Algebra is usually all about writing equations. In this chapter, we’ll show you a fool-proof way to write the equations you need to write, give you a review of the quadratic formula, and teach you how to do simultaneous equations. Plus, you’ll learn two fantastic techniques that will allow you to avoid writing equations on most GMAT questions: Plugging In and Plugging In The Answers (PITA).
Approximately one-fourth of the problems on the computer-adaptive GMAT Math section involve traditional algebra. Your algebra skills can also be tested by some questions in the Integrated Reasoning section.
In this chapter, we’ll show you some powerful techniques that will enable you to solve these problems without using traditional algebra. The first half of this chapter discusses these new techniques. The second half shows you how to do the few algebra problems that must be tackled algebraically.
Algebra is used to come up with general solutions to problems. For example, you might know (for whatever reason) that the price of a new pair of shoes in relation to the cost of a pair of jeans is 3j + 20, where j is the cost of the jeans. Based on this formula, if you know that the jeans cost $50, then you know that the shoes cost $170. But, you also know the cost of the shoes if the jeans cost $75.
In most algebra problems, you need to find an algebraic expression that matches the description of the relationship given in the problem. For the situation above, the problem might say
At a certain store, the price of a pair of shoes is twenty dollars more than three times the price of a pair of jeans. If the price of a pair of jeans is j dollars at this store, then what is the price, in dollars, of a pair of shoes, in terms of j ?
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20 – 3j |
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3j + 20 |
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3j – 20 |
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3j + 60 |
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20j + 3 |
You might consider this question a fairly easy algebra question. To solve it, you might just start translating the phrase “the cost of a pair of shoes is twenty dollars more than three times the cost of a pair of jeans.” But, you need to be careful while doing that. If you get confused about whether you should add 20 or subtract 20, you’ll pick the wrong answer. The test-writers have put a lot of thought into the ways that the key statement in the question can be misinterpreted.
So, if it’s so easy to make a mistake while doing the algebra, is there a more foolproof way to do this question? Of course! Instead of trying to come up with a general solution—the algebraic approach—let’s pick a number and come up with a specific solution. Then, we’ll just pick the answer that matches.
We call this approach Plugging In. It is perhaps our most powerful math technique and will allow you to solve complicated problems more quickly than you might have ever thought possible. Plugging In is easy. There are three steps involved.
Plugging In
1. Pick numbers for the variables in the problem.
2. Using your numbers, find an answer to the problem. At The Princeton Review, we call this the target answer.
3. Plug your numbers into the answer choices to see which choice equals the answer you found in step 2.
Let’s look at the same problem again:
At a certain store, the price of a pair of shoes is twenty dollars more than three times the price of a pair of jeans. If the price of a pair of jeans is j dollars at this store, then what is the price, in dollars, of a pair of shoes, in terms of j ?
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20 – 3j |
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3j + 20 |
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3j – 20 |
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3j + 60 |
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20j + 3 |
Here’s How to Crack It
Let’s pick a number for j. Let’s say that the jeans cost $10. (We don’t need to worry about being realistic!) In your scratch booklet, write down “j = 10”. We’ve now transformed the problem from an algebra problem into an arithmetic problem.
Here’s what the problem now asks:
At a certain store, the price of a pair of shoes is twenty dollars more than three times the price of a pair of jeans. If the price of a pair of jeans is 10 dollars at this store, then what is the price, in dollars, of a pair of shoes, in terms of j ?
You’ll notice that we’ve substituted 10 for the variable j in the original problem. You’ll also notice that we’ve crossed out the phrase “in terms of j”. Once we put a number into the problem, the phrase “in terms of” has no meaning so we can ignore it.
Using $10 for the price of the jeans, the price of the pair of shoes is $50. All we did was translate the phrase “the price of a pair of shoes is twenty dollars more than three times the price of a pair of jeans.” Three times the price of the pair of jeans is 3 × $10 = $30. Twenty dollars more is just $30 + $20 = $50. The numerical answer to our problem is $50. Write that number down on your noteboard and circle it to indicate that it is your target answers.
