Today is the first in our series of blogs that will focus on different problem-solving strategies in mathematics. The strategy that we’ll look at now is simple enough at its core: if you’re stuck on a problem that involves unknowns or variables, try substituting, or “plugging”, some numbers into those letters! What numbers you plug in, what you’re plugging into, and how this helps you find the solution varies from problem to problem.

This strategy is especially useful on standardized tests that involve multiple-choice questions, like the SAT. Substituting specific numbers into unknowns or variables can not only give you the ideas you need to eliminate a few wrong answers, but can often steer you to the right answer right away!

Some SAT questions seem to demand that you solve complicated equations, like this one:

*Question*: What values of \(x\) satisfy the following equation?

$$\sqrt{5x+24}=x+2$$

A. 5 and \(-\)4

B. \(-\)5

C. 4

D. 5

There is a procedure to solve this equation: square both sides, simplify until you get a quadratic equation, factor that quadratic, find the possible solutions, and eliminate extraneous solutions. But what if you’re running out of time, and need an answer, *fast*? Well, given that there are answer choices here, do NOT solve the equation! Rather, just plug in your answer choices to see which ones satisfy that equation. Then you’ll have the solution at your fingertips, while avoiding unnecessary work.

Here, it’s helpful to remember that the square root function is *never* negative. So, if you plug in \(-\)4 or \(-\)5 for *x*, you see that the square root is equal to a negative number, which can’t happen. So that immediately eliminates answer choices A and B. Now try *x* \(=\) 4: \(\sqrt{44}\) is definitely not equal to 6, so C is eliminated as the answer. That leaves D as the only possible option, but check it to be sure: plugging in *x* \(=\) 5 gives us \(\sqrt{49}=7\), correct! So the correct answer is D.

We found the answer not by tediously solving the equation step-by-step, but by taking advantage of the potential answer choices available to us!

Sometimes the multiple-choice answers may look fairly complicated. Then you won’t be able to plug the answer choices into anything, but you may be able to plug into the answer choices themselves! To see what we’re talking about, let’s try another problem.

*Question: *A population of bacteria doubles in size every 3 hours. If the initial population is 500, which of the following functions gives the population after \(t\) hours?

A. \(500\times 2^{3t}\)

B. \(500\times 2^{\frac{t}{3}}\)

C. \(500 + 2^{3t}\)

D. \(3\times 500^{\frac{t}{2}}\)

Well, if you know your exponential models, you’ll be able to figure this out. But suppose you blank out on the test? Then what? Plugging in comes to the rescue!

The initial population is 500; that means that when the experiment starts, at time *t* \(=\) 0, we have 500 bacteria. So, if we plug in *t* \(=\) 0 into the correct answer above, we should get 500 back. Of course we don’t know what the correct answer is yet, so we try all of them! When plugging in *t* \(=\) 0 to choices A and B, we get 500, which means that A and B are *possible* answers. Answer choice C simplifies to 501, and D simplifies to 3, not what we’re looking for, so we can eliminate them immediately.

So which of A and B is correct? Keep using the given information. The bacteria double every 3 hours, so at time *t* \(=\) 3, we should have 2 \(\times\) 500 \(=\) 1000 bacteria. Plug in *t* \(=\) 3 to answer choices A and B; we’re looking for the expression that simplifies to 1000. Choice A fails (check it… it’s way too big!), and only B works out. The answer is B!

The moral here is that when we happen to forget higher level math (how to create exponential functions that model a given physical situation), we might be able to succeed by remembering some more basic math (evaluating functions), and by using our answer choices to our advantage.

Not all SAT questions are multiple choice. But even open-ended questions can be solved by some judicious substitutions. Try this next one.

*Question: *The following equation is true for all values of \(x\). Find the value of \(a\).

$$\frac{3x^2+2x-5}{x-a}=3x+23+\frac{156}{x-a}$$

This one seems pretty tough. You might begin by multiplying both sides by \(x-a\) to clear those fractions, and then see what happens. Doing that isn’t all that terrible, but students tend to be put off by ugly expressions like this. Is there another way to do this that might be a little easier, or at least make us feel like we’re in more comfortable territory?

Well, reread the question: this is true for *all* values of *x*. So, any value you plug in for *x* will give you a true expression. What’s your favorite value of *x*? Our favorite value is 0, because substituting 0 will often simplify things greatly! Let’s do just that:

$$\frac{3\times 0^2+2\times 0-5}{0-a}=3\times 0+23+\frac{156}{0-a}$$

$$\rightarrow \frac{5}{a}=23+\frac{156}{-a}$$Oh wow, now we can just solve for *a*! Start by multiplying both sides by *a.*

$$5=23a-156$$

$$\rightarrow 23a=161$$

$$\rightarrow a=7$$There was still a little bit of work to do after we plugged in 0, but that was simple algebra compared to what we might have been facing otherwise.

Of course, this technique may not be helpful for every problem. But when you have a few variables or unknowns bugging you in a problem, try plugging in some simple numbers. You might be able to eliminate wrong answers, or reduce the problem to something more manageable, and friendlier!

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