How do you solve #1/(r-2)+1/(r^2-7r+10)=6/(r-2)# and check for extraneous solutions?

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Answer 1 · 2016-10-29T23:26:17

#r = 26/5#

Note that #r^2-7r+10 = (r-2)(r-5)#

So the given equation can be written:

#1/(r-2)+1/((r-2)(r-5)) = 6/(r-2)#

Multiply through by #(r-2)# to get:

#1+1/(r-5)=6#

(noting that #r=2# would be a spurious solution, if it occurs)

Subtract #1# from both sides to get:

#1/(r-5) = 5#

Take the reciprocal of both sides to get:

#r-5 = 1/5#

Add #5# to both sides to get:

#r = 26/5#

Since #26/5 != 2# this is not a spurious solution.