How to simplify this fraction?

  • Thread starter kulimer
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  • #1
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Homework Statement



I want the simplify the following equation

Homework Equations



[itex]\frac{x}{{p}^{2}}+\frac{n-x}{{(1-p)}^{2}}[/itex], where p=[itex]\frac{x}{n}[/itex]


The Attempt at a Solution



I got this [itex]\frac{{n}^{2}}{x}-\frac{{n}^{2}}{n-x}[/itex], I don't think this is the right answer.
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
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No, that is not the right answer. How did you get that?
 
  • #3
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No, that is not the right answer. How did you get that?
[itex]=\frac{x{(1-p)}^{2}+{p}^{2}(n-x)}{{p}^{2}{(1-p)}^{2}} [/itex]
[itex]=\frac{x}{{(\frac{x}{n})}^{2}}+\frac{n-x}{{(1-\frac{x}{n})}^{2}}[/itex]
[itex]=x \frac{{n}^{2}}{{x}^{2}}+\frac{n-x}{({\frac{n-x}{n}})^{2}}[/itex]
[itex]=\frac{{n}^{2}}{x}-(n-x)({\frac{n}{n-x}})^{2}[/itex]
[itex]=\frac{{n}^{2}}{x}-\frac{{n}^{2}}{n-x}[/itex]

Did I make a mistake somewhere?
 
Last edited:
  • #4
HallsofIvy
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[itex]=\frac{x{(1-p)}^{2}+{p}^{2}(n-x)}{{p}^{2}{(1-p)}^{2}} [/itex]
[itex]=\frac{x}{{(\frac{x}{n})}^{2}}+\frac{n-x}{{(1-\frac{x}{n})}^{2}}[/itex]
Why did you combine the fractions and then separate them again? In any case, the problem you posted had [itex]1- p^2[/itex], not [itex](1- p)^2[/itex] which you now have.

[itex]=x \frac{{n}^{2}}{{x}^{2}}+\frac{n-x}{({\frac{n-x}{n}})^{2}}[/itex]
[itex]=\frac{{n}^{2}}{x}-(n-x)({\frac{n}{n-x}})^{2}[/itex]
[itex]=\frac{{n}^{2}}{x}-\frac{{n}^{2}}{n-x}[/itex]
 
  • #5
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Why did you combine the fractions and then separate them again? In any case, the problem you posted had [itex]1- p^2[/itex], not [itex](1- p)^2[/itex] which you now have.
I forgot to put parenthesis around 1-p in LaTex. It is fixed now.
 
  • #6
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@HallsofIvy (PF Mentor) Are you still here? What do you think?
 
  • #7
SammyS
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[itex]=\frac{x{(1-p)}^{2}+{p}^{2}(n-x)}{{p}^{2}{(1-p)}^{2}} [/itex]
[itex]=\frac{x}{{(\frac{x}{n})}^{2}}+\frac{n-x}{{(1-\frac{x}{n})}^{2}}[/itex]
[itex]=x \frac{{n}^{2}}{{x}^{2}}+\frac{n-x}{({\frac{n-x}{n}})^{2}}[/itex]
You changed a "+" sign above to a "−" sign below.
[itex]=\frac{{n}^{2}}{x}-(n-x)({\frac{n}{n-x}})^{2}[/itex]
[itex]=\frac{{n}^{2}}{x}-\frac{{n}^{2}}{n-x}[/itex]

Did I make a mistake somewhere?
 

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