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\({2h \over h^{2} - 9} + {h \over h^{2} + 6h + 9} - {3 \over h - 3}\)

 

the textbook answer is \({-3(5h + 9)\over (h + 3)(h + 3)(h - 3)}, h ≠ ± 3\) how should i get it?

Guest Jul 23, 2018
 #1
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2h / (h^2 - 9)  +  h / (h^2 + 6h + 9)   - 3/ ( h - 3)          factor the denominators

 

2h / [ ( h + 3) ( h - 3) ]  +  h / [ ( h + 3) (h + 3) ]  - 3 / (h - 3)

 

The common denominator  is   ( h + 3) ( h + 3) ( h - 3)  ....so we have

 

2h (h + 3)   + h( h -3)   - 3 (h + 3) ( h + 3)

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 (h + 3) ( h + 3)  ( h - 3)

 

Simplify  and note that  x cannot  be either 3  or  -3  because both values make the denominator  = 0

 

So we have

 

2h^2 + 6h + h^2 - 3h  - 3 ( h^2 + 6h + 9)

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( h + 3) ( h + 3) ( h - 3)

 

2h^2 + 6h + h^2 - 3h  - 3h^2 - 18h - 27

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(h + 3) ( h + 3) (h - 3)

 

 

         -15h - 27

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(h + 3) ( h + 3) ( h - 3)

 

 

   -3(5h + 9)

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(h + 3) ( h + 3) ( h - 3)

 

 

 

cool cool cool

CPhill  Jul 23, 2018

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