How do you solve #\frac { x } { x - 1} + 1= \frac { 1} { x ^ { 2} - x }#?

1 Answer
Mrs. R
Dec 1, 2016

x = #-1/2#

Explanation:

Begin by getting a common denominator for all terms. Factor any denominator that can be factored first.
#x/(x-1) + 1 = 1/((x)(x-1))#
The common denominator is #(x)(x-1)#. Multiply each fraction by the appropriate factors to make the denominators the same.
#(x/(x-1))(x/x) +((x)(x-1))/((x)(x-1)) = 1/((x)(x-1))#

Multiplying each term by the common denominator #(x)(x-1)# on both sides of the equation removes the denominator.

Now expand the numerator and rearrange:
#(x)(x) +(x)(x-1) = 1#
#x^2 + x^2 - x = 1#
#2x^2 - x - 1 = 0#

Factoring the quadratic,
#(2x + 1)(x - 1) = 0#

Applying the zero product rule, each factor equals 0.
#2x + 1 = 0# OR #x - 1 = 0#
#2x =-1 # , # x = 1#
#x = -1/2# , #x = 1 #

Now we need to consider the restrictions.
In the original equation, each denominator cannot equal zero.
This means that x cannot equal zero or 1. So the solution #x=1# is inadmissable.
The only solution then is #x = -1/2#.

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