Adding Two Equations in a System

2.5.2 Adding Two Equations in a System

2.5.2 • Adding Two Equations in a System

Activity

A teacher purchased 20 calculators and 10 measuring tapes for her class and paid $495. Later, she realized that she didn’t order enough supplies. She placed another order of 8 of the same calculator and 1 of the same measuring tape and paid $178.50.

A calculator and a tape measure

This system represents the constraints in this situation. Discuss questions 1 – 4 with a partner, then answer the questions.

{20c+10m=4958c+m=178.50{20c+10m=4958c+m=178.50

1.

In this situation, what do the solutions to the first equation mean?

Compare your answer:

Answer:

The solutions to 20c+10m=49520c+10m=495 are all possible combinations of unit prices for the two items that make a total of $495 when buying 20 calculators and 10 measuring tapes.

2.

What do the solutions to the second equation mean?

Compare your answer:

Answer:

The solutions to 8c+m=178.508c+m=178.50 are all possible combinations of unit prices for the two items that make a total of $178.50 when buying 8 calculators and 1 measuring tape.

3.

For each equation, how many possible solutions are there? Explain how you know.

Compare your answer:

Answer:

Your answer may vary, but here is a sample.

In the first equation, c=20c=20 and m=9.50m=9.50 would make the equation true, as would c=15c=15 and m=19.50m=19.50.

4.

In this situation, what does the solution to the system mean?

Compare your answer:

Answer:

The price for a calculator and the price for a measuring tape that would make both equations true (or satisfy both constraints).

5.

Find the value of cc.

21.50

6.

Find the value of mm.

6.50

7.

Explain your reasoning for the solution.

Compare your answer:

Answer:

Your answer may vary, but here is a sample.

Isolate the variable mm.

8c+m=178.50m=178.50−8c8c+m=178.50m=178.50−8c

Substitute

20c+10(178.50−8c)=49520c+1785−80c=495−60c+1785=495−60c=−1290c=21.5020c+10(178.50−8c)=49520c+1785−80c=495−60c+1785=495−60c=−1290c=21.50

m=178.50−8(21.50)m=178.50−172m=6.50m=178.50−8(21.50)m=178.50−172m=6.50

Use the following information to answer questions 8 – 11:

To be reimbursed for the cost of the supplies, the teacher recorded, “Items purchased: 28 calculators and 11 measuring tapes. Amount: $673.50.”

8.

Write an equation to represent the relationship between the numbers of calculators and measuring tapes, the prices of those supplies, and the total amount spent.

Compare your answer:

Answer:

Your answer may vary, but here is a sample. 28c+11m=673.5028c+11m=673.50.

9.

How is this equation related to the first two equations?

Compare your answer:

Answer:

Your answer may vary, but here is a sample. It is the sum of the first two equations.

10.

In this situation, what do the solutions of this equation mean?

Compare your answer:

Answer:

Your answer may vary, but here is a sample. The solutions are all possible combinations of prices for a calculator and for a measuring tape that make a total of $673.50 when 28 calculators and 11 measuring tapes are bought.

11.

How many possible solutions does this equation have? How many solutions make sense in this situation? Be prepared to show your reasoning.

Compare your answer:

Your answer may vary, but here is a sample. It has many possible solutions, but only one solution makes sense. The prices of a calculator and of a measuring tape in this equation are the same as those in the other two equations. The new equation simply shows the total numbers of supplies and the total amount paid.

Self Check

At a toy store, a family bought 3 toy cars, c , and 2 teddy bears, b , for

{ 3 c + 2 b = 46 7 c + 2 b = 70

Which statement must be true?

  1. 10 c + 4 b = 116 is the same line as one of the equations in the system.
  2. The solution of the system is not a solution of 10 c + 4 b = 116 .
  3. The solution of the system is a solution of 10 c + 4 b = 116 .
  4. 10 c + 4 b = 116 represents the solutions of the system.

Additional Resources

Adding Two Equations in a Real-World Situation

Let’s look at the situation described and write a system of equations.

Paulina went to the pool 8 times and the skating rink 4 times this month. She spent a total of $128 on admission to both places. Last month, she went to the pool 3 times and the skating rink 4 times. Last month she spent $83 on admission to both places.

Let pp represent the cost of the pool and ss represent the cost of the skating rink.

This month Last month
8p+4s=1288p+4s=128 3p+4s=833p+4s=83

So, this system represents the constraints in this situation:

{8p+4s=1283p+4s=83{8p+4s=1283p+4s=83

Over both months, she went to the pool 11 times and the skating rink 8 times. She spent a total of $128 + $83, or $211, over both months.

The equation that represents this total is 11p+8s=21111p+8s=211. Notice this is the equivalent of adding the like terms of the two equations in the system of equations.

Let’s solve the original system and determine if the solution also applies to this new equation.

Using substitution, we find that the solution is (9,14)(9,14), or the pool costs $9 and the skating rink costs $14.

Substitute these values into the new equation:

11p+8s=21111(9)+8(14)=21199+112=211211=211✓11p+8s=21111(9)+8(14)=21199+112=211211=211✓

Adding the equations together produces a new equation that has the same solution.

Try it

Adding Two Equations in a Real-World Situation

Alicia is a chef for a restaurant. This week, she ordered 32 steaks and 45 chickens. The total cost this week was $840.

Last week, she ordered 32 steaks and 24 chickens. The total cost was $672.

Let ss represent the cost of a steak and cc represent the cost of a chicken.

Use this information to answer questions 1 – 2.

1.

Write a system of equations to represent the constraints given.

Compare your answer:

{32s+45c=84032s+24c=672{32s+45c=84032s+24c=672

2.

Write a new equation that represents the total cost for both weeks.

Compare your answer:

64s+69c=1,51264s+69c=1,512