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AngelaB97

  • one year ago

please help me simplify this x/y-y/x _________ 1/x^2-1/y^2

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  1. anonymous
    • one year ago
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    \[\frac{ \frac{ x }{ y }-\frac{ y }{ x } }{ \frac{ 1 }{ x^2 } -\frac{ 1 }{ y^2 }}\]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
    • one year ago
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    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
    • one year ago
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    yes and then what happens to the denominator?

  4. anonymous
    • one year ago
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    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
    • one year ago
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    \[(\frac{ x^2 }{ xy }-\frac{ y^2 }{ xy })=(\frac{ x^2-y^2 }{ xy })\]

  6. anonymous
    • one year ago
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    Now the equation looks like this:\[(\frac{ x^2-y^2 }{ xy })\times(x^2-y^2)\]

  7. anonymous
    • one year ago
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    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
    • one year ago
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    when you distribute the fractions will clear out

  9. anonymous
    • one year ago
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    \[\frac{ (x^2-y^2)(x^2-y^2) }{ xy }\]

  10. anonymous
    • one year ago
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    |dw:1438814584619:dw|

  11. anonymous
    • one year ago
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    \[\frac{ (x^4-2x^2y^2+y^4) }{ xy }\]

  12. anonymous
    • one year ago
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    |dw:1438814672337:dw|

  13. anonymous
    • one year ago
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    |dw:1438814727794:dw|

  14. anonymous
    • one year ago
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    \[\frac{ \frac{ x }{ y }-\frac{ y }{ x } }{ \frac{ 1 }{ x^2 }-\frac{ 1 }{ y^2 } }\] \[=\frac{ \frac{ x^2-y^2 }{ xy } }{ \frac{ y^2-x^2 }{ x^2y^2 } }\] \[=\frac{ x^2-y^2 }{ xy }\times \frac{ x^2y^2 }{ -\left( x^2-y^2 \right) }\] =-xy

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