Question

A factory has three machines making energy drinks. When all three machines are running, 520 drinks are made each hour. When only machines A & B are running, 320 drinks are made each hour. Machine C produces 30 more drinks each hour than machine B. Create a system of equations. Solve for how many drinks Machine A makes each hour.

Answer 1

Let A, B and C be the amount machines A, B and C, respectively, produce each hour alone.

With all three machines together produce 520 drinks each hour, we know that the addition of the amount that each produces each hour is equal to 520, that is.

$A+B+C=520_{}$

Now, similarly, machines A and B produce together 320 drinks each hour, so the sum of the amounts A and B produce each hour is equal to 320, so:

$A+B=320$

If machine C produce 30 more drinks than B each hour, than the amount of B plus 30 will be the amount of C:

$C=B+30$

So, we got the system of equations:

$\begin{gathered} A+B+C=520 \\ A+B=320 \\ C=B+30 \end{gathered}$

To solve, notice that A + B is known, because of the second equation, so we can substitute A + B by 320 in the first equation:

$\begin{gathered} (A+B)+C=520 \\ 320+C=520 \\ C=520-320=200 \end{gathered}$

Now, we can substitute C into the third equation and solve for B:

$\begin{gathered} C=B+30 \\ 200=B+30 \\ B+30=200 \\ B=200-30=170 \end{gathered}$

And, finally, we can substitute B into the second equation and solve for A:

$\begin{gathered} A+B=320 \\ A+170=320 \\ A=320-170=150 \end{gathered}$

Answers:

First box:

$A+B+C=520$

Second box:

$A+B=320$

Third box:

$C=B+30$

Fourth box:

$150$