How to simplify this fraction?

[kulimer]

Homework Statement

I want the simplify the following equation

Homework Equations

$\frac{x}{{p}^{2}}+\frac{n-x}{{(1-p)}^{2}}$, where p=$\frac{x}{n}$

The Attempt at a Solution

I got this $\frac{{n}^{2}}{x}-\frac{{n}^{2}}{n-x}$, I don't think this is the right answer.

 

[HallsofIvy]

No, that is not the right answer. How did you get that?

 

[kulimer]

No, that is not the right answer. How did you get that?

$=\frac{x{(1-p)}^{2}+{p}^{2}(n-x)}{{p}^{2}{(1-p)}^{2}}$
$=\frac{x}{{(\frac{x}{n})}^{2}}+\frac{n-x}{{(1-\frac{x}{n})}^{2}}$
$=x \frac{{n}^{2}}{{x}^{2}}+\frac{n-x}{({\frac{n-x}{n}})^{2}}$
$=\frac{{n}^{2}}{x}-(n-x)({\frac{n}{n-x}})^{2}$
$=\frac{{n}^{2}}{x}-\frac{{n}^{2}}{n-x}$

Did I make a mistake somewhere?

 

[HallsofIvy]

$=\frac{x{(1-p)}^{2}+{p}^{2}(n-x)}{{p}^{2}{(1-p)}^{2}}$
$=\frac{x}{{(\frac{x}{n})}^{2}}+\frac{n-x}{{(1-\frac{x}{n})}^{2}}$

Why did you combine the fractions and then separate them again? In any case, the problem you posted had $1- p^2$, not $(1- p)^2$ which you now have.

$=x \frac{{n}^{2}}{{x}^{2}}+\frac{n-x}{({\frac{n-x}{n}})^{2}}$
$=\frac{{n}^{2}}{x}-(n-x)({\frac{n}{n-x}})^{2}$
$=\frac{{n}^{2}}{x}-\frac{{n}^{2}}{n-x}$

 

[kulimer]

Why did you combine the fractions and then separate them again? In any case, the problem you posted had $1- p^2$, not $(1- p)^2$ which you now have.

I forgot to put parenthesis around 1-p in LaTex. It is fixed now.

 

[kulimer]

@HallsofIvy (PF Mentor) Are you still here? What do you think?

 

[SammyS]

$=\frac{x{(1-p)}^{2}+{p}^{2}(n-x)}{{p}^{2}{(1-p)}^{2}}$
$=\frac{x}{{(\frac{x}{n})}^{2}}+\frac{n-x}{{(1-\frac{x}{n})}^{2}}$
$=x \frac{{n}^{2}}{{x}^{2}}+\frac{n-x}{({\frac{n-x}{n}})^{2}}$

You changed a "+" sign above to a "−" sign below.

$=\frac{{n}^{2}}{x}-(n-x)({\frac{n}{n-x}})^{2}$
$=\frac{{n}^{2}}{x}-\frac{{n}^{2}}{n-x}$

Did I make a mistake somewhere?