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

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Answer 1 · 2016-10-24T13:14:22

Use the property #a/b = c/d -> a xx d = b xx c# to start the solving process.

#(x -1)(x + 2) = 2(7x + 6)#

#x^2 - x + 2x - 2 = 14x + 12#

#x^2 + x - 2 = 14x + 12#

#x^2 - 13x - 14 = 0#

#(x - 14)(x+ 1) = 0#

#x = 14 and -1#

None of these solutions are extraneous, since they do not go against the initial restrictions of #-6/7# and #-2#.

Hopefully this helps!