Simplifying Mixed Algebraic ExpressionsDate: 05/08/98 at 20:07:12 From: Erin Subject: Simplyfing mixed expressions(algebra) I have a test on Monday, and I can't get the answers to these problems. I need help! Please help me. x + y 2x - ----- = ? y 8 ------ - 6 = ? 3a - 1 1 (x - 4) - ----- = ? x - 2 x -- - (x + 2) = ? 2y 7 (a + 2) + ----- = ? a - 2 3 1 4 - ----- - ----- = ? y - 1 y + 1 I have a whole lot more, so please help me. Date: 05/12/98 at 14:42:15 From: Doctor Jeremiah Subject: Re: Simplyfing mixed expressions(algebra) To simplify an expression where part is divided by one value and part is divided by another value, you need to find common denominators to make both be divided by the same value. You know that you can modify a piece of an expression by multiplying it by 1: x + y y x + y 2x - ----- = (2x) * - - ----- = y y y See how we multiplied 2x by y/y (which is exactly 1)? Thus: x + y y x + y 2xy x + y 2xy - x - y 2x - ----- = (2x) * - - ----- = --- - ----- = ----------- y y y y y y We need to do the same thing with the next problem: 8 8 (3a - 1) ------ - 6 = -------- - 6 * -------- 3a - 1 (3a - 1) (3a - 1) 8 - 6(3a - 1) 9 - 18a - 6 3 - 18a = ------------- = ----------- = ------- 3a - 1 3a - 1 3a - 1 It's always the same. All you need to do is to match the denominator for all the pieces. The last one you give is the most interesting one: 3 1 4 - ----- - ----- y - 1 y + 1 Here, we have three parts. Obviously, getting the denominator to be (y + 1) or even (y - 1) is not good enough. The only way to get them all to match is to use the product of all the denominators: (y - 1)(y + 1). Thus: 3 1 4(y - 1)(y + 1) 3(y + 1) 1(y - 1) 4 - ----- - ----- = --------------- - -------------- - -------------- y - 1 y + 1 (y - 1)(y + 1) (y - 1)(y + 1) (y - 1)(y + 1) 4(y - 1)(y + 1) - 3(y + 1) - (y - 1) 4y^2 - 4y - 8 = ------------------------------------ = ------------- (y - 1)(y + 1) y^2 - 1 Does that make any sense? -Doctor Jeremiah, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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