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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.
12x+180=15x+210
12x+210=15x+180
12x−210=15x−180
210−12x=180−15x
Movie Theatre A has a popcorn machine with a 100-gallon capacity, and sells 3 gallons of popcorn per hour. Movie Theatre B has a popcorn machine with a 150-gallon capacity, and sells 5 gallons of popcorn per hour. Write an equation to show when the two popcorn machines will have the same amount of popcorn left. Use x
as your variable.(1 point)
Responses

150+3x=100+5x
150 plus 3 x equals 100 plus 5 x

100−3x=150−5x
100 minus 3 x equals 150 minus 5 x

3x−100=5x−150
3 x minus 100 equals 5 x minus 150

100+3x=150+5x
An ice cream shop makes vanilla and chocolate ice cream. The vanilla ice cream machine has a 180-cup capacity, and it sells 20 cups per day. The chocolate ice cream machine has a 140-cup capacity, and it sells 16 cups per day. Write and solve an equation to show when the two machines will have the same amount of ice cream left.(1 point)
Responses

140+16x=180+20x
; x=10
140 plus 16 x equals 180 plus 20 x ; x equals 10

180−20x=140−16x
; x=10
180 minus 20 x equals 140 minus 16 x ; x equals 10

180−20x=140−16x
; x=20
180 minus 20 x equals 140 minus 16 x ; x equals 20

180+16x=140+20x
; x=−10

Two machines at a factory are filling cans of soup. One machine has filled 92 cans of soup. It fills cans at a rate of 16 cans per minute. The second machine has filled 104 cans of soup. It fills cans at a rate of 13 cans per minute. Write and solve an equation to show when the two machines will have filled the same number of cans of soup.(1 point)
Responses

92−16x=104−13x
; x=−4
92 minus 16 x equals 104 minus 13 x ; x equals negative 4

104+16x=92+13x
; x=4
104 plus 16 x equals 92 plus 13 x ; x equals 4

92+16x=104+13x
; x=6
92 plus 16 x equals 104 plus 13 x ; x equals 6

92+16x=104+13x
; x=4

Movie Theatre A has a popcorn machine with a 120-gallon capacity. It sells 7 gallons of popcorn per hour. Movie Theatre B has a popcorn machine with a 150-gallon capacity. It sells 9 gallons of popcorn per hour. Write and solve an equation to show when the two popcorn machines will have the same amount of popcorn left.(1 point)
Responses

120+7x=150+9x
; x=−15
120 plus 7 x equals 150 plus 9 x ; x equals negative 15

120−7x=150−9x
; x=10
120 minus 7 x equals 150 minus 9 x ; x equals 10

120x−7=150x−9
; x=15
120 x minus 7 equals 150 x minus 9 ; x equals 15

120−7x=150−9x
; x=15

5 answers

Let's break down each scenario and verify the equations provided:

1. Two machines at a factory filling cans of soup:

Given:

  • Machine 1 (M1): 180 cans filled, filling rate = 15 cans/min
  • Machine 2 (M2): 210 cans filled, filling rate = 12 cans/min

Let \( x \) represent the number of minutes passed.

Equation:

  • Cans filled by M1 after \( x \) minutes: \( 180 + 15x \)
  • Cans filled by M2 after \( x \) minutes: \( 210 + 12x \)

Set them equal to find when they will have the same number of cans: \[ 180 + 15x = 210 + 12x \]

Solving this leads us to: \[ 15x - 12x = 210 - 180 \] \[ 3x = 30 \] \[ x = 10 \]

So the correct equation is \( 180 + 15x = 210 + 12x \).

2. Movie Theatre A and B with popcorn machines:

Given:

  • Theatre A: 100-gallon capacity, sells 3 gallons/hour
  • Theatre B: 150-gallon capacity, sells 5 gallons/hour

Let \( x \) represent the number of hours passed.

Equation:

  • Popcorn left in Theatre A after \( x \) hours: \( 100 - 3x \)
  • Popcorn left in Theatre B after \( x \) hours: \( 150 - 5x \)

Set them equal to find when they will have the same amount of popcorn left: \[ 100 - 3x = 150 - 5x \]

Solving this leads us to: \[ 5x - 3x = 150 - 100 \] \[ 2x = 50 \] \[ x = 25 \]

So the correct equation is \( 100 - 3x = 150 - 5x \).

3. Ice cream machines (vanilla and chocolate):

Given:

  • Vanilla machine: 180-cup capacity, sells 20 cups/day
  • Chocolate machine: 140-cup capacity, sells 16 cups/day

Let \( x \) represent the number of days passed.

Equation:

  • Ice cream left in vanilla machine after \( x \) days: \( 180 - 20x \)
  • Ice cream left in chocolate machine after \( x \) days: \( 140 - 16x \)

Set them equal: \[ 180 - 20x = 140 - 16x \]

Solving this leads us to: \[ -20x + 16x = 140 - 180 \] \[ -4x = -40 \] \[ x = 10 \]

So the correct equation is \( 180 - 20x = 140 - 16x \).

