# SAT Tip of the Week: How to Solve 3 Types of Algebra Problems

Algebra problems on the SAT may take several forms, but they generally involve some variation on solving for a variable or expression.   In this article, I will break down the process of working through a few variations of those lovely “solve for x” or “what is the value of x?” types of problems.

1.     The standard “what is the value of x” problem:

This kind of problem will generally involve an equation involving the variable whose value you have to find.

Let’s say that you’re given the equation 5x + 14 = 2x +47.

To find the value of x, you have to solve for x in this equation.  Do everything you need to do to isolate x on one side of the equation (making sure to always apply the same operations to both sides of the equation) and get everything else on the other side of the equation.

In the equation 5x + 14 = 2x + 47 , you’ll want to get all your x’s on one side. You can do this by subtracting 5x or 2x from both sides; I’d recommend subtracting the 2x from both sides so you don’t have to deal with negatives.  Then you’ll have 3x + 14 = 47.  Next, you’ll want to subtract that 14 from both sides so that you can have x’s on the left side and numbers on the right side.  Then, you’ll have 3x = 33. Finally, to isolate x, you’ll need to divide both sides by 3, so you end up with x = 11.

CAREFUL for potential SAT trickiness in these kinds of problems. The SAT loves to ask you for the value of 2x or x squared or something along those lines, when your impulse is simply to solve for x.  Be sure to read the question and be sure you’re actually answering it before moving on!

2.     The “in terms of” problem:

This kind of problem will generally involve an equation that contains multiple variables and/or constants and will ask you to solve for one of those variables in terms of the others.

Let’s say you’re given the equation 3A + 5B = 6C + 3 , and you are asked to solve for B in terms of A and C.

Solving for a variable in terms of other variables/constants entails the same process as simply solving for x. It’s a little bit strange because there are multiple variables involved, but don’t be intimidated: the process is the same. You simply need to do whatever you have to do to isolate the variable that you’re solving for in terms of the other variables. In this case, for example, you need to try to isolate B.

To do this, in this case I would subtract 3A from both sides of the equation first. That leaves us with 5B = 6C +3. Then to isolate B, simply divide both sides by 5 and you will get B = (6C + 3) / 5.  Problem solved!

As above, do make sure to be CAREFUL of potential trickiness, and be sure that you’re solving for the variable that the SAT is asking you to solve for.

3.     The “value of an expression” question:

These are the somewhat impossible-seeming questions that give you several equations and then ask for the value of another expression involving some combination of those variables.  These are often trick-based questions and thus can be flummoxing at first but very solvable once you are familiar with them.

Two favorite SAT tricks for these questions: if in doubt, try to multiply (or divide) the given equations by one another, or stack them and add them to one another, and see where that gets you.  Most likely, one of those will get you where you need to be (or at least very close).

Example:  xy = 5 and x+y =7 .  What is the value of  x²y + xy² ?

First observation: there is no easy way to solve this using standard algebra.

Second observation: the two given equations involve x and y, and the question is asking about an expression involving x² and y² .  The only way to get from one to the other would be to multiply the equations together (or to square one of them – but if two are given, it’s more likely that the SAT wants us to multiply them together).

Multiply the two equations together and we have (xy)(x+y) = x²y + xy²  — which is exactly the expression we’re looking for!

If xy = 5 and x+y = 7, and (xy)(x+y) = x²y + xy², then x²y + xy² must equal (5)(7) = 35.  Problem solved!

Example #2: x – y = y – z = z + x = 20. What is the value of x?

First observation: there is no easy way to solve this using standard algebra.

Second observation: there are a few equations here, involving addition of variables. This kind of situation suggests that it might be worth writing out and stacking the equations and then adding them together to see what happens.

Now we have:

x – y = 20

y – z = 20

z + x = 20

If we add them together, we have x – y + y – z + z +x = 20 + 20 + 20

In other words, once we cancel out those y’s and z’s, 2x = 60, so x = 30.  Problem solved!

Now you have a good foundation for a variety of the most common SAT algebra problems. Go forth and practice using SAT problems – and tune in next time for some wise words on functions!

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Alice Rothman-Hicks is a Veritas Prep SAT 2400 instructor. Since graduating from Columbia University (Magna Cum Laude, Phi Beta Kappa), Alice has been teaching and tutoring test prep, helping students achieve their own academic successes. She scored a 2350 on the SAT.