Ever see those weird symbol questions on the SAT? Something to the tune of “if xΩ = 2x-1, then what is the value of 3Ω?” You freeze and think to yourself—we haven’t covered horseshoes yet in precalc!

1) The “weird symbols” questions are really functions in disguise. Whenever you see a weird symbol, its corresponding relationship (here, 2x-1) will ALWAYS be defined immediately after. So don’t worry if you haven’t covered horseshoes yet. In fact, most of the time you can substitute f(x) for the weird symbol. The question above could be rewritten as, “if f(x)=2x-1, what is f(3)?” So take these questions and treat them exactly like functions. Just the same as a function could be called “f(x), g(x) or h(x),” why couldn’t it be called xΩ?

2) The horseshoe, spade, weird brackets, smiley face, triangle and whatever other strange symbols that show up on the exam have no actual defined mathematical meaning. No theoretical mathematician knows what xΩ is either! That’s why it will always be defined for you.

3) Sometimes the SAT is testing your ability to draw visual parallels . For instance, take the following question:

Because there is no triangle in real mathematics, only in SAT land, the SAT wants to see if you reason—“hmm, in the example, the lower left-hand number (a) was multiplied by the top number (b) and then the lower right-hand number (c) gets subtracted from that product. Maybe I should do the same thing to the numbers in the question! In which case, (7(3)-2) = 21-2 = 19.

QUICK TIP: When you see strange symbols, don’t freak out, treat them as functions, and be on the lookout for visual parallels in how the variables are spatially laid out.

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!

When preparing for Algebra questions on the SAT, remember that ETS hates negative powers.  Whenever you see something like x^-2 in an expression, know that you must get rid of the negative exponent in order to solve the problem.  However, you can’t just delete sections of a question to suit your fancy!  In order to “get rid” of something in an expression, you must either simplify, or change the format of that piece.  Remember, negative exponents can be re-written as positive exponents by putting them in the denominator.  X^-2 becomes 1/x^2.  Voila, we have turned our negative power into a positive one.  This rule goes for all negative exponents: x^-n = 1/x^n.  We haven’t really changed the overall quantitative value of the expression, just its format.

Converting negative exponents will quickly set you up to start simplifying.  For example:

(x^2)(x^-4)=
(A)    1/x^8
(B)    1/x^2
(C)    X^8
(D)    –(x^2)
(E)    -4x^2

Let’s try a combination of converting negative powers and simplifying:  (x^2) (x^-4) =  (x^2) x (1/x^4), becoming (x^2)/(x^4).  Two exponents dividing each other with the same base but different powers?  That’s your clue to subtract the powers and simplify the expression to x^-2.  Go one more round of converting negative powers, and we’re left with 1/x^2, choice B.

Alternatively, you can also recognize that the initial expression (x^2)(x^-4) has two exponents being multiplied with the same base and different powers—in this case you add the powers, getting x^-2.  Triumphant, you look at the answer choices, only to find no options with a negative power!  Again, convert it to a positive one, 1/x^2, and choice B becomes apparent.

QUICK TIP: Always convert negative exponents to positive ones.

Algebra is a core subject area of the SAT math section. Master the basics and you will be well positioned for the test.  The first rule of algebra is to ALWAYS BALANCE the equation, meaning “whatever-you-do-to-one-side,” do to the other!  It is important though to not only perform the same operation to both sides of an equation, but to every term as well.

Here’s an example:  3x-7=15.  Simple, right?

Just divide both sides by 3 in order to isolate the variable (get the x on its own), and you get x-7=5.  Perform reverse PEMDAS, add 7 and we find that x=12. Ta-da, done.

But we forgot that we must divide every single TERM by 3, not just both sides of the equation.  So we really end up with x- (7/3) = 5.  Now we’ve made ourselves a fraction, something even more complicated than before, and answer is 7 1/3, or 22/3 –something very different from x=12.

In this particular example, it’d be simplest to start with reverse PEMDAS and add 7, getting 3x=22, and then divide by 3.  But remember the rule of thumb, if you’re manipulating an equation, change every term, not just one term on both sides of the equals sign.

QUICK TIP: One way to avoid careless errors and keep track of your work is to work vertically, and mark ALL YOUR WORK.  Re-copy the entire equation at each step.