Now, you just need to find the answer that matches $50 when you substitute 10 for j. You should write down A, B, C, D, E on your noteboard and work out each answer choice. Here’s what that looks like:
A) 20 – 3j = 20 – 3(10) = –10
B) 3j + 20 = 3(10) + 20 = 50
C) 3j – 20 = 3(10) – 20 = 10
D) 3j + 60 = 3(10) + 60 = 90
E) 20j + 3 = 20(10) + 3 = 203
Answer B is the only answer that matches the target answer so it is the answer to this problem. You’ll note that we checked all five answer choices. We did that to be sure that we were picking the correct answer.
The students in our GMAT course learn to automatically do scratch work. When plugging in, always write down the numbers you are plugging in for each variable. Be sure to clearly label the number for each variable by writing down something like j = 10. Next, do the work for each step in the problem. When you find the numerical answer to the problem, write that down and circle it. Then, try each of the answer choices, crossing them off as you eliminate them. Here’s what your scratch work should have looked like for the last problem:
You might be thinking, “Wait a minute! It was just as easy to solve this problem algebraically. Why should I plug in?” To see why, let’s take a look at another version of the problem.
At a certain store, the price of a pair of shoes is twenty dollars more than three times the price of a pair of jeans and the price of a sweater is fifty percent more than the price of a pair of shoes. If the price of a pair of jeans is j dollars at this store, then what is the price, in dollars, of a pair of shoes, a sweater and a pair of jeans, in terms of j ?
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1.5j + 10 |
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3j + 20 |
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4.5j + 30 |
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5.5j + 30 |
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8.5j + 50 |
This version of the problem is wordier and that helps to make it more confusing. Let’s try plugging in. As before, we’ll start by making j = 10. So, the jeans cost $10. Next, we know the relationship between the price of the jeans and the cost of the shoes. The price of the shoes is “twenty dollars more than three times the price of the jeans.” So, the price of the shoes is $50. Finally, the sweater costs “fifty percent more than the price of the shoes.” So, the sweater is $50 + $25 or $75. So, the cost of all three items is $10 + $50 + $75 = $135. Be sure to circle $135.
Now, it’s time to find the answer that equals 135 when j = 10. Choice A is 25, so cross it off. Choice B is 50, so it’s wrong. Choice C is 75, so it’s also wrong. Choice D is 85, so it can be eliminated. Finally, choice E is 135. Choice E matches the target and is the correct answer.
By the way, choice D is what you get if you misread how to calculate the cost of the sweater. If you were doing the algebra, the cost of the shoes is 3j + 20. It would be pretty easy to think that the price of the sweater is then 1.5j + 10, which is fifty percent of the cost of the shoes rather than fifty percent more. Of course, you could make this mistake while working with the numbers, too. But, what’s easier—to see that $25 is not more than $50 or to see that 1.5j + 10 is not more than 3j + 20?
To recap, there are two reasons why you’d want to plug in even if you are pretty good at algebra.
While you can plug in any number, you’ll find that certain numbers work better than others. Ideally, you want a number that makes it easy to perform the calculations for the problem. For most problems, you can just use small, simple numbers such as 2, 5, or 10. However, you also want numbers that make sense within the context of the problem. For example, for a problem that uses percents, plugging in 100 would be a good idea. Different numbers work for different types of problems. As you practice, you’ll get better at picking good numbers—especially if you keep asking yourself “What number will make this problem easy?” You also shouldn’t be afraid to change your number if the calculations start to get messy.
Sometimes the best way to select a number is to use a little common sense. Here’s an example:
If Jim drives k miles in 50 minutes, how many minutes will it take him to drive 10 miles, at the same rate?
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 |
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 |
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60k |
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10k |
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 |
Here’s How to Crack It
GMAC would like you to use the formula distance = rate × time. There are variables in the answer choices, so we can plug in. Since it’s a good bet that at least a few of the wrong answers are based on using the formula, let’s plug in.
