Find the area (in square units) of the triangle formed by the region bounded by the x-axis, the straight line $3x+7y=42$, and the straight line $4x+9y=72$.

The two lines meet the x-axis at $x=14$ and $x=18$:

$3x=42\Rightarrow x=14$

$4x=72\Rightarrow x=18$

The base of the triangle on the x-axis is $18-14=4$ units.

Solving the two line equations gives their intersection as $(126,-48)$, so the perpendicular height from this point to the x-axis is 48 units.

Area $=\dfrac{1}{2}\times4\times48=96$ square units.