How do you solve #2/(x-1) - 2/3 =4/(x+1)#?

1 Answer
Jul 3, 2016

Answer:

#x = -5 or x = 2#

Explanation:

Make a common denominator of all the fractions. The obvious choice is to make every denominator #3(x-1)(x+1)#. We can do this because multiplying the top and bottom of a fraction does not change the fraction itself.

So, we multiply top and bottom of each fraction by the appropriate part of the required denominator:

#(2)/(x-1) * (3(x+1))/(3(x+1)) - 2/3*((x-1)(x+1))/((x-1)(x+1)) #

#= 4/(x+1)*(3(x-1))/(3(x-1))#

Because everything is over a common denominator, we can ignore it and focus on the numerators.

#6(x+1) - 2(x-1)(x+1) = 12(x-1)#

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

#implies 2x^2 + 6x - 20 = 0#

We've arrived at a simple quadratic, can solve by factorisation but for completeness we'll use the quadratic formula.

Have a = 2, b = 6, c = -20.

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

#x = (-6 +- sqrt(36 + 160))/4 = (-6 +- sqrt(196))/4 = (-6 +- 14)/4#

Hence #x = -5 or x = 2#