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

1 Answer
Oct 13, 2015

Answer:

#x = 0" "# or #" "x = 2#

Explanation:

You know that your ultimate goal here is to isolate #x# on one side of the equation.

Start by getting rid of the denominators. Rewrite the equation as

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

The three denominators you're dealing with are

#(x-1)" "#, #" "(x+1)" "#, and #" "1#

To find the common denominator of the three fractions, you need to find the least common multiple of these expressions. Notice that if you multiply the first one by #(x+1)#, the second one by #(x-1)#, and the third one by #(x-1)(x+1)#, you get

#(x-1) * (x+1) * 1 = (x-1)(x+1)#

This will be your common denominator.

The equation can thus be written as

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

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

This is equivalent to

#x+1 + 3(x-1) = 2(x-1)(x+1)#

Expand the parantheses and rearrange to get all the terms on one side of the equation

#x + 1 + 3x - 3 = 2x^2 - 2#

#2x^2 - 4x - color(red)(cancel(color(black)(2))) + color(red)(cancel(color(black)(2))) = 0#

This is equivalent to

#2x(x-2) = 0#

The two solutions of this equaion will be

#2x = 0 implies x = color(green)(0)#

and

#x - 2 = 0 implies x= color(green)(2)#

Do quick check to make sure the calculations are correct

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

#-1 + 3 = 2color(white)(x)color(green)(sqrt())#

#x = 2 implies 1/(2-1) + 3/(2 + 1) = 2#

#1 + 1 = 2color(white)(x)color(green)(sqrt())#