Equations Involving Fractions

Date: 07/07/2000 at 14:42:35
From: Vivian Lee
Subject: Equations without guess and check

I know how to do equations, but right now it's getting very hard 
because we have to solve problems with fractions and decimals, such as 
17=7-(5/6)t or "If 2x-3=7, evaluate 3x+4."

Please help - I'm really lost.

Date: 07/07/2000 at 23:03:24
From: Doctor Peterson
Subject: Re: Equations without guess and check

Hi, Vivian.

If you really know how to solve the equations with integers, then the 
only real problem when fractions are introduced is a greater chance of 
making mistakes. I have two opposite ways to deal with that. One is to 
try to eliminate fractions as soon as possible, so you don't have to 
worry about them; the other is to keep them as long as possible so you 
can do only algebra at first, then do all the arithmetic at once when 
your mind isn't cluttered with algebra. I'll demonstrate on your first 
problem. (The second doesn't involve fractions, so if there's a 
different issue there, write back and tell me what trouble you had 
with it.)

First, let's eliminate fractions early. We can do that by multiplying 
by the denominator, 6:

         17 = 7 - (5/6) t
     17 * 6 = (7 - (5/6) t) * 6
     17 * 6 = 7 * 6 - (5/6) t * 6
        102 = 42 - 5 t

Do you see what I did? On the right I had to multiply the whole 
expression by 6, so I first put it in parentheses, and then used the 
distributive property so that I multiplied each term by 6. (In reality 
I would do it in one step, but you probably aren't used to it yet.)

Now the equation looks simpler, and you should be able to solve it.

If there were two fractions in the problem, you could do this twice, 
once with each denominator.

The other way is basically to forget that the numbers are fractions 
until the end; just treat each fraction as a number and don't try to 
do anything to it yet. If the 5/6 were replaced by, say, 3, we would 
normally do this:

     17 = 7 - 3t       subtract 7
     17 - 7 = -3t      simplify
     10 = -3t          divide by -3
     -10/3 = t

Notice that I calculated 17 - 7 in the middle. I'm going to do 
absolutely no calculations this time, as I would if the 7 were a 
fraction, just so you can see the whole technique:

     17 = 7 - 3t       subtract 7
     17 - 7 = -3t      divide by -3
     (17 - 7)/(-3) = t

Now all the arithmetic is left to the end; I subtract to get 10, then 
divide to get -3 1/3. (This is also what you would have to do if the 
numbers were variables, so this is good practice.)

Now let's do it with the fraction in place:

     17 = 7 - 5/6 t    subtract 7
     17 - 7 = -5/6 t   divide by -5/6
     (17 - 7)/(-5/6) = t

Now again we have to do the arithmetic last. Subtract to get 10, then 
divide by -5/6 (that is, multiply by -6/5, which I could have done on 
that second step) to get the answer.

You probably won't find this second way terribly easy, but if you are 
inclined to make arithmetic errors, this can save some confusion while 
you do the algebra. The point is, fractions are just numbers, so the 
algebra is identical; only the arithmetic is harder.

- Doctor Peterson, The Math Forum
  http://www.mathforum.org/dr.math/