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### Simplifying Fractional Equations

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Date: 4/4/96 at 16:52:37
From:  RDIANNE GREEN

Dear Doctor Math,

I am home schooled, and I have been unable to answer the following
algebra questions. I am hoping you can help give me insight into
these problems:

I am supposed to perform the indicated operations and express
results in lowest terms.

1.  x^2 - 2xy + y^2          3x^2y^2
__________________   *  ____________   =   ?

6xy               x^2 - y^2

2.    5           4x         2 - x
________  -  _______  -  ________  =  ?

x - 1        x - 1       x - 1

Hope to get your response soon,

jodi
```

```
Date: 4/5/96 at 11:24:1
From: Doctor Aaron

Hi Jodi,

Working with fractions of algebraic expressions is just like
working with fractions of integers.  One thing we have to do is
find a common denominator.  Since you were asked to express the
result in lowest terms, we also have to worry about that.

In problem 1 you have a product of two expressions.  If it were

5   6
- * -
3  15

you'd want to do some canceling, but with algebraic expressions
it's a little harder to see what cancels with what and what
will be left over.

With algebraic expressions we just have to define what we mean by
a factor.

Well 6xy isn't too bad - it's just 6*x*y so the factors are 6, x
and y.

Now we have to figure out x^2-y^2.  Well, we can rewrite this as
x^2 - xy + xy - y^2, which may seem silly at first, but it will be
useful.  Then we can factor out an x from the first two pieces to
rewrite it again as:

x*(x-y) + xy - y^2

We can do the same for y and the last two pieces to get
x*(x-y) + y*(x-y), which we can rewrite as:

(x+y)*(x-y).

Then the factors, or multiples, are (x+y) and (x-y)

This is the process known as factoring.  We have broken up a
sum of variables to the second power to a product of sums of
variables to the first power.

Try to work through x^2 - 2xy - y^2 by yourself.  You should get
(x-y)*(x-y).

Once you believe that x^2-2xy-y^2 = (x-y)*(x-y), we can rewrite

(x-y)*(x-y)      3*x^2*y^2
----------  *  -----------
6*x*y        (x+y)*(x-y)

we can cancel an x-y to get

(x-y)   3*x^2*y^2
----- * ---------
6*x*y      (x+y)

but we can still simplify because we can write x^2 as x*x and y^2
as y*y and then cancel to get:

(x-y)    3*x*y                                      (x-y)*x*y
----- *  ----  but 3*2 = 6 so we can rewrite again: ---------
6       (x+y)                                       2*(x+y)

The second problem is a little different.  Instead of a product,
you have to calculate a difference.  Luckily you already have a
common denominator, x-1.  You can multiply both sides of the
equation by x-1 just as you would multiply by 3 if you saw:

4   +    5   -   x     22
-        -       -  =  -
3        1       3     3

After you multiply both sides by x-1, the denominators will cancel
and you'll be left with 5 - 4x = 2 - x.

Good luck.

-Doctor Aaron,  The Math Forum

```
Associated Topics:
Middle School Fractions

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