Brainteaserfan Posted August 23, 2011 Share Posted August 23, 2011 I got this wrong in a test. Now I'm sitting and puzzling. The initial fraction is what the answer key said, the last, mine. What did I do wrong in: 2x/(1+4x) = (2x/x)/(1/x+4x/x) = 2/(1/x+4) = 2(x/1)+2(1/4) = 2x+1/2 When I substitute 10 for x, the answers are different. If there is a link or source, it would be very helpful too. Thanks! Link to comment Share on other sites More sharing options...
DrRocket Posted August 23, 2011 Share Posted August 23, 2011 I got this wrong in a test. Now I'm sitting and puzzling. The initial fraction is what the answer key said, the last, mine. What did I do wrong in: 2x/(1+4x) = (2x/x)/(1/x+4x/x) = 2/(1/x+4) = 2(x/1)+2(1/4) = 2x+1/2 When I substitute 10 for x, the answers are different. If there is a link or source, it would be very helpful too. Thanks! [math]2x/(1+4x) = (2x/x)/(1/x+4x/x) = 2/(1/x+4) \ne 2(x/1)+2(1/4) = 2x+1/2 [/math] 1 Link to comment Share on other sites More sharing options...
Brainteaserfan Posted August 23, 2011 Author Share Posted August 23, 2011 [math]2x/(1+4x) = (2x/x)/(1/x+4x/x) = 2/(1/x+4) \ne 2(x/1)+2(1/4) = 2x+1/2 [/math] Thanks! Link to comment Share on other sites More sharing options...
Fuzzwood Posted August 23, 2011 Share Posted August 23, 2011 A trick to solving this is to multiply by 1: (1/x + 4)/(1/x + 4), then you have a way to split this equation into (2/x)/(1/x + 4)² and 8/(1/x + 4)² Link to comment Share on other sites More sharing options...
Brainteaserfan Posted August 23, 2011 Author Share Posted August 23, 2011 (edited) A trick to solving this is to multiply by 1: (1/x + 4)/(1/x + 4), then you have a way to split this equation into (2/x)/(1/x + 4)² and 8/(1/x + 4)² I don't understand why I should split it. I didn't need to simplify past the first fraction on the initial question. Dr Rocket answered my question perfectly. I have another quick question: if you are finding the (sqrt of 2x)/(the sqrt of x), can you just cancel the x's and get sqrt of 2, x positive for an answer? And if the numerator was x squared, you'd have sqrt of x for your answer? Thanks! I have another question too. If the conjugate of a + b is a - b, is -a + b also a conjugate? Edited August 23, 2011 by Brainteaserfan Link to comment Share on other sites More sharing options...
Fuzzwood Posted August 23, 2011 Share Posted August 23, 2011 Ah very well. That trick can sometimes simplify to the removal of the entire fraction. This was what I was aiming at but failed a bit with it: http://en.wikipedia.org/wiki/Conjugate_(algebra) And yes, because you can split that into ((sqrt of 2)*(sqrt of x))/(sqrt of x) = sqrt of 2 1 Link to comment Share on other sites More sharing options...
DrRocket Posted August 23, 2011 Share Posted August 23, 2011 I have another quick question: if you are finding the (sqrt of 2x)/(the sqrt of x), can you just cancel the x's and get sqrt of 2, x positive for an answer? And if the numerator was x squared, you'd have sqrt of x for your answer? [math] \frac {\sqrt {2x}}{\sqrt x} = \frac {\sqrt 2 \sqrt x}{\sqrt x} = \sqrt 2 [/math] I have another question too. If the conjugate of a + b is a - b, is -a + b also a conjugate? [math] \overline {a + bi} = a-bi[/math] and [math] \overline {a+b} = a+b[/math] where [math]a,b \in \mathbb R[/math] and [math] i = \sqrt {-1}[/math] Link to comment Share on other sites More sharing options...
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