Simplifying compound fraction not producing answer provided by book












1












$begingroup$


I am working on a problem in a textbook(Precalculus Mathematics for Calculus, by James Stewart) and the answer in the back of the book for the problem(1.4 #67) is -xy but I am not getting that. Here is the problem:



begin{align}
frac{frac{x}{y}-frac{y}{x}}{frac{1}{x^2}-frac{1}{y^2}}
end{align}



Here is how I worked it out:



begin{align}
frac{frac{x}{y}-frac{y}{x}}{frac{1}{x^2}-frac{1}{y^2}}=frac{frac{x^2-y^2}{xy}}{frac{y^2-x^2}{x^2y^2}}=frac{x^2-y^2}{xy}*frac{x^2y^2}{y^2-x^2}
=frac{xy(x^2-y^2)}{y^2-x^2}=frac{(xy)(x+y)(x-y)}{(y+x)(y-x)}=frac{xy(x-y)}{y-x}=frac{x^2y-xy^2}{y-x}
end{align}



Can someone please explain what I am doing wrong here?






algebra-precalculus fractions







share|cite|improve this question









$endgroup$












  • $begingroup$
    You are not cancelling the factor $y-x$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 30 at 5:49










  • $begingroup$
    @LordSharktheUnknown You can cancel (y-x) with (x-y)?
    $endgroup$
    – SoleCore
    Jan 30 at 5:55










  • $begingroup$
    Yes, since $(y-x)=-(x-y)$, it leaves $-1$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 30 at 5:55










  • $begingroup$
    Oh snap. I am so oblivious. Thanks!
    $endgroup$
    – SoleCore
    Jan 30 at 5:56
















1












$begingroup$


I am working on a problem in a textbook(Precalculus Mathematics for Calculus, by James Stewart) and the answer in the back of the book for the problem(1.4 #67) is -xy but I am not getting that. Here is the problem:



begin{align}
frac{frac{x}{y}-frac{y}{x}}{frac{1}{x^2}-frac{1}{y^2}}
end{align}



Here is how I worked it out:



begin{align}
frac{frac{x}{y}-frac{y}{x}}{frac{1}{x^2}-frac{1}{y^2}}=frac{frac{x^2-y^2}{xy}}{frac{y^2-x^2}{x^2y^2}}=frac{x^2-y^2}{xy}*frac{x^2y^2}{y^2-x^2}
=frac{xy(x^2-y^2)}{y^2-x^2}=frac{(xy)(x+y)(x-y)}{(y+x)(y-x)}=frac{xy(x-y)}{y-x}=frac{x^2y-xy^2}{y-x}
end{align}



Can someone please explain what I am doing wrong here?






algebra-precalculus fractions







share|cite|improve this question









$endgroup$












  • $begingroup$
    You are not cancelling the factor $y-x$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 30 at 5:49










  • $begingroup$
    @LordSharktheUnknown You can cancel (y-x) with (x-y)?
    $endgroup$
    – SoleCore
    Jan 30 at 5:55










  • $begingroup$
    Yes, since $(y-x)=-(x-y)$, it leaves $-1$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 30 at 5:55










  • $begingroup$
    Oh snap. I am so oblivious. Thanks!
    $endgroup$
    – SoleCore
    Jan 30 at 5:56














1












1








1





$begingroup$


I am working on a problem in a textbook(Precalculus Mathematics for Calculus, by James Stewart) and the answer in the back of the book for the problem(1.4 #67) is -xy but I am not getting that. Here is the problem:



begin{align}
frac{frac{x}{y}-frac{y}{x}}{frac{1}{x^2}-frac{1}{y^2}}
end{align}



Here is how I worked it out:



begin{align}
frac{frac{x}{y}-frac{y}{x}}{frac{1}{x^2}-frac{1}{y^2}}=frac{frac{x^2-y^2}{xy}}{frac{y^2-x^2}{x^2y^2}}=frac{x^2-y^2}{xy}*frac{x^2y^2}{y^2-x^2}
=frac{xy(x^2-y^2)}{y^2-x^2}=frac{(xy)(x+y)(x-y)}{(y+x)(y-x)}=frac{xy(x-y)}{y-x}=frac{x^2y-xy^2}{y-x}
end{align}



Can someone please explain what I am doing wrong here?






algebra-precalculus fractions







share|cite|improve this question









$endgroup$




I am working on a problem in a textbook(Precalculus Mathematics for Calculus, by James Stewart) and the answer in the back of the book for the problem(1.4 #67) is -xy but I am not getting that. Here is the problem:



begin{align}
frac{frac{x}{y}-frac{y}{x}}{frac{1}{x^2}-frac{1}{y^2}}
end{align}



Here is how I worked it out:



begin{align}
frac{frac{x}{y}-frac{y}{x}}{frac{1}{x^2}-frac{1}{y^2}}=frac{frac{x^2-y^2}{xy}}{frac{y^2-x^2}{x^2y^2}}=frac{x^2-y^2}{xy}*frac{x^2y^2}{y^2-x^2}
=frac{xy(x^2-y^2)}{y^2-x^2}=frac{(xy)(x+y)(x-y)}{(y+x)(y-x)}=frac{xy(x-y)}{y-x}=frac{x^2y-xy^2}{y-x}
end{align}



