MommyThrice Posted April 22, 2010 Share Posted April 22, 2010 Saxon Alg. I, Lesson #117. The question was... Add: (x/x * y^-2)-(3/y^3 * x^2) - (2/x+y) The answer says: x^3 * y^5 + x^2 * y^6 - 3x - 3y - 2x^2 * y^3/x^2y^3(x+y) We were able to get this answer, too, but my ds originally used "1" as a common denominator to get... y^2 - 3y^-3 * x^-2 - 2(x^-1 + y^-1) Is this the same answer? Would this be correct? Quote Link to comment Share on other sites More sharing options...
Teachin'Mine Posted April 22, 2010 Share Posted April 22, 2010 I can't find this problem in lesson 117. What edition are you using, and what problem or practice number is it? It's harder to understand when it's typed in a post. :) Quote Link to comment Share on other sites More sharing options...
MommyThrice Posted April 22, 2010 Author Share Posted April 22, 2010 I'm not very good at typing algebra problems. We're using 2nd edition - problem #27. Quote Link to comment Share on other sites More sharing options...
Teachin'Mine Posted April 22, 2010 Share Posted April 22, 2010 I'm sorry - we have the 3rd edition and the problems are different - there's an a, b and c problem. Quote Link to comment Share on other sites More sharing options...
forty-two Posted April 22, 2010 Share Posted April 22, 2010 Saxon Alg. I, Lesson #117. The question was... Add: (x/x * y^-2)-(3/y^3 * x^2) - (2/x+y) The answer says: x^3 * y^5 + x^2 * y^6 - 3x - 3y - 2x^2 * y^3/x^2y^3(x+y) We were able to get this answer, too, but my ds originally used "1" as a common denominator to get... y^2 - 3y^-3 * x^-2 - 2(x^-1 + y^-1) Is this the same answer? Would this be correct? I'm unclear about what is/isn't part of the denominators: By (x/x * y^-2)-(3/y^3 * x^2) - (2/x+y), do you mean x/[x*y^(-2)] - 3/[(y^3)*(x^2)] - 2/(x+y) OR (x/x)*y^(-2) - (3/y^3)*(x^2) - (2/x) + y And by x^3 * y^5 + x^2 * y^6 - 3x - 3y - 2x^2 * y^3/x^2y^3(x+y), do you mean (x^3)*(y^5) + (x^2)*(y^6) - (3x) - (3y) - (2*x^2)*(y^3)/[(x^2)(y^3)(x+y)] or something else? Quote Link to comment Share on other sites More sharing options...
AngieW in Texas Posted April 22, 2010 Share Posted April 22, 2010 (edited) For the answer in the answer key to be correct, the problem must be: x/[x(y^-2)] - 3/(y^3x^2) - 2/(x+y) If you divide all of that out into separate fractions, you get: (xy^2+y^3)/(x+y) - 3/[xy^3(x+y)] - 3/[x^2y^2(x+y)] - 2/(x+y) Edited April 22, 2010 by AngieW in Texas Quote Link to comment Share on other sites More sharing options...
MommyThrice Posted April 22, 2010 Author Share Posted April 22, 2010 For the answer in the answer key to be correct, the problem must be:x/[x(y^-2)] - 3/(y^3x^2) - 2/(x+y) If you divide all of that out into separate fractions, you get: (xy^2+y^3)/(x+y) - 3/[xy^3(x+y)] - 3/[x^2y^2(x+y)] - 2/(x+y) Yes, that is the correct problem. And I just broke the answer into seperate fractions and got the same answer as you did. So, his solution to using "1" as a common denominator cannot be correct. Thanks for the help. Sure would be nice if I could just DRAW an equation here! Quote Link to comment Share on other sites More sharing options...
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