Date: 25 Apr 1995 17:23:39 -0400 From: Sandip Mody Subject: Fractions and Fractional Equations Hello, I was having trouble with the following problems: 4/x-3 + 6/x+3 = 2/3 and solving for x F= w/x+1 The other problem was a^2 + b^2/a^2-b^2 / a-b/a+b - a+b/a-b Thank You for your help.
Date: 26 Apr 1995 11:15:09 -0400 From: Dr. Ethan Subject: Re: Fractions and Fractional Equations Hello, Good questions. I will assume that you know how to work with fractions when there are just numbers. If so, then this will be easy because here we do the same thing. What I will do is work the first and last problem; then I will leave the second problem for you to think about. Okay? Great. The first problem is: 4/x-3 + 6/x+3 = 2/3 Well the thing that we need to do is find a common denominator for the first two fractions. That will be (x-3)(x+3), so we need to multiply the first fraction by (x+3)/(x+3) and the second by (x-3)/(x-3). This will not change the value of the equation because both of these reduce to one. Now we have: 4(x+3)/(x+3)(x-3) + 6(x-3)/(x+3)(x-3) = 2/3 which becomes via addition: (4(x+3) + 6(x-3))/(x^2 - 9) = 2/3 Now, we multiply both sides by (x^2 - 9) to get: 4(x+3) + 6(x-3) = 2(x^2 - 9)/3 Now simplify the left hand side and we have: 10x - 6 = 2(x^2 - 9)/3 Now multiply both sides by 3 and we have: 30x - 18 = 2x^2 - 18 the 18's cancel and we are left with: 30x = 2x^2 Divide by two on both sides: 15x = x^2 So both 0 and 15 are solutions to this problem. Yea! Now to your last problem. Here it is. I am adding () where I think that you meant to put them. If I am wrong then write back and I will rework the problem. ((a^2 + b^2)/(a^2 - b^2))/((a-b)/(a+b) - (a+b)/(a-b)) Let's work with the bottom first. Since we have subtraction we need to find a common denominator. Again it will be (a+b)(a-b), so the first term on the bottom needs to be multiplied by (a-b)/(a-b) and the second by (a+b)/(a+b) so it looks like this: ((a^2 + b^2)/(a^2 - b^2))/((a-b)(a-b)/(a-b)(a+b) - (a+b)(a+b)/(a+b)(a-b)) this adds together to be: ((a^2 + b^2)/(a^2 - b^2))/(((a-b)^2 - (a+b)^2)/a^2 - b^2 Now square the subtracted terms and combine them and we have: ((a^2 + b^2)/(a^2 - b^2))/(-4ab/(a^2-b^2)) Now we can consider the whole fraction, and we can do the old invert and multiply trick for dividing fractions so the division problem above looks like this now: (a^2 + b^2) a^2 - b^2 ------------- * ----------- a^2 - b^2 -4ab Well the a^2-b^2 terms cancel and we are left with: -(a^2 + b^2)/4ab as our answer. I hope that I haven't made a careless mistake. I have tried to explain each step. If you need more explanation or you think that I have made a mistake, then you can write back. Also, if you need help on that other problem or have other questions please write to Dr. Math. Ethan, Doctor On Call
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum