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a teacher purchased 20 calculators and 10 measuring tapes for her class…

Question

a teacher purchased 20 calculators and 10 measuring tapes for her class and paid $495. later, she realized that she didnt order enough supplies. she placed another order of 8 of the same calculators and 1 more of the same measuring tape and paid $178.50. this system represents the constraints: {20c + 10m = 495, 8c + m = 178.50} 1. discuss with a partner: a. in this situation, what do the solutions to the first equation mean? b. what do the solutions to the second equation mean? c. for each equation, how many possible solutions are there? explain how you know. d. in this situation, what does the solution to the system mean? 2. find the solution to the system. explain or show your reasoning.

Explanation:

Step1: Isolate $m$ in the second equation

From $8c + m=178.50$, we get $m = 178.50 - 8c$.

Step2: Substitute $m$ into the first equation

Substitute $m = 178.50 - 8c$ into $20c + 10m=495$.
We have $20c+10(178.50 - 8c)=495$.
Expand the left - hand side: $20c + 1785-80c = 495$.
Combine like terms: $- 60c=495 - 1785$.
$-60c=-1290$.

Step3: Solve for $c$

Divide both sides of $-60c=-1290$ by $-60$.
$c=\frac{-1290}{-60}=21.5$.

Step4: Solve for $m$

Substitute $c = 21.5$ into $m = 178.50 - 8c$.
$m=178.50-8\times21.5$.
$m=178.50 - 172$.
$m = 6.5$.

Answer:

$c = 21.5$, $m = 6.5$