How do you solve #\frac { r } { r - 1} + \frac { 3} { r ^ { 2} - 1} = \frac { 9} { r + 1}#?

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Answer 1 · 2017-07-22T20:00:47

#r = 2" "# or #" "r = 6#

Given:

#r/(r-1)+3/(r^2-1) = 9/(r+1)#

Note that:

#r^2-1 = (r-1)(r+1)#

So we can make this rational equation into a polynomial one by multiplying both sides by #(r^2-1)# to get:

#r(r+1)+3=9(r-1)#

which multiplies out to:

#r^2+r+3 = 9r-9#

Subtract #9r-9# from both sides to get:

#0 = r^2-8r+12 = (r-6)(r-2)#

(found by finding a pair of factors of #12# with sum #8#)

So:

#r = 2" "# or #" "r = 6#

Note that neither of these values cause #r^2-1 = 0#, so both are valid solutions of the original rational equation.