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anonymous

  • 5 years ago

Perform the indicated operations and simplify y-2/y-4-y+1/y+4+y-20/y^2-16 I got answer: 6/y+4

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  1. anonymous
    • 5 years ago
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    is this correct???

  2. anonymous
    • 5 years ago
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    can u put the brackets?

  3. anonymous
    • 5 years ago
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    (y-2/y-4)-(y+1/y+4)+(y-20/y^2-16) then ur ans is correct

  4. jhonyy9
    • 5 years ago
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    (y-2/y-4)-(y+1/y+4)+(y-20/y*2-16)=((y-2)(y+4)/(y*2-16)-((y+1)(y-4)/(y*2-16)+ +(y-20)/(y*2-16)=((y-2)(y+4)-(y+1)(y-4)+(y-20))/(y*2-16)=(y*2+2y-8- y*2+3y+4+y-20)/(y*2-16)=(6y-24)/(y-4)(y+4)=6(y-4)/(y-4)(y+4)=6/(y+4)

  5. anonymous
    • 5 years ago
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    What's really missing here are the correct paranthesis for this problem :)

  6. anonymous
    • 5 years ago
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    Finally, figured out how to get the equation editor to show the expression we are assuming properly: \[\frac{y-2}{y-4}-\frac{y+1}{y+4}+\frac{y-20}{y^2-16}\] = \[\frac{y-2}{y-4}-\frac{y+1}{y+4}+\frac{y-20}{(y+4)(y-4)}\] = \[\frac{(y-2)(y+4)}{(y-4)(y+4)}-\frac{(y+1)(y-4)}{(y+4)(y-4)}+\frac{y-20}{(y-4)(y+4)}\] = \[\frac{(y^2+4y-2y-8)-(y^2-4y+y-4)+(y-20)}{(y-4)(y+4)}\] = \[\frac{6y-24}{(y-4)(y+4)}\] = \[\frac{6(y-4)}{(y-4)(y+4)}\] = \[\frac{6}{y+4}\]

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