Two machines at a factory are filling cans of soup. One machine has filled 92 cans of soup. It fills cans at a rate of 16 cans per minute. The second machine has filled 104 cans of soup. It fills cans at a rate of 13 cans per minute. Write and solve an equation to show when the two machines will have filled the same number of cans of soup.(1 point)

Question

Responses

92+16x=104+13x
; x=6
92 plus 16 x equals 104 plus 13 x ; x equals 6

92−16x=104−13x
; x=−4
92 minus 16 x equals 104 minus 13 x ; x equals negative 4

92+16x=104+13x
; x=4
92 plus 16 x equals 104 plus 13 x ; x equals 4

104+16x=92+13x
; x=4

Answer 1

To find when the two machines will have filled the same number of cans of soup, we can set up the following equation:

Let \( x \) be the number of minutes that pass after the point when the machines have already filled the cans.

For the first machine, it has already filled 92 cans and continues to fill cans at a rate of 16 cans per minute. Thus, the number of cans filled by the first machine after \( x \) minutes is: \[ 92 + 16x \]

For the second machine, it has already filled 104 cans and fills at a rate of 13 cans per minute. Therefore, the number of cans filled by the second machine after \( x \) minutes is: \[ 104 + 13x \]

We want to set these two expressions equal to each other to find when both machines have filled the same number of cans: \[ 92 + 16x = 104 + 13x \]

Now, let's solve for \( x \):

1. Subtract \( 13x \) from both sides: \[ 92 + 3x = 104 \]

2. Subtract 92 from both sides: \[ 3x = 12 \]

3. Divide by 3: \[ x = 4 \]

Thus, the two machines will have filled the same number of cans after 4 minutes.

The correct response is: \[ 92 + 16x = 104 + 13x; , x = 4 \]