## Search found 4 matches

Mon Sep 12, 2011 6:57 am
Forum: Beginning Algebra
Topic: Could Someone Explain the Intermediary Steps?
Replies: 3
Views: 1761

### Re: Could Someone Explain the Intermediary Steps?

Ok, I now understand their simplification. (x^-1 – y^-1) / (x-y)^-1 = (1/x - 1/y) / (1/x-y) = (y-x/xy) / (1/x-y) = (y-x/xy)(x-y/1) ... But then wouldn't you continue, and go: = 2xy - y^2 - x^2 / xy = -y^2 - x^2 +2 ? My book stops at the ellipse. Why? What makes that simplification better? Oh, and th...
Sun Sep 11, 2011 11:16 pm
Forum: Beginning Algebra
Topic: Could Someone Explain the Intermediary Steps?
Replies: 3
Views: 1761

### Could Someone Explain the Intermediary Steps?

Ok, so this review packet I am working from does not fully explain their answers, and I am having difficulty following some of them. Any help would be greatly appreciated. Here is the expression: (x^-1 – y^-1) / (x-y)^-1 Now, I understand the next step: (1/x - 1/y) / (1/x-y) But how do they get to t...
Sun Sep 11, 2011 11:07 pm
Forum: Beginning Algebra
Topic: Having trouble factoring this expression...
Replies: 2
Views: 2042

### Re: Having trouble factoring this expression...

Oh, thanks . It makes sense now, haha.
Fri Sep 09, 2011 11:22 pm
Forum: Beginning Algebra
Topic: Having trouble factoring this expression...
Replies: 2
Views: 2042

### Having trouble factoring this expression...

I am trying to review through self-study, but am having difficulty factoring this expression (despite having some help). The Expression: 2x^2(x+3) + 4x(x+3)^2 Their method of factoring is as follows: 2x(x+3)[x+2(x+3)] = 2x(x+3)(3x+6) = 2x(x+3)3(x+2) = 6x(x+3)(x+2) Any suggestions or links I can foll...