Question

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)
Responses

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

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

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

92−16x=104−13x ; x=−4
92 minus 16 x equals 104 minus 13 x ; x equals negative 4
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Answer 1

To find out when the two machines will have filled the same number of cans of soup, we can establish an equation based on the cans they have already filled and their respective filling rates.

Let \( x \) be the number of minutes that pass.

**For Machine 1:**
- Initial filled cans: 92
- Rate of filling: 16 cans per minute

So, the total number of cans filled by Machine 1 after \( x \) minutes is:
\[ 92 + 16x \]

**For Machine 2:**
- Initial filled cans: 104
- Rate of filling: 13 cans per minute

So, the total number of cans filled by Machine 2 after \( x \) minutes is:
\[ 104 + 13x \]

Now we can set up the equation where the total cans filled will be equal for both machines:
\[ 92 + 16x = 104 + 13x \]

Now, let's solve the equation step by step:

1. Subtract \( 13x \) from both sides:
\[ 92 + 16x - 13x = 104 \]
\[ 92 + 3x = 104 \]

2. Subtract 92 from both sides:
\[ 3x = 104 - 92 \]
\[ 3x = 12 \]

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

Thus, \( x = 4 \) minutes.

The machines will have filled the same number of cans of soup after 4 minutes.

So the correct response is:
**92 + 16x = 104 + 13x; x = 4**.