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

1 Answer
Jul 22, 2016

Answer:

#x=2" "# or #" "x=-5#

Explanation:

The first thing to note here is that the solution set cannot include #x = +-1#, since those values would make the denominators equal to zero.

Now, rearrange the equation to get the all the terms on one side

#2/(x-1) - 4/(x+1) - 2/3 = 0#

The common denominator here will be

#3(x-1)(x+1)#

which means that you must multiply the first fraction by #1 = (3(x+1))/(3(x+1))#, the second fraction by #1 = (3(x-1))/(3(x-1))#, and the third fraction by #1 = ((x-1)(x+1))/((x-1)(x+1))# to get

#2/(x-1) * (3(x+1))/(3(x+1)) - 4/(x+1) * (3(x-1))/(3(x-1)) - 2/3 * ((x-1)(x+1))/((x-1)(x+1)) = 0#

This will be equivalent to

#(6(x+1) - 12(x-1) - 2(x-1)(x+1))/(3(x-1)(x+1)) = 0#

Next, focus on the numerator. Expand the parentheses and group like terms to get

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

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

You can divide all the terms by #-2# to get equivalent form

#x^2 + 3x - 10 = 0#

Two numbers that add up to give #3# and multiply to give #-10# are #-2# and #5#, which means that the above quadratic equation can be factored as

#(x-2)(x+5) = 0 implies {(x_1 = 2), (x_2 = -5) :}#

Since

#x_1, x_2 !=+-1#

you can say that the original equation has two possible solutions

#x = 2" " "or" " "x = -5#

Do a quick check to make sure that the calculations are correct

#x = 2:" " 2/(2-1) - 2/3 = 4/(2+1) <=> 2 - 2/3 = 4/3 " "color(green)(sqrt())#

#x = -5: " " 2/(-5 -2) -2/3 = 4/(-5 + 1) <=> -1/3 - 2/3 = -1 " "color(green)(sqrt())#