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

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Answer 1 · 2017-10-30T16:14:10

ans : x = 2 and x=7

you perform cross multiplication

that is,

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

simplifying to obtain :
#x^2 - x - 2x + 2# = #6x - 12#

further simplify

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

sending everything on the left hand side;

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

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

simplify into brackets

( x- 2) ( x - 7)

therefore ans are :

x= 2 and x= 7

:)