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# Systems of Equations on the GRE

## The Power of Elimination

How would you solve the system of linear equations below?

xy = 5

2x + y = 13

There are two primary approaches for solving systems of linear equations:

1)   Substitution Method

2)   Elimination Method

## The Substitution Method

With this method, we take one of the equations and solve for a certain variable. For example, we might take xy = 5 and add y to both sides to get x = y + 5.

Then we take the second equation (2x + y = 13) and replace x with y + 5 to get: 2(y + 5)+ y = 13

From here, we have an equation we can solve for y: y = 1

Now that we know the value of y, we can take one of the equations and replace y with 1 to find the value of x: x = 6

So, the solution is x = 6 and y = 1.

## The Elimination Method

With this method, we notice that, if we add the two original equations (xy = 5  and 2x + y = 13), the y’s cancel out (i.e., they are eliminated), leaving us with: 3x = 18.

From here, when we divide both sides by 3, we get: x = 6, and from here we can find the value of y: y = 1.

Okay, so that’s how the two methods work. What’s my point?

The point I want to make is that, although both methods get the job done, the Elimination method is superior to the Substitution method. And by “superior,” I mean “faster.”

First, the Elimination method can often help us avoid using fractions. Consider this system:

5x – 2y = 7

3x + 2y = 17

To use the Substitution method here, we’d have to deal with messy fractions. For example, if we take the equation 5x – 2y = 7 and solve for x, we get x = (2/5)y + 7/5. Then when we take the second equation (3x + 2y = 17) and replace x with (2/5)y + 7/5, we get: 3[(2/5)y + 7/5] + 2y = 17. Yikes!!

Alternatively, we can use the Elimination method and add the two original equations (5x – 2y = 7 and 3x + 2y = 17). When we do this, the y’s cancel, leaving us with: 8x = 24, which means x = 3. No messy fractions.

It has been my experience that many students rely solely on the Substitution method to solve systems of equations, and this can potentially eat up a lot of time on test day. So, be sure to learn the Elimination method soon. In fact, if I were you, I’d drop the Substitution method from my repertoire; it isn’t very useful.

If you still believe that the Substitution method is just as good as the Elimination method, try solving this question using the Substitution method.

If 5x – 8y = 11, and 4x – 9y = 4, what is the value of x + y?

1. 3
2. 4
3. 5
4. 6
5. 7

To view the solution to the above question, see my post, The Reasonable Test-Maker.

By the way, students who use Magoosh GRE improve their scores by an average of 8 points on the new scale (150 points on the old scale.) Click here to learn more.

### 5 Responses to Systems of Equations on the GRE

1. George August 21, 2013 at 11:02 pm #

Both are straight line eq’s-a even faster way would be just to graph both mentally and get an intercept.

• Brent Hanneson August 24, 2013 at 7:41 am #

Hi George,

Yes, mentally graphing the lines and finding their point of intersection will, indeed, work. I’m just not sure how many people are able to do this. It would certainly be difficult to do so with this system:
5x – 8y = 11
4x – 9y = 4

Cheers,
Brent

2. sanket July 10, 2012 at 7:21 am #

In example 1; the value of x is 6.
as the calculated value of y is 1, x-1=5
which makes x=6.

• Brent Hanneson July 10, 2012 at 6:17 pm #

You’re absolutely right, sanket.
We’ll fix that right away.

Cheers,
Brent

3. Ahmad December 28, 2011 at 10:56 am #

thanks

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