $x + 3y = 12$ intersect, I can substitute $x + 4$ for $y$ in the third equation and get $x + 3(x + 4) = 12$, which simplifies to $4x + 12 = 12$, which implies $x = 0$. So the third vertex is $(0, 4)$. Now I have the three vertices of the triangle: $(-4, 0)$, $(12, 0)$ and $(0, 4)$. I see that the base of the triangle is the segment from $(-4, 0)$ to $(12, 0)$, which has a length of $16$ units. I also see that the height of the triangle is the segment from $(0, 4)$ to $(0, 0)$, which has a length of $4$ units. Therefore, the area of the triangle is half the product of the base and the height, which is $\frac{1}{2} \times 16 \times 4$, which is $32$ square units. 