Use the situation below to answer all parts of the question.

The number of calories in a piece of strawberry pie is 20 less than 3 times the number of calories in a scoop of strawberry ice cream. The pie and ice cream together have 500 calories.

Part 1:

Identify the two equations that represent this real-world situation where p represents the number of calories in a piece of strawberry pie and c represents the number of calories in a scoop of strawberry ice cream.

Responses

p−20=3c
p minus 20 is equal to 3 c

p−c=500
p minus c is equal to 500

p=3c−20
p is equal to 3 c minus 20

pc=500
p c is equal to 500

3p=c−20
3 p is equal to c minus 20

p+c=500
Part 2:

How many calories are in a piece of strawberry pie?

Calories

Part 3:

How many calories are in a scoop of strawberry ice cream?

Calories:

3 answers

To solve the problem, let's start with Part 1 by setting up the equations based on the given information.

We know that:

  1. The number of calories in a piece of strawberry pie (p) is 20 less than 3 times the number of calories in a scoop of strawberry ice cream (c). This translates to the equation: \[ p = 3c - 20 \]

  2. The total calories from the pie and the ice cream is 500. This translates to the equation: \[ p + c = 500 \]

So, the two equations representing the situation are:

  • Equation 1: \( p = 3c - 20 \)
  • Equation 2: \( p + c = 500 \)

Part 2: How many calories are in a piece of strawberry pie?

Now we can substitute \( p \) from Equation 1 into Equation 2.

Substituting: \[ (3c - 20) + c = 500 \]

Combining like terms: \[ 4c - 20 = 500 \]

Adding 20 to both sides: \[ 4c = 520 \]

Dividing both sides by 4: \[ c = 130 \]

Now, substitute \( c \) back into Equation 1 to find \( p \): \[ p = 3(130) - 20 \] \[ p = 390 - 20 \] \[ p = 370 \]

So, the number of calories in a piece of strawberry pie is: Calories: 370


Part 3: How many calories are in a scoop of strawberry ice cream?

From our calculations above, we found: Calories: 130

So to summarize:

  • Calories in a piece of strawberry pie: 370
  • Calories in a scoop of strawberry ice cream: 130
Use the real-world situation to answer all part of the question.

Best Buy received a shipment of 20 TVs. Some TVs weigh 30 lbs. and other weigh 50 lbs. The total weight of the shipment is 880 lbs.

Part 1:

Write an equation to represent the total number of TVs.

Use x to represent the TVs weighing 30 lbs. Use y to represent the TVs weighing 50 lbs.

Question 2
Part 2:

Write an equation to represent the weight of the TVs.

Use x to represent the TVs weighing 30 lbs. Use y to represent the TVs weighing 50 lbs.

Question 3
Part 3:

How many 30 lb. TVs were in the shipment?

Number of 30 lb. TVs =

Let's break down the problem step by step to answer all parts of the question.

Part 1: Write an equation to represent the total number of TVs.

Let:

  • \( x \) = number of 30 lb TVs
  • \( y \) = number of 50 lb TVs

The total number of TVs received is 20. Therefore, the equation representing the total number of TVs is: \[ x + y = 20 \]

Part 2: Write an equation to represent the weight of the TVs.

The total weight of the shipment is given as 880 lbs. For the weights of the TVs, we have:

  • Each 30 lb TV contributes \( 30x \) to the total weight.
  • Each 50 lb TV contributes \( 50y \) to the total weight.

Thus the equation representing the total weight is: \[ 30x + 50y = 880 \]

Part 3: How many 30 lb. TVs were in the shipment?

Now we will solve the system of equations formed by:

  1. \( x + y = 20 \)
  2. \( 30x + 50y = 880 \)

From the first equation, we can express \( y \) in terms of \( x \): \[ y = 20 - x \]

Now, substitute \( y \) into the second equation: \[ 30x + 50(20 - x) = 880 \]

Distributing the 50: \[ 30x + 1000 - 50x = 880 \]

Combining like terms: \[ -20x + 1000 = 880 \]

Subtracting 1000 from both sides: \[ -20x = -120 \]

Dividing both sides by -20: \[ x = 6 \]

So, the number of 30 lb. TVs in the shipment is: Number of 30 lb. TVs = 6

Summary:

  • Part 1: \( x + y = 20 \)
  • Part 2: \( 30x + 50y = 880 \)
  • Part 3: 6 30 lb. TVs.