Two machines at a factory are filling cans of soup. One machine is already filled 180 cans of soup and fill cans at the rate of 15 per minute the second machine has already filled 210 cans of soup and fill cans at the rate of 12 per minute equation to show the two machines will have filled cans of soup. Use X as your variable.

Question

• 12 X +210 =15 X +180
• 12 X +180 = 15 X +210
• 210-12 X =180-15 X
• 12 X -210= 15 X -180

Answer 1

To model the situation where two machines will have filled the same number of cans of soup, we can set up an equation based on the cans each machine has already filled and the rate at which they fill additional cans.

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

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

We want to find the point in time when both machines have filled the same number of cans. This gives us the equation: \[ 180 + 15X = 210 + 12X \]

Rearranging this equation to isolate \( X \) leads to:

1. Subtract \( 12X \) from both sides: \[ 180 + 15X - 12X = 210 \] \[ 180 + 3X = 210 \]

2. Subtract 180 from both sides: \[ 3X = 30 \]

3. Divide by 3: \[ X = 10 \]

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

From the answer options you provided, the correct equation for setting up the problem is: \[ \text{None of the provided equations directly match } 180 + 15X = 210 + 12X\text{ but it can be rearranged to: } 12X + 210 = 15X + 180. \]

So, the correct answer is:

• 12 X + 210 = 15 X + 180.