How do you solve #6/x - 4/(x-3) = 1#?

1 Answer
Roella W.
Dec 31, 2015

#x=(5+-sqrt(-47))/2#

Explanation:

First put everything on the LHS on a common denominator
#6/x - 4/(x-3) = 1#
#(6(x-3) -4x)/(x(x-3)) = 1#
#6x -18 -4x = 1*(x(x-3))#
#2x - 18 = x^2 - 3x#
#x^2 - 5x +18 = 0#
Then use the quadratic formula to solve
#x = (-b +- sqrt(b^2 - 4ac))/(2a)#
#x = (-(-5) +-sqrt((-5)^2 -4*1*18))/2#
#x =(5+-sqrt(25-72))/2#
#x=(5+-sqrt(-47))/2#
There are only imaginary numbers as solutions to this equation.

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