## Precalculus (6th Edition) Blitzer

$13$ triangles, $21$ rectangles, and $6$ pentagons
Step 1. Assume there are $x$ triangles, $y$ rectangles, and $z$ pentagons Step 2. Based on the given conditions, we have $\begin{cases} x+y+z=40\\ 3x+4y+5z=153 \\ 2y+5z=72 \end{cases}$ Step 3. Multiply -3 to the first equation and add it to the second equation; we have $y+2z=33$ or $y=33-2z$ Step 4. Substitute $y$ in the third equation; we have $2(33-2z)+5z=72$; thus $z=6$ Step 5. Using back-substitution, we have $y=33-12=21$ and $x=40-21-6=13$ Step 6. We conclude that there are $13$ triangles, $21$ rectangles, and $6$ pentagons