Pre-Algebra: Equations and Other Tips

Date: 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/