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

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

Answer 1

To solve the problem, we first need to establish how many cans each machine will have filled over time.

Let's denote \( x \) as the number of minutes that have passed.

• The first machine has already filled 180 cans and fills at a rate of 15 cans per minute, so after \( x \) minutes, it will have filled: \[ 180 + 15x \]

• The second machine has already filled 210 cans and fills at a rate of 12 cans per minute, so after \( x \) minutes, it will have filled: \[ 210 + 12x \]

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

This can be rearranged to match one of the forms given in your responses. Rearranging gives: \[ 15x - 12x = 210 - 180 \] \[ 3x = 30 \]

This does not directly fit the forms you've shared since the equation involves simplification. However, if we rearrange the original form \( 180 + 15x = 210 + 12x \), we can also express it as:

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

So, the correct equation from your choices is: 12x + 210 = 15x + 180.