In the last math post we talked about how the new GRE is putting more emphasis on word problems. However, there are plenty of pure algebra problems on the new test. Here’s a couple of good examples of the types of algebra problems you might see.

1. This first problem involves two equations and two variables. You may have learned how to do a problem like this in school with your graphing calculator. Unfortunately, you won’t have that available to you for the GRE.

Usually, the best way to solve a problem like this is to use a technique called elimination. The idea is to eliminate a variable so that you only have one unknown instead of two. What you will do is subtract one equation from the other, but you have to make sure you are getting rid of a variable.

What I’m going to do in this case is multiply the first equation by 7 and the second equation by 3. You’ll see why very shortly. It’s important to remember when you’re doing this to multiply on both sides of the equation.

7 * (7x + 3y = 12) ——–> 49x + 21y = 84

3 * (3x + 7y = 6) ——–> 9x + 21y = 18

Now, I’m going to subtract one equation from the other. Notice that because we have 21y in both equations, were going to be left with 0y. The y variable has been eliminated!

49x + 21y = 84

– (9x + 21y = 18)

40x +0y = 66

40x = 66

Now, solve for x by dividing both sides by 40. Notice that since there are no decimals in the answer choices, it’s probably a good idea to keep this as a fraction.

x = 66/40 = 33/20

Now that we have a value for x, plug this into one of the equations. It usually doesn’t matter which one, but I’m going to choose the second one.

3x + 7y = 6

3(33/20) + 7y = 6

99/20 + 7y = 6

Subtract 99/20 from both sides.

7y = 6 – 99/20

Convert 6 into a fraction. 6 is equal to 120/20.

7y = 120/20 – 99/20

7y = 21/20

Divide both sides by 7.

y = 3/20

Ok, last step. We have values for x and y, we just need to subtract them.

33/20 – 3/20 = 30/20 = 3/2

Therefore, the correct answer is (B).

2. Here’s a good problem involving exponents.

There’s a few different ways to think about this problem, but here’s how I solved it. On the left side of the equation you have 1/2^m + 1/2^m. Since the term 1/2^m is being added to itself, you can think of it as 1/2^m times 2. To give another example, let’s say you had x + x. That would be equal to 2x. So 1/2^m + 1/2^m is equal to 2 times 1/2^m.

2(1/2^m) = 1/2^x

This can be rewritten as the following.

2/2^m = 1/2^x

Here is still another way to write this equation.

2^1/2^m = 1/2^x

Now, remember that when you are dividing two numbers that have the same base but different exponents, you subtract the exponents. For example, 3^5 divided by 3^2 is equal to 3^(5-2), or 3^3. So that means that the following is true.

2^(1 – m) = 1/2^x

The right side of the equation can also be rewritten. Because 2^x is in the denominator, you can think of the exponent as negative. 1/2^x is really 2^-x. If it helps, just think of the numerator as equal to 2^0, because any number to the zero power is equal to 1. So, we have 2^0 divided by 2^x, which is equal to 2^(0 – x) or 2^-x.

2^(1 – m)= 2^–x

Now, notice that the base is now the same on both sides. For a lot of these problems with exponents, this is really your goal. Get the base on both sides to be the same. Because the base is the same, the exponents on each side must be equal to each other.

1 – m = -x

Lastly, you can move things around until you get x on one side, or just multiply both sides by -1.

-1(1 – m) = -1(-x)

m – 1 = x

Therefore, the correct answer is (B).

I don’t understand how you got from 2^1/2^m = 1/2^x to 2^(1 – m) = 1/2^x

Plug in 2 for m.

1/4 + 1/4 = 1/2^x

Solve for x.

1/4 + 1/4 = 2/4 which is simplified to 1/2.

Since 2/4 = 1/2, 1/2^1 = 1/2 also.

Since m = 2, the correct answer is B. 2-1 = 1.

I think there’s a much easier way to do the first one.

7x + 3y = 12

3x + 7y = 6

Subtract the second equation from the first one

4x – 4y = 6

Divide both sides by 4

x – y = 1.5

Yup

There’s an easier way to do the second one too

Plug in m = 1

1/2 + 1/2 = 1/(2^x)

1 = 1/(2^x)

x = 0

So x = m-1