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
The correct system of equations is:
A + B + C = 287
A + C = 197
A + B = 202
The correct system of equations that represents this situation is:
A + B + C = 287
A + C = 197
A + B = 202
The correct system of equations that represents this situation is:
A + B + C = 287
A + C = 197
A + B = 202
To determine the correct system of equations, we need to analyze the given information. The problem states that when all three machines (A, B, and C) are working, they produce 287 bolts per hour. This translates to the equation A + B + C = 287.
The problem also states that when only machines A and C are working, they produce 197 bolts per hour. This gives us the equation A + C = 197.
Finally, the problem states that when only machines A and B are working, they produce 202 bolts per hour. This gives us the equation A + B = 202.
By combining these equations, we can represent the situation accurately.