PSAT Math Sample Test
Systems of Equations
The linear equations detailed in the previous chapter are well suited for modeling a variety of scenarios and for solving for a single variable in terms of another that is clearly defined (e.g., what is the cost of a data plan if you consume 4 GB of data in a month). However, sometimes you will be given a set of multiple equations with multiple variables that are interdependent. For example, suppose a $50/month cell phone plan includes $0.05 text messages and $0.40 voice calls, with a cap of 1,000 combined text messages and voice calls.
This scenario can be represented by the following system of equations:
Solving such a system would enable you to determine the maximum number of text messages and voice calls you could make under this plan, while optimizing total usage. To solve systems of equations, you’ll need to rely on a different set of tools that builds on the algebra you’re already familiar with. The following question shows an example of such a system in the context of a testlike question.

If 3r + 2s = 24 and r + s = 12, what is the value of r + 6 ?
 0
 4
 6
 12
Difficulty: Medium
Category: Heart of Algebra / Linear Equations
Getting to the Answer: Although substitution will certainly work here, there is a quicker way. Rewrite the first equation so that (r + s) is a component. First, express 3r as r + 2r. Then factor a 2 out of the lefthand side to get r + 2(r + s) = 24. Now you can substitute 12 in for (r + s), simplify, and solve for r. Be careful. The question is asking for r + 6, so A is a trap corresponding to the value of r alone. Instead, (C) is correct.
You might be tempted to switch on math autopilot at this point and employ substitution, solving the second equation for s in terms of r:
s = 12 – r
You could plug the resulting expression back into the other equation and eventually solve for r, but remember, the PSAT tests your ability to solve math problems in the most efficient way. The following table contains some strategic thinking designed to help you find the most efficient way to solve this problem on Test Day, along with some suggested scratchwork.
Strategic Thinking  Math Scratchwork 

Step 1: Read the question, identifying and organizing important information as you go In this case, you’re looking for the value of r. There are two equations that involve r and s. 
3r + 2s = 24 r + s = 12 
Step 2: Choose the best strategy to answer the question Is there any way you can make the first equation look like the second one? Does the quantity r + s exist in the first equation in some form? How can you effectively use both equations? Once you’ve written the first equation in terms of r + s, substitute the value of r + s (which is 12) into the second equation and solve for r. 
3r + 2s = 24 r + 2r + 2s = 24 r + 2(r + s) = 24 r + 2(12) = 24 r = 0 
Step 3: Check that you answered the right question Be careful! The question isn’t asking for the value of r. Add 6 to your result and you should see that (C) is the correct answer. 
r + 6 = 0 + 6 r + 6 = 6 
While you could use “bruteforce” and substitution to solve this problem, looking for patterns first results in a process that is much quicker and relies on simpler, less errorprone calculations. This chapter will focus on showing you the most efficient strategies to solve systems of equations so that you can get through these problems as quickly as possible on the PSAT.
Independent versus Dependent Equations
Generally, when you have a system involving n variables, you need n independent equations to solve for those variables. Thus, if you have a system of two variables, you need two independent equations. Three variables would require three independent equations, and so on.
Systems of equations are extremely useful in modeling and simulation. Complex mathematical problems such as weather forecasting or crowd control predictions often require 10 or more equations to be simultaneously solved for multiple variables. Fortunately, you won’t encounter anything this daunting on Test Day.
Before we outline the process for solving twovariable systems of equations, let’s clarify one of the key requirements. Earlier, it was stated that you need two independent equations to solve for two variables, but what exactly is an independent equation? Consider the equation 4x + 2y = 8. You could use properties of equality to transform this equation in a number of different ways. For example, you could multiply both sides by 2, resulting in the equation 8x + 4y = 16.
While it seems as though we’ve just created an additional equation, this is misleading, as the second equation has the same core variables and relationships as the first equation. This is termed a dependent equation, and two dependent equations cannot be used to solve for two variables. Look what happens when we try to use substitution. Start by isolating y in the original equation; the result is y = 4 – 2x.
Substituting that into the second equation, notice what happens:
Although 16 does in fact equal 16, this doesn’t bring us any closer to solving for either of the variables. In fact, if you arrive at a result like this when solving a system of equations, then the two equations are dependent. In this case, the system has infinitely many solutions because you could choose any number of possible values for x and y.
At other times, you’ll encounter equations that are fundamentally incompatible with each other. For example, if you have the two equations 4x + 2y = 8 and 4x + 2y = 9, it should be obvious that there are no values for x and y that will satisfy both equations at the same time. Doing so would violate fundamental laws of math. In this case, you would have a system of equations that has no solution. These two equations define parallel lines, which by definition never intersect.
Knowing how many solutions a system of equations has will tell you how graphing them in the same coordinate plane should look. Remember, the solution of a system of equations consists of the point or points where their graphs intersect.
If your system has …  …then it will graph as:  Reasoning 

no solution  two parallel lines  Parallel lines never intersect. 
one solution  two lines intersecting at a single point  Two straight lines have only one intersection. 
infinitely many solutions  a single line (one line directly on top of the other)  One equation is a manipulation of the other—their graphs are the same line. 
Because you could encounter any of these three situations on Test Day, make sure you are familiar with all of them.
Let’s examine a sample problem to investigate the requirements for solving a system of equations:

$\left\{\begin{array}{l}\frac{1}{8}q+\frac{1}{5}s=40\\ zq+8s=1,600\end{array}\right.$
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 ?
 $\frac{1}{8}$
 5
 8
 40
Difficulty: Medium
Category: Heart of Algebra / Linear Equations
Getting to the Answer: A system of equations that has infinitely many solutions results when you can algebraically manipulate one to arrive at the other. Examining the right sides of the equations, you see that 40 × 40 = 1,600; therefore, multiplying the top 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).
Work through the Kaplan Method for Math stepbystep to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.
Strategic Thinking  Math Scratchwork 

Step 1: Read the question, identifying and organizing important information as you go You are looking for the value of z, given that the equation has infinitely many solutions. This means that the second equation should be the same as the first equation after some kind of algebraic manipulation. 

Step 2: Choose the best strategy to answer the question Look for ways to make the first equation resemble the second. Are there any clues? The constant on the right gives you a strong clue: Multiplying 40 by 40 gives 1,600. This also works well with the $\frac{1}{5}$s term because multiplying by 40 yields 8s. 
$\begin{array}{c}\frac{1}{8}q+\frac{1}{5}s=40\\ zq+8s=1,600\end{array}$
$40\left(\frac{1}{8}q+\frac{1}{5}s\right)=40(40)$ 5q + 8s = 1,600 
Step 3: Check that you answered the right question Be careful! Choice D is a trap. The question isn’t looking for the number you’d multiply the first equation by. Instead, it’s looking for z, the coefficient of q. Because the number in front of q in your transformed equation is 5, you know that z must also be equal to 5, making (B) the correct answer. 
5q = zq z = 5 