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)%0D%0AResponses%0D%0A%0D%0AA+B+C=197A+C=202A+B=287%0D%0A%0D%0A+%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A197%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A202%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A287%0D%0A%0D%0AA+B+C=250A+C=197A+B=287%0D%0A%0D%0A+%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A250%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A197%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A287%0D%0A%0D%0AA+B+C=287A+C=197A+B=202%0D%0A%0D%0A+%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A287%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A197%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A202%0D%0A%0D%0AA+B+D=197A+C=202A+B=287

Answer 1

The correct system of equations is:

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