SOLUTION: Solve the equation (b+17)/(b^2 –1) – 1/(b+1) = (b-2)/(b-1)

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Question 23939: Solve the equation
(b+17)/(b^2 –1) – 1/(b+1) = (b-2)/(b-1)

Found 2 solutions by longjonsilver, AnlytcPhil:
Answer by longjonsilver(2297)   (Show Source): You can put this solution on YOUR website!



mutliply all terms by (b+1)(b-1) and then cancel the denominators. You will be left with:





(b-5)(b+4) = 0
so either b-5=0 OR b+4=0
so either b=5 OR b=-4

jon
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
(b+17)/(b^2 –1) – 1/(b+1) = (b-2)/(b-1) 

 b+17      1      b-2
—————— - ————— = —————
 b2-1     b+1     b-1

We clear of fractions.  Factor the denominator of
the first fraction

    b+17         1      b-2
———————————— - ————— = —————
 (b-1)(b+1)     b+1     b-1

The LCD of these denominators is (b-1)(b+2)

Multiply each term by (b-1)(b+1)/1

(b-1)(b+1)    b+17        (b-1)(b+1)   1     (b-1)(b+1)  b-2
——————————·———————————— - ——————————·————— = ——————————·—————
    1       (b-1)(b+1)        1       b+1        1       b-1

Now we cancel


(b-1)(b+1)    b+17        (b-1)(b+1)   1     (b-1)(b+1)  b-2
——————————·———————————— - ——————————·————— = ——————————·—————
    1       (b-1)(b+1)        1       b+1        1       b-1

That leaves just

                             b+17 - (b-1) = (b+1)(b-2)
                            b + 17 - b + 1 = b2 - 2b + b - 2
                                        18 = b2 - b - 2
                                         0 = b2 - b - 20
                                         0 = (b-5)(b+4)
                                
              Set each factor = 0
                                 

                                   b-5 = 0 gives b = 5                          
                                   b+4 = 0 gives b = -4

Edwin
AnlytcPhil@aol.com
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