How do you solve #(x-2)/6 = (x-2)/(x-1)#?

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Answer 1 · 2016-10-02T15:50:49

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

#x^2 - 2x - x + 2 = 6x- 12#

#x^2 - 3x + 2 - 6x + 12 = 0#

#x^2 - 9x + 14 = 0#

#(x- 7)(x - 2) = 0#

#x = 7 and 2#

Hopefully this helps!