How do you solve the rational equation 1/(x-1)+3/(x+1)=21x1+3x+1=2?

1 Answer
Dec 18, 2015

x = 0 , x= 2x=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= 1x1=0x=1
x+1 = 0 <=> x = -2 x+1=0x=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)LCD

1/(x-1) + 3/(x+1) = 21x1+3x+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)(x1)(x+1)(1x1)+(x1)(x+1)(3x+1)=2(x1)(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