PSAT/NMSQT Prep 2022 - Eggert M.D., Strelka A. 2022

How much do you know?
Systems of linear equations
The heart of algebra

LEARNING OBJECTIVES

After completing this chapter, you will be able to:

· Solve systems of linear equations by substitution

· Solve systems of linear equations by combination

· Determine the number of possible solutions for a system of linear equations, if any

45/600 SmartPoints^® (Medium Yield)

How much do you know?

Directions: Try the questions that follow. Show your work so that you can compare your solutions to the ones found in the Check Your Work section immediately after this question set. The “Category” heading in the explanation for each question gives the title of the lesson that covers how to solve it. If you answered the question(s) for a given lesson correctly, and if your scratchwork looks like ours, you may be able to move quickly through that lesson. If you answered incorrectly or used a different approach, you may want to take your time on that lesson.

1. What is the value of x for the given equations above?

A. −3

B.0

C.3

D. 5

2. A television set costs $25 less than twice the cost of a radio. If the television and radio together cost $200, how much more does the television cost than the radio?

A. $50

B.$75

C.$100

D. $125

3. At a snack stand, hot dogs cost $3.50 and hamburgers cost $5.00. If the snack stand sold 27 snacks and made $118.50 in revenue, how many hot dogs and how many hamburgers were sold?

A. 16 hot dogs; 11 hamburgers

B.16 hot dogs; 16 hamburgers

C.11 hot dogs; 14 hamburgers

D. 11 hot dogs; 16 hamburgers

4. A certain student cell phone plan charges $0.10 per text and $0.15 per picture, with no additional monthly fee. If a student sends a total of 75 texts and pictures in one month and is billed $8.90 for that month, how many more texts did he send than pictures?

A. 19

B.28

C.36

D. 47

5. In the system of linear equations shown, z represents a constant. If the system of equations has infinitely many solutions, what is the value of z ?

A.

B.5

C.8

D. 40

Check Your Work

1. A

Difficulty: Easy

Category: Substitution

Getting to the Answer: Solve the second equation for y in terms of x (which yields y = −x), then substitute into the first equation and solve:

Choice (A) is correct.

2. A

Difficulty: Medium

Category: Substitution

Getting to the Answer: Translate English into math to write a system of equations with r as the cost of the radio in dollars and t as the cost of the television in dollars. A television costs $25 less than twice the cost of the radio, or t = 2r − 25. Together, a radio and a television cost $200, so r + t = 200.

The system of equations is:

The top equation is already solved for t, so substitute 2r − 25 into the second equation for t:

The radio costs $75, so the television costs 2(75) − 25 = 150 − 25 = $125. This means the television costs $125 − $75 = $50 more than the radio, which is (A).

3. D

Difficulty: Medium

Category: Combination

Getting to the Answer: Begin by translating English into math. Define the variables logically: d for hot dogs, b for hamburgers. You’re given the cost of each, as well as the number of snacks sold and the total revenue generated. Next, write the system of equations that represents the information given:

Multiplying the top equation by −5 allows you to solve for d using combination:

Dividing both sides by −1.5 gives d = 11, which eliminates (A) and (B). Plugging 11 in for d in the first equation in the system gives you 11 + b = 27. Subtract 11 from both sides to find that b = 16.(D) is correct.

4. A

Difficulty: Medium

Category: Combination

Getting to the Answer: Translate English into math to make sense of the situation. First, define your variables: t for texts and p for pictures are good choices. You know that this student sent a total of 75 texts and pictures. You’re also told each text costs $0.10 and each picture is $0.15, and that the bill is $8.90. You’ll have two equations: one relating the numbers of texts and pictures and a second relating the costs associated with each:

Multiplying the second equation by 10 allows you to solve for p using combination:

Subtract the second equation from the first to find that −0.5p = −14 and p = 28. But you’re not done yet; you’re asked for the difference between the text and picture count. Substitute 28 for p in the first equation and then solve for t to get t = 47. Subtracting 28 from 47 yields 19, which is (A).

5. B

Difficulty: Medium

Category: Number of Possible Solutions

Getting to the Answer: A system of equations that has infinitely many solutions results when you can algebraically manipulate one equation to arrive at the other. Examining the right sides of the equations, you see that 40 × 40 = 1,600; therefore, multiplying the first ­equation by 40 will give 1,600 on the right: 5q + 8s = 1,600. The first equation is now identical to the second equation, meaning z must be 5, which is (B).