wo 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 as your variable.(1 point)

Question

Responses

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

12x+210=15x+180

Answer 1

To find the equation that shows when the two machines will have filled the same number of cans of soup, we need to express the number of cans filled by each machine as a function of time (in minutes).

1. For the first machine:

• It has already filled 180 cans and fills at a rate of 15 cans per minute.
• Therefore, after \( x \) minutes, the first machine will have filled \( 180 + 15x \) cans.

2. For the second machine:

• It has already filled 210 cans and fills at a rate of 12 cans per minute.
• Therefore, after \( x \) minutes, the second machine will have filled \( 210 + 12x \) cans.

Now we can set the two expressions equal to each other:

\[ 180 + 15x = 210 + 12x \]

Rearranging this gives:

\[ 15x - 12x = 210 - 180 \] \[ 3x = 30 \] \[ x = 10 \]

So the equation we wrote above, \( 180 + 15x = 210 + 12x \), is correct. However, from your response options, the correct equation can be represented as:

\[ 12x + 210 = 15x + 180 \]

This is an equivalent formulation of the same equation derived from balancing both machines' filled cans.

So the correct choice is:

12x + 210 = 15x + 180