How do you solve #\frac { 2r } { r - 4} - \frac { 4r } { r - 5} = - \frac { 5r } { r ^ { 2} - 9r + 20}# ?

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Answer 1 · 2017-07-01T11:50:19

Solution : #r=0 or r =11/2#

#(2r)/(r-4) - (4r)/(r-5) = - (5r)/(r^2-9r+20)# or

#(2r)/(r-4) - (4r)/(r-5) = - (5r)/((r-4)(r-5)) # . Mutiplying by

(r-4)(r-5) on both sides , we get,

#2r(r-5) - 4r(r-4) = -5r or 2r^2 -10r -4r^2 +16r = -5r# or

#-2r^2+6r = -5r or 2r^2 -11 r = 0 or r(2r-11) =0 # . Either

#r=0 or 2r-11 = 0 :. 2r =11 :. r =11/2#

Solution : #r=0 , r =11/2# [Ans]