Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Exercise Set - Page 832: 53


$13$ triangles, $21$ rectangles, and $6$ pentagons

Work Step by Step

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
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.