# Thinking Algebraically, Part 1

Algebra is a tool. Practice helps us learn how to use it. Here are some ways to practice Algebra in ordinary activities.

"How much is this with the discount?"

You are shopping at Penney's. The sign on a rack says, "St. Johns Bay Tops 40% Off Marked Price." Your sister looks at a top and asks, "How much is this with the discount?" You want a way to figure the sale price no matter the marked (original) price or the discount rate.

What do you want to find? The sale price. Let's label that s.

What is the marked price? Let's label that p.

What is the discount? Let's label that d.

So now, how do we use the discount to find the sale price from the marked price?
s = ?
s = p - d

We have to remember that the discount rate is a percentage of p. Let's label that r. We also want to simplify the arithmetic, make it easier if we are figuring this in our heads in the store. We know that p equals 100% of itself. And we are subtracting a smaller percentage, the discount rate, which is r. So what is left is (100 - r)%, or (100 - r)% of p, which is p(100 - r).

We know that x% of n = xn/100, for instance, 20% of \$50 = 20*\$50/100 = 2*\$50/10 = \$100/10 = \$10.

So s = p(100 - r)/100.

Now back to Penney's. You ask your sister, "What is the marked price?" She says, "\$34.99." We want to make this easier, so first we add one cent to round the price to \$35. We'll take that cent back off at the end. And we're taking 40% off.

p = 35
r = 40
s = p(100 - r)/100 = \$35(100 - 40)/100 = \$35*60/100 = \$35*6/10 = \$210/10 = \$21.
Now we take the added cent off: \$21.00 - \$0.01 = \$20.99.

You answer your sister, "That will be twenty dollars and ninety-nine cents."

More to follow

I really like teaching a quick, cheap way to do this problem that doesn't get caught up in variables, substitutions, and formula, which we both know can be quite daunting. The basic idea is realizing that 40% is 4 times 10%. 10% can be found with no "real" math, and multiplying be 4 is a cinch. After that it's just a matter of adjusting the original number appropriately. 10% of \$34.99 is \$3.499, or 3.50 if we simplify, which can be gotten simply by moving the decimal. Since we're looking for 40%, then we just have to take whatever 10% was, \$3.50, and multiply it by 4. So we'd have 3.50 x 4 = 14. So... 40% of \$34.99 is \$14, and since we're reducing it by that much for a 40% discount, \$35 - \$14 = \$21. Generally after I teach this way I'll go back and explain why the formulas work to keep everybody happy. The big advantage to teaching this is students can then apply it without pulling out a calculator in the middle of a mall--maybe even without realizing it. I also like it since it helps ensure the concepts of what a percent actually is and makes for a good tip-off whether or not the answer they end up with even makes sense. All the best.
Eve R: Thank you for your comment. I think discussion really helps us to think about a subject. I understand what you are saying. That is a quick way to find the answer. But finding the quick answer apart from Algebra was not my immediate concern. The thrust of this article was to show that an everyday problem can be expressed algebraically. The point is to get a student to see the variables (letters) as labels for real things. Then after practicing this, it can be applied to more complex problems. Personally, when I'm figuring the sale price, I first subtract the discount rate from 100% (this does not always end in zeroes). Then I multiply the marked price by the remaining percentage in decimal format.

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