Multiplying and Dividing FractionsDate: 8/18/96 at 8:31:12 From: D.S.K. Subject: Multiplying and Dividing Fractions Hi Dr. Math! I am a first year algebra student and I was hoping you could help me with a few problems. The instructions are to perform the indicated operations and express the results in lowest terms. Here is the first one: x + 6 x ----- * ------- 3xy 2x + 12 Is this the right answer? 2 (x + 6)x x + 6x x(x + 6) 1 ----------- = ------------ = ----------- = ---- 3xy(2x + 12) 2 6xy(x + 6) 6y 6x y + 36xy _________________________________________________ Here is another problem: 1 9x ----- / ------ 2 x + 3 x - 9 Here is how I tried to do it: 1 x + 3 1 x + 3 ------ * ------- = --------------- * ------- = 2 9x (x - 3)(x + 3) 9x x - 9 1(x + 3) 1 --------------- = --------- . Is this correct? 9x(x + 3)(x - 3) 9x(x - 3) _________________________________________________ How about this one? 2 2 2 2 x - 2xy + y / x - y ----------- --------- 6xy 2 2 3x y Is this how to do it? 2 2 2 2 2 2 2 x - 2xy + y 3x y (x - y) 3x y ----------- * -------- = ------- * ------------- = 6xy 2 2 6xy (x + y)(x - y) x - y 2 2 2 3x y (x - y) xy(x - y) ----------------- = --------- 6xy(x - y)(x + y) 2(x + y) _________________________________________________ I would really appreciate it if you could help me. Algebra can be kind of confusing sometimes. D.S.K. Date: 8/30/96 at 15:40:46 From: Doctor James Subject: Re: Multiplying and Dividing Fractions All your answers are right! The first one you could have also done this way, which saves a little bit of work: x + 6 x ----- * ------- = ( [x + 6]/3xy) * ( x/[2x + 12] ) 3xy 2x + 12 = ( [x + 6]/3xy) * ( x/2*[x + 6] ) = ( [x + 6]*x )/(2*3xy*[x + 6]) = x/6xy = 1/6y, which gives you the same answer, but saves you from having to multiply so much. However, your answer was still just as good. See, algebra isn't that hard! They can make it bigger and more messy, but it doesn't get much harder than that. -Doctor James, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/