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

Question

Kennedy Adams · Accepted Answer

#x = 3#

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#

Genesis Allen · Answer

#x=3#

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#

Eva Adamson · Answer

undefined To solve the rational equation (x+1)/(x-1) = x/3 + 2/(x-1), we can start by multiplying both sides of the equation by the common denominator, which is (x-1)(3). This will eliminate the denominators in the equation. Simplifying the equation after multiplying, we get 3(x+1) = x(x-1) + 2(3). Expanding and simplifying further, we have 3x + 3 = x^2 - x + 6. Rearranging the equation, we get x^2 - 4x + 3 = 0. Factoring the quadratic equation, we have (x-3)(x-1) = 0. Setting each factor equal to zero, we find x = 3 and x = 1 as the solutions to the equation. Therefore, the solution to the rational equation is x = 3 and x = 1.