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1/(w+3) + 1/(w-3) = (w^2-3)/(w^2-9)

User Kohske
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2 Answers

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1/(w+3) + 1/(w-3) = (w^2-3)/(w^2-9)

Answer w = -1
1/(w+3) + 1/(w-3) = (w^2-3)/(w^2-9)-example-1
User Bitsy
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We want all denominators to be (w^2-9)

[(w-3)+(x+3)]/(w^2-9)=(w^2-3)/(w^2-9) multiply both sides by (w^2-9)

w-3+w+3=w^2-3 combine like terms on left side

2w=w^2-3 subtract 2w from both sides

w^2-2w-3=0 factor

w^2+w-3w-3=0

w(w+1)-3(w+1)=0

(w+1)(w-3)=0

w=-1 and 3

HOWEVER 3 is an extraneous solution as it would make a couple of our original fractions undefined because of division by zero, so there really is only one soluton:

w=-1
User Marc Fowler
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