Any number you choose to plug in for k will eventually give you the answer to this problem, but there are some numbers that will make your task even easier.
We need to find a good number for k. Notice that we know that Jim is going to drive 10 miles. Suppose we just made k equal to half of 10? The question now reads:
If Jim drives 5 miles in 50 minutes, how many minutes will it take him to drive 10 miles, at the same rate?
Now the problem is pretty easy. Since Jim is going to drive twice the distance, it’s going to take him twice the time. So, the target is 100 minutes. Now, all we need to do is find the answer that matches the target when k = 5. Start with answer choice A. Divide 500 by 5 to get 100. Bingo! To double check, plug 5 into the other answer choices as well. None of them match the target.
The answer to this question is choice A.
So far, we’ve been looking at questions that have explicit variables in both the problem and the answer choices. When you see variables in the problem or answer choices, that’s one of the signs to plug in.
However, sometimes GMAC expects you to use algebra to answer a question that doesn’t have explicit variables. You can still plug in on these problems. In fact, you should consider plugging in whenever you feel the urge to do algebra. The urge to do algebra is the ultimate sign that the problem can be cracked using some form of Plugging In.
Let’s take a look at some other ways to Plug In.
Some problems have a hidden variable. For these problems, all the calculations are usually based off one item but you don’t know the value of that item. There’s a simple solution—just plug in a value for the item!
Let’s look at an example:
A merchant reduces the original price of a coat by 20 percent
for a spring sale. Finding that the coat did not sell, the merchant reduces the spring price by a further 15 percent at the start of the summer. The coat’s summer price is what percent of its original price?
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35% |
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64% |
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65% |
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68% |
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80% |
Here’s How to Crack It
You may have noticed that this problem never gave us the coat’s original price. You may have felt the urge to do algebra starting to kick in as you read the problem. For example, you might have started to say to yourself “Well, if the price of the coat is x, then the spring price is …” That’s your sign that you can do this problem as a Plug In! All you need to do is pick a price for the coat.
Let’s make the original price of the coat $100. When you are solving a percent problem, 100 is a great number to plug in. The merchant discounts the price of the coat by 20% for the spring sale. Twenty percent of $100 is $20, so the spring price is $80. For the summer, the spring price of the coat is discounted by another 15%. Fifteen percent of $80 is $12, making the summer price $68.
The question asks for the summer price as a percent of the original price. In other words, $68 is what percent of $100? 68%. The answer is choice D.
Easy Eliminations
If you were running out of time, you’d need to make a guess. But, you could eliminate some obviously wrong answers first. When you read the problem fast, you might be tempted to think that the overall reduction is 20% + 15% = 35% but that’s too easy. So, cross off choice A. Choice C is just 100 − 35 = 65. Also, too easy. Finally, choice E is what you get after the first reduction. The answer is probably B or D.
Another way to tell that you can probably solve a question as a hidden plug in is to look at the answer choices. The answer choices for hidden plug in questions are typically percents, fractions, or sometimes ratios. How can you tell that this problem is a hidden plug in?
At College P, one–fourth of the students are seniors and one–fifth of the seniors major in business. If two–fifths of all students at the college major in business, the business majors who are not seniors are what fraction of all business majors?
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 |
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 |
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 |
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 |
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 |
Here’s How to Crack It
You may have noticed that while this problem provides lots of fractions to work with, it never reveals how many students attend College P. Of course, knowing the total enrollment at the college would make solving the problem fairly straightforward. So, before you start setting up equations based on x students at College P, let’s plug in a number for the total students.
An easy way to come up with a good number when the problem has fractions is to simply multiply the denominators of the fractions together. In this case, that’s 4 × 5 × 5 = 100. Don’t worry, by the way, that some of the denominators are not distinct. Multiplying by the extra 5 may help us to avoid getting a result that leads to some fractional part of a student!
It Works for Ratios, Too!
You can also plug in when the answer choices are expressed as ratios. Here’s how the same question would be rewritten as a ratio problem.
