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 as your variable. (1 point)

• 122 - 210 = 15х - 180
• 12x + 210 = 15х + 180
• 12х + 180 = 15х + 210
• 210 — 12х = 180 - 15х

1 answer

To set up the equation, we need to express the total number of cans filled by both machines in terms of the variable \(x\), which represents the number of minutes that have passed.

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

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

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

Rearranging this equation leads to: \[ 180 - 210 = 12x - 15x \] \[ -30 = -3x \] Dividing both sides by -3 gives: \[ x = 10 \]

The correct equation from the options provided that represents this is: \[ 12x + 210 = 15x + 180 \]

Thus, the correct choice is: 12x + 210 = 15x + 180