Fractions and Lowest TermsDate: 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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