If you want to do well on the SAT Math test, there are certain topics (functions, right triangles involving Pythagorean Theorem, 30-60-90 and 45-45-90 triangles, problems that can be solved by plugging in numbers) that you simply must master. If you’re looking to score exceptionally well (think 700+), then you’ll also need to be very good at the types of problems that appear less frequently but often in the latter part of a section. These are often the problems that can be the difference between a * very good* score and a *great* score.

One type of problem that falls into this category is the systems of equation problem. You remember these from Algebra 1 (and probably again in Algebra 2) — multiple equations with more than one variable share a common solution that you are asked to find. A typical problem of this type that you would have seen in math class would go something like this:

Find the solution to the system 3*x* – 2*y* = 22 and 7*x* + 2*y* = 18.

In this particular problem you’d be happy to see that the two terms containing the variable *y *are opposites. When you add these two equations, the *y* terms cancel and it becomes pretty straightforward to find the value of *x*. Your work would look something like this:

3*x* – 2*y* = 22

7*x* + 2*y* = 18

———————

10*x* = 40

*x* = 4

From there you would substitute the 4 in for *x* in one of the equations and solve for *y,* which turns out to equal -5. So the solution to the system is the ordered pair (4, -5).

While you might see such a problem on the SAT, it’s more like that you’ll encounter a problem that requires you to be “clever” in order to find the solution …. or at least in order to solve the problem efficiently so that you have enough time to finish the rest of the section. Consider the following problem:

Given the two equations 10*a* – 5*b* = 17 and -2*a* + 13*b* = 15, what is the value of *a* + *b*?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 7

This problem looks different than what you’ve seen in math class. There you might be asked to find the value of *a* or the value of *b *or the values of both *a* and *b*, but you wouldn’t be asked to find their sum. When people tell you that the SAT is a *reasoning* test, this is the type of problem that they have in mind. Why? Because there’s a “clever” way to solve this problem that looks very different than what you were taught in math class. Go ahead and add the two equations ….. don’t worry, I know what you’re thinking: “We can’t do that — there are no opposites!” Trust me on this one.

10*a* – 5*b* = 17

-2*a* + 13*b* = 15

———————-

8*a* + 8*b* = 32

But wait, if 8*a* + 8*b* is equal to 32, can’t we just divide both sides of the equation by 8 to get what we want? Doing that shows us that *a* + *b *is equal to 4. The correct answer is Choice (B). Notice, by the way, that we never found the values of either *a* or *b*. Could we have solved the problem by finding each of these as you would have in Algebra 1? Sure, but the time we would have used to do that can now be put to better use answering more questions. The “clever” way takes much less time!

Here’s another systems problem that can be solved in this sort of “clever” way.

Jan bought 3 small candles and 5 large candles at the craft fair and spent a total of $42. Her friend, Sarah, bought 9 small candles and 7 large candles and spent a total of $78. What would be the total cost, in dollars, of 1 small candle and 1 large candle?

(A) 8

(B) 9

(C) 10

(D) 11

(E) 12

Our “clever” solution involves writing two equations, one for Jan’s purchase and one for Sarah’s purchase, and then seeing what happens when we add them:

3*x* + 5*y* = 42

9*x* + 7*y* = 78

________________

12*x* + 12*y* = 120

Now divide both sides of that equation by 12 and we see that the total cost of 1 small and 1 large candle would be $10. The correct answer is Choice (C).

A couple of final thoughts:

- This technique works just as well if there are more than two equations with more than two variables
- If
*adding*the two equations doesn’t work (you don’t get the same coefficient in front of each variable), then try*subtracting*the two equations.

For more practice on this type of problem, go here.

If you have questions about the System of Equations problems or anything else to do with the SAT, leave a comment or send me an email at info@cardinalec.com .