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−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

210−12x=180−15x

Answer 1

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

For the first machine:

• Initially filled: 180 cans
• Rate: 15 cans per minute
• Cans filled after \(x\) minutes: \(180 + 15x\)

For the second machine:

• Initially filled: 210 cans
• Rate: 12 cans per minute
• Cans filled after \(x\) minutes: \(210 + 12x\)

To find when both machines will have filled the same number of cans, we set the two expressions equal to each other:

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

Now, simplifying the equation for the response options:

Rearranging gives:

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

Now, let's re-write the original equation form from the list given:

The correct equation based on our derivation would be:

• 12x + 210 = 15x + 180

This matches the first response option.