Question

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 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 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 92−16x=104−13x ; x=−4

Answer 1

To determine when the two machines will have filled the same number of cans of soup, we can start by setting up the equations based on their rates.

Let \( x \) represent the number of minutes that have passed.

The first machine starts with 92 cans and fills at a rate of 16 cans per minute, so after \( x \) minutes, it will have filled: \[ 92 + 16x \]

The second machine starts with 104 cans and fills at a rate of 13 cans per minute, so after \( x \) minutes, it will have filled: \[ 104 + 13x \]

We want to find the point at which both machines have filled the same number of cans: \[ 92 + 16x = 104 + 13x \]

Now, we will solve for \( x \).

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 both sides by 3: \[ x = 4 \]

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

The correct answer choice is: 104 + 16x = 92 + 13x ; x = 4 ( which was repeated in the options)