…what is the ratio of the senior business majors to the business majors who are not seniors?
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1:3 |
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1:5 |
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1:7 |
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1:8 |
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7:8 |
As before, just plug in 100 for the number of students. There are 5 seniors who are business majors and 35 business majors who are not seniors. The ratio is 5:35, which reduces to 1:7. The answer is B.
So, if there are 100 students at the college, there are 25 seniors and 75 students who are not seniors. Next, you know that one-fifth of the seniors major in business, so that’s 5 seniors who are business majors. Since two-fifths of all students are business majors, the college has 40 business majors. Of those 40 business majors, 40 − 5 = 35 of the business majors are not seniors. So, the fraction of business majors who are not seniors is 
. The correct answer is E.
Some algebra questions ask for a numerical answer. GMAC expects you to write an equation, solve it and then pick the answer. However, the person who wrote the problem had to do all of that hard work. So, rather than duplicate all of the work the question writer did, why not just test out the answers to see which one works?
We call this method of solving the problem Plugging In the Answers (or PITA for short). You’ll find that solving problems this way can save you a lot of time and help you to avoid common algebra mistakes. There are three steps involved when you Plug In the Answers.
Plugging in the Answers
Let’s take a look at an example.
At a certain restaurant, the price of a sandwich is $4.00 more than the price of a cup of coffee. If the price of a sandwich and a cup of coffee is $7.35, including a sales tax of 5%, what is the price of a cup of coffee, EXCLUDING the sales tax?
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$1.50 |
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$3.00 |
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$4.00 |
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$5.50 |
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$7.00 |
If there were no answer choices, you’d be forced to write an equation and solve it. However, because you know the answer must be one of the five provided choices, it will be easier and faster to simply try the answers.
Start by writing down the answers and labeling them as “coffee.” Now, start with choice C. If the cup of coffee costs $4.00, what can you figure out? The problem states that the sandwich costs $4.00 more than the cup of coffee, so the sandwich is $8.00. Make a column for “sandwich” and write down $8.00 next to choice C. You now know that the cost for the sandwich and the cup of coffee is $12.00. Choice C is too big because the cost for the sandwich and the cup of coffee is only supposed to be $7.35 including the sales tax. Go ahead and cross off choice C.
So far, your work should look like this:
Now, its time to decide if you need a bigger number or a smaller numbers. It’s a pretty easy choice for this problem. Since $4.00 turned out to be too big, it makes sense to try a smaller number. There’s no need to debate over choice A or B. One of them must be correct, so just try choice B. If it works, you’re done. If it doesn’t, choose choice A and you’re still done.
If the cup of coffee costs $3.00 (choice B), then the sandwich is $7.00. That would make the total cost for both items $10.00. That’s still too big! The answer must be A.
Here’s what the work should look like at this point:
There’s really no need to test choice A. However, if the coffee costs $1.50, then the sandwich costs $5.50. Together the two items cost $7.00. The tax on the two items is 5% of $7.00 or $0.35. So, the total with tax is $7.35, which meets the condition stated in the problem. The answer is choice A.
When you solve a problem using Plugging In the Answers (PITA), you’ll usually know if you need a bigger or smaller number if choice C didn’t work. However, there are times that you won’t be sure. Rather than wasting time trying to decide if you need a bigger or smaller number, just pick a new number and try it. You’ll actually waste less time testing an extra answer choice or two than you will trying to decide which type of number to try next.
Jim is now twice as old as Fred, who is and two years older than Sam. Four years ago, Jim was four times as old as Sam. How old is Jim now?
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8 |
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12 |
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16 |
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20 |
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24 |
Here’s How to Crack It
This question has numbers in the answers and you probably started to think “Well, if Jim is x years old, then Fred is …” The urge to do algebra means that it’s time to plug in the answers. Start with choice C.