4. Two machines filling cans of soup:

Given:

  • Machine 1: 92 cans filled, filling rate = 16 cans/min
  • Machine 2: 104 cans filled, filling rate = 13 cans/min

Let \( x \) represent the number of minutes passed.

Equation:

  • Cans filled by M1 after \( x \) minutes: \( 92 + 16x \)
  • Cans filled by M2 after \( x \) minutes: \( 104 + 13x \)

Set them equal: \[ 92 + 16x = 104 + 13x \]

Solving this leads us to: \[ 16x - 13x = 104 - 92 \] \[ 3x = 12 \] \[ x = 4 \]

So the correct equation is \( 92 + 16x = 104 + 13x \).

5. Movie Theatre A and B with popcorn machines:

Given:

  • Theatre A: 120-gallon capacity, sells 7 gallons/hour
  • Theatre B: 150-gallon capacity, sells 9 gallons/hour

Let \( x \) represent the number of hours passed.

Equation:

  • Popcorn left in Theatre A after \( x \) hours: \( 120 - 7x \)
  • Popcorn left in Theatre B after \( x \) hours: \( 150 - 9x \)

Set them equal: \[ 120 - 7x = 150 - 9x \]

Solving this leads us to: \[ 9x - 7x = 150 - 120 \] \[ 2x = 30 \] \[ x = 15 \]

So the correct equation is \( 120 - 7x = 150 - 9x \).

In summary, here’s the correct approach for each situation and the derived equations when the machines would have the same amount:

  1. From soup machines: \( 180 + 15x = 210 + 12x \)
  2. From popcorn machines: \( 100 - 3x = 150 - 5x \)
  3. From ice cream machines: \( 180 - 20x = 140 - 16x \)
  4. From the second soup machines: \( 92 + 16x = 104 + 13x \)
  5. From the second popcorn machines: \( 120 - 7x = 150 - 9x \)

The corresponding solutions are:

  1. \( x = 10 \)
  2. \( x = 25 \)
  3. \( x = 10 \)
  4. \( x = 4 \)
  5. \( x = 15 \)
Are you sure about the first answer?

Let's re-evaluate the first scenario involving the two machines filling cans of soup.

Given:

  1. Machine 1 (M1):

    • Already filled: 180 cans
    • Filling rate: 15 cans/min

  2. Machine 2 (M2):

    • Already filled: 210 cans
    • Filling rate: 12 cans/min

Let:

  • \( x \): represent the number of minutes that have passed.

Equation Setup:

  1. Total cans filled by Machine 1 after \( x \) minutes:
    \( 180 + 15x \)

  2. Total cans filled by Machine 2 after \( x \) minutes:
    \( 210 + 12x \)

Set them equal to find when they will have filled the same number of cans:

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

Solving the Equation:

  1. Rearranging gives:

    • \( 15x - 12x = 210 - 180 \)
    • \( 3x = 30 \)

  2. Dividing both sides by 3:

    • \( x = 10 \)

Conclusion:

The machines will have filled the same number of cans after 10 minutes.

So, the initial equation \( 180 + 15x = 210 + 12x \) is indeed correct. The solution \( x = 10 \) also holds.

If there's still any specific aspect of the solution you would like me to clarify or review further, please let me know!

Answer choices:
A. 12x+180=15x+210
B. 12x+210=15x+180
C. 12x−210=15x−180
D. 210−12x=180−15x

To determine the correct equation from the provided answer choices, we must revisit the setup for when the two machines will have the same total number of cans filled:

Given:

  1. Machine 1 (M1):

    • Already filled: 180 cans
    • Filling rate: 15 cans/min

  2. Machine 2 (M2):

    • Already filled: 210 cans
    • Filling rate: 12 cans/min

Setup:

  • Cans filled by M1 after \( x \) minutes:
    \( 180 + 15x \)

  • Cans filled by M2 after \( x \) minutes:
    \( 210 + 12x \)

Equation:

Set the two expressions equal to each other: \[ 180 + 15x = 210 + 12x \]

Now, let's rearrange this equation to see if it matches any of the given choices:

  1. Move \( 12x \) to the left: \[ 180 + 15x - 12x = 210 \] \[ 180 + 3x = 210 \]

  2. Move 180 to the right: \[ 3x = 210 - 180 \] \[ 3x = 30 \]

Rewriting the original equation:

From \( 180 + 15x = 210 + 12x \), we can directly express it in the options format:

  • Rearranging gives \( 12x + 180 = 15x + 210 \), which matches option A.

Thus:

The answer choice that correctly represents the situation is: A. \( 12x + 180 = 15x + 210 \).