# GMAT Tricks with Systems of Equations: Part 3

A lot of GMAT test-takers vaguely remember a rule from high school, that it’s possible to solve for two variables if and only if you’re given two equations, and generally that it’s possible to solve for n variables if and only if you’re given n equations. Unfortunately, that rule isn’t quite correct as written, and even the correct rule isn’t always relevant.

Applying the rule incorrectly causes quite a few errors on the quant section, particularly with Data Sufficiency questions. This is the third post in our series on avoiding those errors:

In my last two posts, we saw that sometimes two equations aren’t enough to allow us to solve for two variables, or even for one. In those posts we saw that sometimes a Data Sufficiency question that looks like a (C) turns out to be an (E), because information that seemed to be sufficient wasn’t.

It turns out, though, that even when apparently sufficient information turns out to be insufficient, the answer isn’t always (E). Let’s see why.

## A Sample Problem

Take a minute or two to answer this problem:

If 2x + y = 16, what is the value of x – y? (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

In this case, if we too hastily apply the rule described above, we’ll probably choose (A) right away. After all, x and y are two variables and the question stem itself provided one equation; Statement (1) provides the second equation.  However, the correct answer is (C).  Why?

We’ve already seen that the when an equation includes any of the following, it’s not linear, and so a system of two equations may not be sufficient to solve for two variables.

• a variable raised to a power
• two or more variables multiplied together
• a variable in the denominator of a fraction and a variable in the numerator of another fraction or outside any fraction

Notice that in our problem the equation in Statement (1) has one variable (x) in a denominator and another variable (y) outside of any fraction.

Let’s try to solve the problem as we would a system of linear equations and see what happens. We could solve this by simultaneous solution, but I’m going to take the apparently simpler route of substitution, solving for x in terms of y, and then substituting the y-expression for x.

## Consider Statement (1) by Itself

Begin with the equation from Statement (1). Transpose to isolate x. Substitute the right-hand expression for x in the equation from the question stem. Distribute 2 to clear the parentheses. Multiply both sides of the equation by y to clear the fraction. Combine like terms and put everything on the left-hand side of the equation. Because this is a quadratic but not a quadratic square (not of the form or of the form ) it has two solutions (and on the GMAT those will always be two real solutions).

If we’re just considering Statement (1) for now, and ignoring Statement (2), then we don’t really care what exactly those solutions are, just that there are two of them, and that if we determine the two solutions and the x values that correspond to each, we’ll have two different answer to the question “What is the value of  x – y?” This means that Statement (1) is not sufficient by itself, and we can eliminate answers (A) and (D).

## Consider Statement (2) by Itself

Statement (2) is probably easier to deal with so long as we remember to look at it by itself first, ignoring Statement (1). We can see that even with the equation from the question stem, Statement (2) isn’t sufficient. There is more than one pair of values x and y such that 2x + y = 16 and such that x>2y, and each of those pairs of values gives a different answer to the question, “What is the value of x-y?”

If it’s not immediately obvious that Statement (2) is not sufficient, test values in a disciplined way. Stipulate legal values for x and y, and see whether they give the same answer to the question. Coming up with legal values can itself be tricky. In this case, it’s probably simplest to begin with a small even value for y, say y=0. This will yield a relatively large integer value for x, x=8. For that pair of values x – y = 8. Try again with y=2 and x=7. For that pair of values x – y = 5. That’s two different answers to the question, “What is the value of x – y?”

So we can eliminate (B), leaving just (C) and (E).

## Consider Statements (1) and (2) Together

Choosing between these requires actually solving for y in the equation . Begin by solving for y in the equation .

We need two negative numbers whose product is 12 and whose sum is -8. (As a practical matter we’ll probably think first of two positive numbers whose product is 12 and whose sum is 8, and then change their signs.) 12 has just a few factors, so it won’t take long to arrive at -6 and -2.

(y-6)(y-2) = 1, y = 6 and y = 2.

Plug 2 in for y in the equation 2x + y = 16

2x + 2 = 16

2x = 14

x = 7

This pair of values means that x – y = 7 – 2 = 5

Plug 6 in for y in the equation 2x + y = 16

2x + 6 = 16

2x = 10

x = 5

But this pair of values, x=5 and y=6, fails Statement (2), x>2y, so we needn’t consider it.

Neither Statement was sufficient to answer the question, but Statements (1) and (2) together allow just one answer to the question “What is the value of x – y?”, so we mark (C).

## The Takeaway

Just as we saw in simper questions in my last two blog posts, it’s not unusual for a Data Sufficiency question to require more information than we might expect from that often poorly remembered rule that we started with. In the earlier posts that meant that the answer appeared to be (C) but was in fact (E); in this post that meant that the answer appeared to be (A) but was in fact (C).

## Next

In my next few posts we’ll look at an even more common trap: Data Sufficiency questions that require less information than we might expect.

## Author

• Michael Schwartz is really good at standardized tests. He’s earned multiple perfect scores on the GRE, GMAT, and LSAT. He’d rather have perfect pitch or be able to run low 1:40s for the 800 meters, but you take what you get. He has decades of teaching and curriculum-development experience. One of these days he might finish his dissertation and collect that Ph.D. in philosophy. Might.

### 4 Responses to GMAT Tricks with Systems of Equations: Part 3

1. Mayank February 20, 2015 at 11:32 pm #

isn’t the solution to {y^2} – 8y + 12 = 0, supposed to be y=2 and y=6 instead of the one provided above? Overall answer will still be C after substituting 2 and 6 in the original equation. Please correct me if I am wrong.

• Rita Kreig February 24, 2015 at 12:03 pm #

Hi Mayank!

Yes, you are absolutely right – thank you for catching the error! If you factor y^2-8y+12=0, then your solutions would be y=2 and y=6, and the solution to the problem would be (C). Thanks for reading, and for letting us know when we make mistakes. 🙂

Best of luck with your prep! Please let me know if I can help with anything.

Cheers,
Rita

• Mayank February 24, 2015 at 7:51 pm #

Thanks You and appreciate all the fantastic work you guys do on this blog :-).

• Rita Kreig February 25, 2015 at 10:36 am #

Magoosh blog comment policy: To create the best experience for our readers, we will only approve comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! 😄 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors.

We highly encourage students to help each other out and respond to other students' comments if you can!

If you are a Premium Magoosh student and would like more personalized service from our instructors, you can use the Help tab on the Magoosh dashboard. Thanks!