Two machines at a factory are filling cans of soup. One machine has already filled 180 cans of soup, and fills cans at a rate of 15 per minute. The second machine has already filled 210 cans of soup, and fills cans at a rate of 12 per minute. Write an equation to show when the two machines will have filled the same number of cans of soup. Use x

Question

as your variable.(1 point)
Responses

12x+180=15x+210
12 x plus 180 equals 15 x plus 210

210−12x=180−15x
210 minus 12 x equals 180 minus 15 x

12x−210=15x−180
12 x minus 210 equals 15 x minus 180

12x+210=15x+180

Answer 1

To find when the two machines will have filled the same number of cans of soup, we need to set up an equation based on the information provided.

Let \( x \) represent the number of minutes that have passed since we started counting.

1. The first machine has already filled 180 cans and fills at a rate of 15 cans per minute. Therefore, after \( x \) minutes, the total number of cans filled by the first machine will be: \[ 180 + 15x \]

2. The second machine has already filled 210 cans and fills at a rate of 12 cans per minute. Therefore, after \( x \) minutes, the total number of cans filled by the second machine will be: \[ 210 + 12x \]

To find when they will have filled the same number of cans, we set the two expressions equal to each other: \[ 180 + 15x = 210 + 12x \]

Therefore, the correct equation is: \[ 180 + 15x = 210 + 12x \]

This matches the provided options, so the answer is: \[ 15x + 180 = 12x + 210 \]