The answers represent possible ages for Jim. If Jim is 16 years old, then Fred is 8 because Jim is twice as old as Fred. Next, you know that Fred is two years older than Sam, so Sam is 6. Now, you need to compare Jim’s age and Sam’s age four years ago. Four years ago, Jim was 12 and Sam was 2. Remember that there’s always a condition in the problem that must be met by the correct answer.
In this case, Jim’s age four years ago must be four times that of Sam. Is 12 four times 2? No. So, choice C can be eliminated.
Here’s what your work should look like up to this point.
Now, it’s time to choose a bigger or smaller number. But, it isn’t really clear which direction to go, is it? So, rather than waste a lot of time trying to figure that out, just pick an answer and try it. If Jim is 20 years old now (choice D), then Fred is 10 and Sam is 8. Four years ago, Jim was 16 and Sam was 4.
Here’s what your work should look like.
Since 4 × 4 = 16, the condition in the problem is met. Choice D is the correct answer.
Most of the time, starting at C when doing PITA makes the most sense. By starting at C, you can usually cut down on the number of answer choices that you need to test. However, there can be times when it makes sense to start with a different answer choice.
If x is a positive integer such that x2 + 5x − 14 = 0, what is the value of x ?
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–7 |
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–5 |
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0 |
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2 |
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5 |
Here’s How to Crack It
Notice that the question states that x is a positive integer. Don’t waste time with choices A, B, or C—cross them off immediately. For this problem, it makes sense to start with choice D. Plug 2 into the equation for x to get (2)2 + 5(2) − 14 = 0. Since the equation is true when x = 2, choice D is the correct answer.
Plugging in the Answers Advanced Tips
Some questions will use the words “must be.” For example, the question may ask which of the expressions in the answers must be even or must be divisible by 3. These questions can be solved easily by plugging in. However, you’ll probably need to plug in at least twice to find the answer.
Must Be Plugging In
If n is a positive integer, which of the following must be even?
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(n − 1)(n + 1) |
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(n − 2)(n + 1) |
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(n − 2)(n + 4) |
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(n − 3)(n + 1) |
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(n − 3)(n + 5) |
Here’s How to Crack It
Start by picking a value for n. How about n = 2? Now, evaluate each answer choice. For choice A, the expression equals 3, so cross it off. For choice B, the expression equals 0. Remember that 0 is an even number so keep this answer choice. Keep checking the answers. Choice C is also equal to 0, so keep it as well. Choice D equals −3. Don’t be fooled by the negative sign. Negative integers can also be even or odd. Since −3 is odd, cross this answer off. Choice E equals −7 and can also be eliminated.
Next, try a new number. Since this problem is about even and odd numbers, it makes sense to try an odd number next. How about n = 3? You only need to check the two answers that remain. Choice B equals 4, which is still even. However, choice C now equals 7, so can be eliminated. The correct answer is choice B.
When doing a must be question, it’s very helpful to set up your scratchwork so that it looks like a chart. Write down A, B, C, D, E and the actual expressions. Make a new column for each number that you plug in. Be sure to cross off answers as you go.
Here’s what your work should look like once you have worked through the entire problem.
You can solve most GMAT algebra problems using some form of Plugging In. However, there are some questions that you may need to solve using some relatively basic algebra.
Before we review some very specific types of algebra that GMAC likes to test, let’s review the two basic rules that are true for any algebraic situation. These two rules are used pretty much any time that you solve a problem algebraically.
Even the simplest equalities can be solved by Plugging In the Answers, but it’s probably easier to solve a simple equation algebraically. If there is one variable in an equation, isolate the variable on one side of the equation and solve it. Let’s try an example of this type, although a question this easy wouldn’t actually be seen on the GMAT. This one is just for practice.
If x – 5 = 3x + 2, then x =
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–8 |
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 |
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–7 |
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 |
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 |
Here’s How to Crack It
Get all of the x’s on one side of the equation. If we subtract x from both sides we have:
Now subtract 2 from both sides:
Finally, divide both sides by 2:
The answer is choice B.
