Okay, so now you know about equations, inequalities, and are familiar with some of the operations that you can perform if you are given an algebra problem on the GRE exam. But the exam is not THAT easy. It measures basic skills, but there are tricks thrown into the mix to make sure that you really do understand the content of each question.

For example, although the GRE exam is intended to evaluate your ability to perform simple mathematical operations, it won't ask you to compute 3 + 5 or 0.1 times 10. Rather, the GRE will ask you to solve for x when (1+4)(0.1)(2x) = 10. This leads us to a very important point:

Test Tip!

If you find yourself doing a long calculation involving multiplication, addition, or subtraction, then STOP. This means that there most likely is a shortcut to the problem because on the exam you will not be asked to perform tedious calculations. Look at the problem and see if you can spot a shortcut or if you can simplify the problem.

Using the above example where you are asked to solve for x when (1+4)(0.1)(2x) = 10, notice that (5)(0.1) = 0.5, which is one half, and one half of 2x is x, so the answer is simply 10. You should be competent enough to do these sorts of calculations in your head, or at least be able to spot such shortcuts.

Related to the topic of simplification is the topic of rewriting. Questions on the GRE will require you to be familiar with elementary concepts in order to arrive at the right answer, but often the questions will be presented in disguise, using either short-hand notations or strange symbols -- don't let this fool you. For example, the following question:

For which values of @ is (@+3)(@-3) = -5^{2} +16 always true?

is actually a straight-forward product of squares question that uses a strange symbol for the variable and, once simplified, the problem is quite easy. Here we solve the problem and point out the "tricks" and shortcuts that you should notice when doing a similar problem.

Step 0 | Original question: When is the following equation true? | For which values of @ is (@+3)(@-3) = -5^{2} +16 |

Step 1 | Notice that the left-hand side is the product of squares, so rewrite it | @^{2}-9 = -5^{2} +16 |

Step 2 | Simplify the right hand side, noting that 5 squared is 25, and then you take the negation of that, so -5^{2} +16 = -25 + 16 = -9.
Notice here that -5^{2} is not the same as (-5)^{2}. |
@^{2}-9 = -9 |

Step 3 | Add 9 to each side | @^{2} = 0 |

Step 4 | What value raised to the second power equals 0? Only 0, so that's the answer | @ = 0 |