Simplifying Fractional Equations
Date: 4/4/96 at 16:52:37 From: RDIANNE GREEN Subject: Need help please 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 Thanks a lot in advance for your help. Hope to get your response soon, jodi
Date: 4/5/96 at 11:24:1 From: Doctor Aaron Subject: Re: Need help please 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 your equation as: (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
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