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

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

210−12x=180−15x

12x−210=15x−180

Answer 1

To determine when the two machines will have filled the same number of cans of soup, you can set up equations based on the number of cans they've filled so far, combined with their filling rates.

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

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

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

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

This equation simplifies to: \[ 15x - 12x = 210 - 180 \] \[ 3x = 30 \] Thus, \(x = 10\) minutes will be the point at which both machines have filled the same number of cans.

Among the provided responses, the correct equation is: \[ 15x + 180 = 12x + 210 \]

However, your options don't have that specific format. The closest and correct representation is: \[ 15x + 180 = 12x + 210 \]