Algebra: Common DenominatorsDate: 5/14/96 at 15:43:56 From: RDIANNE GREEN Subject: Algebra help needed, please :) Hi. I am having problems working out the following two problems because I don't know how to get all of them to have equal denominators. 1. 2 a + 1 4 ------- + ---------- = ----- a - 3 a^2 - 9 a + 3 2. x 1 ----- = ----- x + 2 x Please help. Thanks in advance. :) Date: 5/14/96 at 21:12:59 From: Doctor Jodi Subject: Re: Algebra help needed, please :) Hi there! You may remember that one common denominator (though not always the lowest) is found by multiplying the two (or more) denominators together. For example, for the fractions 2/7 and 1/3, you can multiply 7 * 3 to get 21, one common denominator. For your first problem, a^2 - 9 will be the lowest common denominator. [I know this because of the "difference of squares" factoring rule. Since a^2 - 9 is the difference of squares, we can factor it as (a - 3)(a+3)] So for the first problem, we will need to multiply fractions with the denominator of a-3 by a+3, (which equals 1) and fractions with the --- a+3 denominator of a+3 by a-3 (also equal to 1). Does this make sense? --- a-3 2 a + 3 = 2(a+3) = 2a+6 -- * ------ ------ ----- a-3 a + 3 a^2 -9 a^2 -9 We can leave a+1 as it is for the moment. --- a^2 -9 We will need to multiply 4 a-3 4 (a-3) 4a -12 - * --- to get ------- = ------ a+3 a-3 a^2 -9 a^2 -9 So now the problem reads 2a +6 +a+1 = 4a -12 ----------- ------ a^2 -9 a^2 -9 Does that look better? You can now multiply through by a^2 -9 to remove the denominators if you like and continue solving this like any other algebra problem. Are you comfortable with the rest of the problem? If not, write us back and we'll give you a hand. Here's just a clue for the second problem. Again, if you want more help, let us know. For the second problem, the common denominator is x(x+2) (aka x^2 + 2x) Good luck and thanks for your question! -Doctor Jodi, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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