1. 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.$(\frac{ x }{ y }-\frac{ y }{ x })\div (\frac{ 1 }{ x^2 }-\frac{ 1 }{ y^2 })$

2. anonymous

When you flip the second equation, the division sign will change to a multiplication sign.$(\frac{ x }{ y }-\frac{ y }{ x })\times(x^2-y^2)$

3. AngelaB97

yes and then what happens to the denominator?

4. anonymous

Look at the first equation now and try to get the denominator's the same by cross multiplying.$(\frac{ x(x) }{ y(x) }-\frac{ y(y) }{ x(y) })$

5. anonymous

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

6. anonymous

Now the equation looks like this:$(\frac{ x^2-y^2 }{ xy })\times(x^2-y^2)$

7. 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|

8. anonymous

when you distribute the fractions will clear out

9. anonymous

$\frac{ (x^2-y^2)(x^2-y^2) }{ xy }$

10. anonymous

|dw:1438814584619:dw|

11. anonymous

$\frac{ (x^4-2x^2y^2+y^4) }{ xy }$

12. anonymous

|dw:1438814672337:dw|

13. anonymous

|dw:1438814727794:dw|

14. 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