In a factory there are three machines, A, B, and C. When all three machines are working, they produce 287 bolts per hour. When only machines A and C are working, they produce 197 bolts per hour. When only machines A and B are working, they produce 202 bolts per hour. Which system of equations represents this situation?(1 point)

Question

Responses

A+B+C=287A+C=197A+B=202
𝐴
+
𝐵
+
𝐶
=
287
𝐴
+
𝐶
=
197
𝐴
+
𝐵
=
202

A+B+C=197A+C=202A+B=287
𝐴
+
𝐵
+
𝐶
=
197
𝐴
+
𝐶
=
202
𝐴
+
𝐵
=
287

A+B+D=197A+C=202A+B=287
𝐴
+
𝐵
+
𝐷
=
197
𝐴
+
𝐶
=
202
𝐴
+
𝐵
=
287

A+B+C=250A+C=197A+B=287

Answer 1

The correct system of equations that represents the situation is:

\[ \begin{align*} A + B + C & = 287 \ A + C & = 197 \ A + B & = 202 \end{align*} \]

This corresponds to the first option:

A + B + C = 287, A + C = 197, A + B = 202.