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Fractions and Lowest Terms
Date: 07/19/2001 at 20:34:42
From: Bethany Boruta
Subject: Algebra lowest terms
Problem:
2
6a 14b
__ . __
2
7b 2a
2
3 14b
___ . ___
7b a
How do you finish cancelling and then what do you multiply?
What is the lowest term?
Date: 07/20/2001 at 16:40:59
From: Doctor Ian
Subject: Re: Algebra lowest terms
Hi Bethany,
To see what's going on, sometimes it helps to expand the exponential
terms:
6a 14b^2 6 * a * 14 * b * b
-- * ----- = ------------------
7b 2a^2 7 * b * 2 * a * a
Now you can rearrange the factors to get
6 * 14 * a * b * b
= ------------------
2 * 7 * a * a * b
Now, every time you see something in both the numerator and
denominator, you can get rid of it, e.g.,
6 * 14 * a * b b
= --------------- * -
2 * 7 * a * a b
6 * 14 * a * b
= --------------- * 1
2 * 7 * a * a
because multiplying by something and then dividing by the same thing
has no effect. (More precisely, it has the same effect as multiplying
by 1.)
An easy way to do this is to line similar terms up and cross them off
in pairs:
x x
6 * 14 * a * b * b
= ----------------------
2 * 7 * a * a * b
x x
6 * 14 * b
= ----------
2 * 7 * a
What do you do about the numbers? Well, here's where all that stuff
you learned about prime factors becomes useful. If you break down the
numbers into their prime factors, you can cross off common factors in
the same way that we just crossed off common variables:
x x
(2 * 3) * (2 * 7) * b
= ---------------------
2 * 7 * a
x x
3 * 2 * b
= ---------
a
6b
= --
a
Now, this is kind of a pain, although when you're just starting to
learn a new technique, it's really helpful to go through all the
steps. But here's how I'd do it now:
6a 14b^2 <--- drop this exponent by 1
-- * -----
remove this b ---> 7b 2a^2
remove this a ---> 6a 14b^1 <--- drop this 1
-- * -----
7 2a^2 <--- drop this exponent by 1
6 14b <--- divide 14 by 7
- * ---
divide 7 by 7 ---> 7 2a
6 2b <---
- * -- Cancel the 2's
1 2a <---
6 b
- * - = 6b/a
1 a
Can you follow the same steps to simplify the second expression?
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
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