The indoor climbing gym is a popular field trip destination. This year the senior class at High School A and the senior class at High School B both planned trips there. The senior class at High School A rented and filled 8 vans, v, and 3 buses, b, with 242 students. High School B rented and filled 2 vans and 6 buses with 260 students. Every van had the same number of students in it as did the buses. How many students can a van carry? How many students can a bus carry?

The indoor climbing gym is a popular field trip destination This year the senior class at High School A and the senior class at High School B both planned trips class=

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Solution:

We would start by analyzing the statements;

High School A rented and filled 8 vans, v, and 3 buses, b, with 242 students. Mathematically;

[tex]8v+3b=242.........equation1[/tex]

High School B rented and filled 2 vans, v, and 6 buses, b, with 260 students. Mathematically;

[tex]2v+6b=260...............equation2[/tex]

Now, we would solve equation 1 and equation 2 simultaneously.

From equation 2;

[tex]\begin{gathered} 2v+6b=260 \\ \\ \text{ Divide through by 2;} \\ \frac{2v}{2}+\frac{6b}{2}=\frac{260}{2} \\ \\ v+3b=130 \\ \\ v=130-3b..........equation3 \end{gathered}[/tex]

Substitute equation 3 in equation 1;

[tex]\begin{gathered} 8(130-3b)+3b=242 \\ \\ 1040-24b+3b=242 \\ \\ 1040-242=24b-3b \\ \\ 798=21b \\ \\ b=\frac{798}{21} \\ \\ b=38 \end{gathered}[/tex]

Substitute the value of b in equation3;

[tex]\begin{gathered} v=130-3(38) \\ \\ v=16 \end{gathered}[/tex]

ANSWERS:

[tex]\begin{gathered} \text{ High School A: }8v+3b=242 \\ \\ \text{H}\imaginaryI\text{gh School B: 2}v+6b=260 \\ \\ \text{ Van: \$}16 \\ \\ \text{ Bus: \$}38 \end{gathered}[/tex]

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