To solve inequalities, you must be able to recognize the following symbols:
| > | is greater than |
| < | is less than |
| ≥ | is greater than or equal to |
| ≤ | is less than or equal to |
As with an equation, you can add a number to or subtract a number from both sides of an inequality without changing it; you can collect similar terms and simplify them. In fact, an inequality behaves just like a regular equation except in one way:
If you multiply or divide both sides of an inequality by a negative number, the direction of the inequality symbol changes.
For example,
–2x > 5
To solve for x, you would divide both sides by –2, just as you would in an equality. But when you do, the sign flips:
Simultaneous equations are almost always tested in data sufficiency format on the GMAT. It’s impossible to solve one equation with two variables. But if there are two equations, both of which have the same two variables, then it is possible to solve for both variables. An easy problem might look like this:
If 3x + 2y = 6 and 5x – 2y = 10, then x = ?
To solve simultaneous equations, add or subtract the equations so that one of the variables disappears.
In more difficult simultaneous equations, you’ll find that neither of the variables will disappear when you try to add or subtract the two equations. In such cases you must multiply both sides of one of the equations by some number in order to get the coefficient in front of the variable that you want to disappear to be the same in both equations. This sounds more complicated than it is. A difficult problem might look like this:
If 3x + 2y = 6 and 5x – y = 10, then x = ?
Let’s set it up the same way:
3x + 2y = 6
5x – y = 10
Unfortunately, in this example, neither adding nor subtracting the two equations gets rid of either variable. But look what happens when we multiply the bottom equation by 2:
On the GMAT, quadratic equations always come in one of two forms: factored or expanded. Here’s an example:
The first thing to do when solving a problem that involves a quadratic equation is to see which form the equation is in. If the quadratic equation is in an unfactored form, factor it immediately. If the quadratic equation is in a factored form, unfactor it. The test writers like to see whether you know how to do these things.
To unfactor a factored expression, just multiply it out using FOIL (First, Outer, Inner, Last):
To factor an unfactored expression, put it into the following format and start by looking for the factors of the first and last terms.
For the first term of the unfactored expression to be x2, the first term of each parentheses of the factored expression has to be x.
For the last term of the unfactored expression to be 15, the last term in each parentheses of the factored expression must be either 5 and 3 or 15 and 1. Since there is no way to get a middle term for the unfactored expression with a coefficient of 2 if the terms were 15 and 1, we are left with
To decide where to put the pluses and minuses in the factored expression, look to see how the inner and outer terms of the factored equation would combine to form the middle term. If we put a minus in front of the 5 and a plus in front of the 3, then the middle term would be –2x (not what we wanted). Therefore, the final factored expression looks like this
Quadratic equations are usually set equal to 0. Here’s an example:
What are all the values of x that satisfy the equation x2 + 4x + 3 = 0?
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–3 |
| |
–1 |
| |
–3 and –1 |
| |
3 and 4 |
| |
4 |
Here’s How to Crack It
This problem contains an unfactored equation, so let’s factor it.
In order for this equation to be correct, x must be either –3 or –1. The correct answer is choice C.
Note: This problem would also have been easy to solve by Plugging In The Answers. It asked a specific question, and there were five specific answer choices. One of them was correct. All you had to do was try the choices until you found the right one. Bear in mind, however, that in a quadratic equation there are usually two values that will make the equation work.
There are three types of quadratic equations the GMAT test writers find endlessly fascinating. These equations appear on the GMAT with great regularity in both the problem-solving format and the data-sufficiency format:
(x + y)2 = x2 + 2xy + y2
(x + y)(x – y) = x2 – y2
(x – y)2 = x2 – 2xy + y2
Memorize all three of these. As with all quadratic equations, if you see the equation in factored form, you should immediately unfactor it; if it’s unfactored, factor it immediately. Here’s an example:
If 
, then x =
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3 |
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5 |
| |
6 |
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7 |
| |
9 |
Here’s How to Crack It
It is unfactored, so let’s factor it:
The (x + 2)s cancel out, leaving us with (x – 2) = 5. So x = 7, and the answer is choice D.
