Systems of Equations – SAT Algebra Tip 3

Some SAT Algebra questions will not only have one variable, but two or three for you to keep in mind. On occasion you’ll be given a set of equations, a “system” of equations, and asked to solve for one or all of the many variables in that system. You will also encounter questions like these that will give you a fifth option, choice E: This question cannot be solved with the information given. Remember this general rule of thumb: you must have the SAME NUMBER of equations as unknowns. If there are three variables and only two equations provided, chances are you will not be able to solve the problem. If there are two variables AND two equations given, chances are you will be able to solve it.

However, the SAT has a few tricks up its sleeve—here are two of the most common ones:

2x-y=14, and 4x-2y=28.

Two equations, two variables—should be simple, right? Let’s try our first tactic, substitution, and see what happens. Solve the first equation for y, and we get 2x-14=y. Fantastic—plug that expression in for y in the second equation. 4x-2(2x-14)=28. Distribute: 4x-4x+28=28. Reverse PEMDAS, and we’re left with… 0=0.

Well, while 0=0 is certainly true, it isn’t very helpful! Where was the trick? 2x-y=14 and 4x-2y=28 are the same equation. The second is simply the first multiplied by two. None of the relationships have changed between the variables and thus no new information is provided. This is really a two-variable, one-equation problem—it can’t be solved with the information provided.
Here’s a second trick, one that may make you give up before you even examine the numbers.

2m+z+5p=10, and 2m-z+3p=6. Find the value of p.

Shoot, only two equations, but three variables; it must be unsolvable… Let’s try another technique before we write this one off and move on. This problem requires a little bit of heavy lifting in the world of systems of equations: linear combination (combining multiple equations by either addition, subtraction, multiplication or division in order to cancel out variables). In this case, subtract the second equation from the first (setting up the problem vertically):

2m+z+5p=10
+ 2m-z+3p=6
0 + 0 +8p = 16 becomes 8p=16. Divide both sides by 8, and p=2.

Using linear combination, you were able to cancel out two variables, leaving you with 8p=16, a one-variable, one-equation problem!

QUICK TIP: Use the general rule of thumb that you must have the same number of equations as unknowns, but keep in mind the two ways the SAT can disguise these problems!