GRE : Systems of Equations on the revised GRE

2:58 PM

GRE

Systems of Equations on the revised GRE

The Power of Elimination

How would you solve the system of linear equations below?

x – y = 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 x – y = 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 (x – y = 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?

Upon seeing this question, it would be perfectly natural for you to immediately begin applying one of the techniques you learned in high school for solving systems of linear equations. After all, this is a system of linear equation, and you probably solved dozens of similar systems in your life. So, with great gusto and confidence, you begin solving the system.

If you happen to like the substitution method, you might take the first equation and solve it for x to get x = 8y/5 + 11/5. Then, you take the second equation and replace x with 8y/5 + 11/5 to get the not-so-pleasant equation 4(8y/5 + 11/5) – 9y = 4.

At this point, the original question is not testing your expertise with systems of equations - it’s testing whether or not you’re one of those people who latches onto a certain approach, and refuses to consider alternative approaches. If you happen to be one of those people, you may continue with these calculations and needlessly waste a lot of time in the process.

So, what should you do instead?

You should remember our tip and ask yourself, “Are the test-makers really interested in whether or not I can solve the equation 4(8y/5 + 11/5) – 9y = 4”?

The answer to this is a resounding NO. The test-makers have very little interest in this. In fact, they’ve given you an onscreen calculator to show how little they care about your computational skills.

So, if you truly believe (and embrace) the idea that the test-makers have little interest in your computational skills, you can be certain that there MUST be another approach to this question that does not involve extremely messy equations.

Now, what is that approach?

Well, the trick here is to recognize that the question does not ask you to find the value ofx and/or the value of y. Instead, you are asked to find the sum of x and y. This is an important clue, since it tells us that we do not need to find the individual values of each variable.

From here, if we recognize that the x-coefficients (5 and 4) differ by 1, and the y-coefficients differ by 1, we might see that something convenient happens when we subtract the second equation from the first equation.

5x – 8y = 11

- 4x – 9y = 4

x + y = 7

When we do this, we get x + y = 7. So, the answer is E.

Now, this is a tricky question, so you may not have spotted that particular shortcut. If that were the case, you would be forced to tediously solve the system of equations. So, just knowing that the test-makers are reasonable people with little interest in your computational skills does not necessarily mean that you’ll be able to spot shortcuts. However, by accepting the fact that the test-makers are reasonable people, you will be better able to assess the practicality of certain approaches, and you will be able to identify instances where an easier approach must exist.