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

2 Answers
May 4, 2018

Answer:

#x = 3#

Explanation:

Here we go...

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

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

#(x + 1)(3x - 3)# = #(x-1)(x^2 - x + 6)#

#3x^2 - 3# = #x^3 - 2x^2 + 7x - 6#

#x^3 - 5x^2 + 7x - 3 = 0#

By factorizing this equation we get :

#(x-1)(x-1)(x-3)# = 0

#:.# #x = 1 or 3#

But as observed in the given question #x - 1# is in denominator, So #x# cannot be equal to 1.

Hence, #x = 3#

May 4, 2018

Answer:

#x=3#

Explanation:

We have an equation with fractions.

It is possible to get rid of the denominators immediately by multiplying each term by the LCM of the denominators, which in this case is #color(blue)(3(x-1))#

#(color(blue)(3cancel((x-1)))xx(x+1))/cancel((x-1))=(color(blue)(cancel3(x-1))xx x)/cancel3 + (2color(blue)(xx3cancel((x-1))))/cancel((x-1))#

This leaves us with:

#3(x+1) = (x-1)x +6" "larr# no fractions

#3x+3 = x^2-x+6" "larr# remove the brackets

#0 = x^2-4x+3" "larr# make the quadratic equal to #0#

#(x-3)(x-1)=0" "larr# factorise

If #x-3 = 0," " rArr x=3#

If #x-1=0," "rArr x = 1#

However, #x=1# is an extraneous solution.
In this equation, #x !=1# because that will make the denominators #0#.

Therefore there is only one solution.

#x =3#