AngelaB97
  • AngelaB97
please help me simplify this x/y-y/x _________ 1/x^2-1/y^2
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{ \frac{ x }{ y }-\frac{ y }{ x } }{ \frac{ 1 }{ x^2 } -\frac{ 1 }{ y^2 }}\]First separate the equations so it can look more simpler.\[(\frac{ x }{ y }-\frac{ y }{ x })\div (\frac{ 1 }{ x^2 }-\frac{ 1 }{ y^2 })\]
anonymous
  • anonymous
When you flip the second equation, the division sign will change to a multiplication sign.\[(\frac{ x }{ y }-\frac{ y }{ x })\times(x^2-y^2)\]
AngelaB97
  • AngelaB97
yes and then what happens to the denominator?

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anonymous
  • anonymous
Look at the first equation now and try to get the denominator's the same by cross multiplying.\[(\frac{ x(x) }{ y(x) }-\frac{ y(y) }{ x(y) })\]
anonymous
  • anonymous
\[(\frac{ x^2 }{ xy }-\frac{ y^2 }{ xy })=(\frac{ x^2-y^2 }{ xy })\]
anonymous
  • anonymous
Now the equation looks like this:\[(\frac{ x^2-y^2 }{ xy })\times(x^2-y^2)\]
anonymous
  • anonymous
The original denominator is made up of 2 different fractions. So they don't have a single reciprocal. Solve these problems by multiplying by 1in the form of the least common denominator divided by itself. The least common denominator of x, y, x², and y² is x²y². |dw:1438814492460:dw|
anonymous
  • anonymous
when you distribute the fractions will clear out
anonymous
  • anonymous
\[\frac{ (x^2-y^2)(x^2-y^2) }{ xy }\]
anonymous
  • anonymous
|dw:1438814584619:dw|
anonymous
  • anonymous
\[\frac{ (x^4-2x^2y^2+y^4) }{ xy }\]
anonymous
  • anonymous
|dw:1438814672337:dw|
anonymous
  • anonymous
|dw:1438814727794:dw|
anonymous
  • anonymous
\[\frac{ \frac{ x }{ y }-\frac{ y }{ x } }{ \frac{ 1 }{ x^2 }-\frac{ 1 }{ y^2 } }\] \[=\frac{ \frac{ x^2-y^2 }{ xy } }{ \frac{ y^2-x^2 }{ x^2y^2 } }\] \[=\frac{ x^2-y^2 }{ xy }\times \frac{ x^2y^2 }{ -\left( x^2-y^2 \right) }\] =-xy

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