Can someone please explain what I am doing wrong here?






algebra-precalculus fractions




algebra-precalculus fractions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 30 at 5:45









SoleCoreSoleCore

105




105












  • $begingroup$
    You are not cancelling the factor $y-x$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 30 at 5:49










  • $begingroup$
    @LordSharktheUnknown You can cancel (y-x) with (x-y)?
    $endgroup$
    – SoleCore
    Jan 30 at 5:55










  • $begingroup$
    Yes, since $(y-x)=-(x-y)$, it leaves $-1$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 30 at 5:55










  • $begingroup$
    Oh snap. I am so oblivious. Thanks!
    $endgroup$
    – SoleCore
    Jan 30 at 5:56


















  • $begingroup$
    You are not cancelling the factor $y-x$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 30 at 5:49










  • $begingroup$
    @LordSharktheUnknown You can cancel (y-x) with (x-y)?
    $endgroup$
    – SoleCore
    Jan 30 at 5:55










  • $begingroup$
    Yes, since $(y-x)=-(x-y)$, it leaves $-1$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 30 at 5:55










  • $begingroup$
    Oh snap. I am so oblivious. Thanks!
    $endgroup$
    – SoleCore
    Jan 30 at 5:56
















$begingroup$
You are not cancelling the factor $y-x$.
$endgroup$
– Lord Shark the Unknown
Jan 30 at 5:49




$begingroup$
You are not cancelling the factor $y-x$.
$endgroup$
– Lord Shark the Unknown
Jan 30 at 5:49












$begingroup$
@LordSharktheUnknown You can cancel (y-x) with (x-y)?
$endgroup$
– SoleCore
Jan 30 at 5:55




$begingroup$
@LordSharktheUnknown You can cancel (y-x) with (x-y)?
$endgroup$
– SoleCore
Jan 30 at 5:55












$begingroup$
Yes, since $(y-x)=-(x-y)$, it leaves $-1$.
$endgroup$
– Lord Shark the Unknown
Jan 30 at 5:55




$begingroup$
Yes, since $(y-x)=-(x-y)$, it leaves $-1$.
$endgroup$
– Lord Shark the Unknown
Jan 30 at 5:55












$begingroup$
Oh snap. I am so oblivious. Thanks!
$endgroup$
– SoleCore
Jan 30 at 5:56




$begingroup$
Oh snap. I am so oblivious. Thanks!
$endgroup$
– SoleCore
Jan 30 at 5:56










1 Answer
1






active

oldest

votes


















1












$begingroup$

You were almost there. $frac{xy(x-y)}{y-x}$(your second last one)$=-xy$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How did you get that result?
    $endgroup$
    – SoleCore
    Jan 30 at 5:50










  • $begingroup$
    Nevermind. LordSharktheUnknown explained it.
    $endgroup$
    – SoleCore
    Jan 30 at 5:59












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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You were almost there. $frac{xy(x-y)}{y-x}$(your second last one)$=-xy$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How did you get that result?
    $endgroup$
    – SoleCore
    Jan 30 at 5:50










  • $begingroup$
    Nevermind. LordSharktheUnknown explained it.
    $endgroup$
    – SoleCore
    Jan 30 at 5:59
















1












$begingroup$

You were almost there. $frac{xy(x-y)}{y-x}$(your second last one)$=-xy$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How did you get that result?
    $endgroup$
    – SoleCore
    Jan 30 at 5:50










  • $begingroup$
    Nevermind. LordSharktheUnknown explained it.
    $endgroup$
    – SoleCore
    Jan 30 at 5:59














1












1








1





$begingroup$

You were almost there. $frac{xy(x-y)}{y-x}$(your second last one)$=-xy$






share|cite|improve this answer









$endgroup$



You were almost there. $frac{xy(x-y)}{y-x}$(your second last one)$=-xy$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 30 at 5:47









abc...abc...

3,237739




3,237739












  • $begingroup$
    How did you get that result?
    $endgroup$
    – SoleCore
    Jan 30 at 5:50










  • $begingroup$
    Nevermind. LordSharktheUnknown explained it.
    $endgroup$
    – SoleCore
    Jan 30 at 5:59


















  • $begingroup$
    How did you get that result?
    $endgroup$
    – SoleCore
    Jan 30 at 5:50










  • $begingroup$
    Nevermind. LordSharktheUnknown explained it.
    $endgroup$
    – SoleCore
    Jan 30 at 5:59
















$begingroup$
How did you get that result?
$endgroup$
– SoleCore
Jan 30 at 5:50




$begingroup$
How did you get that result?
$endgroup$
– SoleCore
Jan 30 at 5:50












$begingroup$
Nevermind. LordSharktheUnknown explained it.
$endgroup$
– SoleCore
Jan 30 at 5:59




$begingroup$
Nevermind. LordSharktheUnknown explained it.
$endgroup$
– SoleCore
Jan 30 at 5:59


















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