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

1 Answer
Dec 18, 2015

Answer:

#x = 0 , x= 2#

Explanation:

Step 1 : Identify the restricted value.

This is done by set the denominator equal to zero like this

# x-1= 0 <=> x= 1#
#x+1 = 0 <=> x = -2 #

The idea of restricted value, is to narrow down what value our variable can't be (aka domain)

Step 2: Multiply the equation by #color(red)(LCD)#

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

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

#color(red)(cancel(x-1)(x+1))(1/cancel(x-1)) +color(red)( (x-1)cancel(x+1))(3/cancel(x+1)) = 2color(red)((x-1)(x+1)#

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

Step 3: Multiply and combine like terms

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

#4x -2 = 2(x^2 -1)#

#4x -2 = 2x^2 -2#

#0 =2x^2-4x#

Step 4: Solve the quadratic equation

# 2x^2 -4x = 0 #
#2x(x-2) = 0#

#2x = 0 => color(blue)(x = 0) #

#x-2 = 0 => color(blue)(x = 2)#

Step 5 Check your solution ..

Check to see if the answer from Step 4 is the same a restricted value.

If it's not, the the solution is #x = 0, x= 2#