Two machines at a factory are filling cans of soup. Once 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 can at a rate of 12 per minutes. writes an equation to show when the two machines will have filles the same number of cans of soup. Use x as your variable

Let's define \( x \) as the number of minutes that have passed since we started measuring the filling process for both machines.

1. For the first machine:

   • It has already filled 180 cans.
   • It fills cans at a rate of 15 cans per minute.
   • The total number of cans filled by the first machine after \( x \) minutes can be expressed as: \[ \text{Cans filled by first machine} = 180 + 15x \]

2. For the second machine:

   • It has already filled 210 cans.
   • It fills cans at a rate of 12 cans per minute.
   • The total number of cans filled by the second machine after \( x \) minutes can be expressed as: \[ \text{Cans filled by second machine} = 210 + 12x \]

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

Now you can solve for \( x \).

1. Rearranging the equation gives: \[ 15x - 12x = 210 - 180 \] \[ 3x = 30 \]

2. Dividing both sides by 3: \[ x = 10 \]

So, after 10 minutes, both machines will have filled the same number of cans of soup.