Learn to Solve Equations for the PSAT/NMSQT
An equation is a balancing act. Think of the equal sign as the support for a seesaw. When you work on an equation, you want to keep the sides of the seesaw balanced. So whatever you do to one side of the equation, you have to do to the other side also. Suppose you open Section 2 of the PSAT/NMSQT and see this question:
That equation may look big and nasty, but it isn’t. You know that the side to the right of the equal sign is 45, so the other side must also be 45. The left side of the equation is just written differently. You can solve this problem in a few ways:
Backsolving: Take one of the answers — Choice (C) is often a good place to start — and substitute that answer in the equation. So place 6 into the spot where
appears. Okay, 5 x 6 = 30. Add 5 and you have 35 — too little. Try something bigger, such as Choice (D). Now you have 5 x 8, which gives you 40. Add 5 and you have 45.
You’re there! The answer is Choice (D).
Plugging in: Suppose this problem is a grid-in, with no multiple-choice answers. No worries! Pick a likely number. You can’t go too high, because you have to multiply by 5 and then add 5. How about 4? No, 5 x 4 = 20, and when you add 5 more, you’re only at 25.
You need something bigger. How about 7? Now you have 5 x 7, which is 35, but adding 5 takes you only to 40. The next number up, 8, is what you want.
Isolate what you need to solve: In this question, you’re supposed to find the value of
So your goal is to isolate it on one side of the equation. Remember, you have to do the same thing to each side. First, subtract 5 from each side. Now you have
Now divide each side by 5, and you get
Notice that you never have to deal with the fact that the element you’re asked to find is a fraction. The test-makers like to throw some extra information at you to see whether you can focus on the real question.
On the PSAT/NMSQT, expect to see equations with absolute value, exponents, and square roots in them.
Toughen up your equation-solving muscles with these problems:
Solve for x. 2x2 + 6 =38
Solve for x. |x + 5| = 2x + 5
(A) –10/3 only
(B) 0 only
(C) 5 only
(D) 0 or –10/3
(E) 0 or 5
Solve the equation for
(A) 4 – 2x
(B) 4 – 6x
(D) 6 – 3x
(E) 12 – 6x
Now check your answers:
Don’t let the x2 make you nervous; just isolate it from the rest of the equation, and then worry about it. First, subtract 6 from both sides: 2x2 = 32. Then divide both sides by 2: x2 = 16. You can see that x needs to be 4 or –4 to become 16 when squared. Choice (A) is your answer. Remember, you can always choose to backsolve with questions like these!
B. 0 only
Backsolving is a great idea here. You have only three numbers to try! Plug in 0 first. That gives you 5 = 5, so you know that 0 must be a possible answer. You can eliminate Choices (A) and (C). Now plug in 5, and you get 10 = 15. Nope!
Eliminate Choice (E). If you plug in –10/3, you get 5/3 = –5/3, which is clearly wrong. Choice (B) is the answer.
Your first step should be to square both sides so that the square root goes away. After squaring, your equation is 81 = x – 3, so you add 3 to both sides, and voilà! Choice (E), 84, is your answer!
A. 4 – 2x
The goal here is to isolate
So first subtract 6x from each side of the equation. Now you have
Now divide each side by 3, which gives you Choice (A):