Pre-Algebra: Equations and Other TipsDate: 08/30/97 at 11:13:13 From: Erik Sull Subject: Math Help This year I am in Pre Algebra. I'm in 8th Grade; can you give me tips on equations and other things that will be useful for me to know? Date: 08/31/97 at 15:12:31 From: Doctor Guy Subject: Re: Math Help I'll try to give you a few tips about equations. Remember that mathematics is, for most people, a way of understanding the world better and solving real problems that come up. To me, the essence of algebra is translating what appear to be messy problems involving, perhaps, entire paragraphs of explanations and/or diagrams, into short, easily-understood symbols (equations). These equations can be manipulated, using fairly simple rules, allowing you to solve them and to come up with answers to problems that would be otherwise extremely difficult to answer. This is why, as a society, we want just about everybody to have some understanding of algebra at some level. Of course, it is also necessary that you understand the rules of arithmetic involving whole numbers, decimals, fractions, and percents. I'm sure that your pre-algebra class will involve that as well. I do not know how well you perform in arithmetic, but if you have problems with that, then be sure to pay attention to the explanations that your teacher and your textbook give in that area. It may help if I give an example of a relatively easy problem, to show how it can be handled via algebra. Here is the problem: Alfeda Baxter wants to rent a car in order to visit some business clients in the San Francisco Bay Area next week when she flies in for a week-long business trip. She calls up two auto-rental agencies and gets price quotes from each. Dependable Rent-a-Car will rent her a mid-sized sedan for $143.00 for the week, and 13 cents per mile driven; she has to pay for her own gasoline. Reliable Automobile Rentals will rent her the same sized car for $129.00 for the week, and charges 28 cents per mile driven. They are equally convenient companies about which she has heard equally good things. Ms. Baxter is not quite sure how many miles she is going to drive during the week, but it might be a lot. Since all of these transportation costs are going to come out of her own pocket, she wants to hold the cost as low as possible. The question is, at how many miles does the car from Reliable Auto Rentals become as expensive as the one from Dependable Rent-a-Car? Based on that, Ms. Baxter can make a decision on which company to rent from. That's a pretty long problem, but it can be translated into symbols pretty easily. The first step is to ask, "what is the question?" The question is, how many miles would she have to drive so that the RAR car would cost the same as the DRAC car. Since the question involves miles, we may as well now do one of the next important steps, choosing a variable. Let's use M for miles. Now let's look back at the information about RAR: the cost there is $129.00 plus 28 cents per mile, or $0.28 times the number of miles. We can translate that into an expression, [Cost at RAR, in dollars] = 129.00 + 0.28 * M Note that I use "*" to mean multiplication. I could also have writtten 129 + .28M (notice that I dropped unnecessary zeroes and omitted the times sign; that is legal--it still means 28 cents times M). Now at DRAC, the cost is $143.00 plus 13 cents per mile, or [Cost at DRAC, in dollars]=143.00 + 0.13 * M. Remember, the question was, at how many miles are these two costs equal? Well, to find that out, all we need to do is to write an equation that states that they ARE equal, and then we use some rules of algebra (which I will explain as we go along) to find out what M is, i.e. the number of miles. Here goes: [Cost at RAR, in dollars] = [Cost at DRAC, in dollars] Now we substitute equal things for equal things (important rule) and get 129 + .28 * M = 143 + .13 * M The next general idea is to get M on one side of the equation and all the numbers on the other, by multiplying, dividing, subtracting, and adding the same thing to both sides. The main rule to remember is that you have to do the same thing to both sides of the equation. I think I want to get M on the left-hand side of the equation, and get everything else on the right. First I will subtract 129 from both sides. 129 - 129 + .28*M = 143 - 129 + .13 * M which simplifies to 0 + .28 * M = 14 + .13 * M When you add 0 to anything, it stays the same: .28 * M = 14 + .13 * M. Another important rule you need to know: Multiplication and division come before addition and subtraction when evaluating an expression. Since we do not know what M is, we cannot simply add the 14 and the .13*M in the previous line to get 14.13*M. I want to get all the M's on the lefthand side of the equation, so we CAN subtract .13*M from both sides of the equation: .28 * M - .13 * M = 14 + .13 * M - .13 * M We can now use the distributive property (have you learned it?) and get (.28-.13)*M = 14 + 0 or .15 * M = 14 Now we want to find M, so we can divide both sides by .15 .15 * M / .15 = 14/.15 On the left, .15/.15 equals 1 and 1 times M is M; so M = 14 / .15, and I will now use a calculator to find 14/.15, or M = 93.33333333 miles. What does this mean? It means that if Ms. Baxter drives 93 and 1/3 miles that week, then the two car companies' prices are the same. If she drives less than that, then Reliable will be cheaper. If she drives more than that, then Dependable will be cheaper. So all she has to do is to estimate how many miles she thinks she will drive, and she can take her pick. Now you may not have understood each step; I went rather fast. Perhaps you see why we take several months teaching this stuff to students-- there is a lot to learn. Good luck this year, and if you have a specific question you need help with, don't hesitate to e-mail us. You may also be able to find specific answers to questions in our archives. -Doctor Guy, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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