How do you solve #2/(x-1) - 2/3 =4/(x+1)#?

1 Answer
Jun 26, 2016

Answer:

#x = 10/7 or x = -1#

Explanation:

When we have an equation with fractions it is possible to get rid of the denominators.

Instead of finding a common denominator and converting all the numerators, find the LCM and multiply each term by the LCM of the denominators. The denominators can cancel.

LCM = #color(red)(3(x+1)(x-1))#

#(color(red)(3(x+1)(x-1))xx2)/(x-1) -(2xxcolor(red)(3(x+1)(x-1)))/3 " " = (4xxcolor(red)(3(x+1)(x-1)))/(x+1)#

#(color(red)(3(x+1)cancel((x-1)))xx2)/cancel((x-1) ) -(2xxcolor(red)(cancel3(x+1)(x-1)))/cancel3 " "=4xxcolor(red)(3cancel((x+1))(x-1))/cancel((x+1))#

#3(x+1)xx2 -2(x+1)(x-1) = 4xx3(x+1)(x-1)#

#6(x + 1) -2(x^2 -1) = 12(x^2 -1)#

#6x + 6 -2x^2 + 2 = 12x^2 -12#

#14x^2-6x-20 =0 " divide by 2"#

#7x^2 -3x -10 = 0#

#(7x - 10)(x+1) = 0" factorise"#

#if 7x -10 = 0" then "x = 10/7#

#if x+1 = 0" then "x = -1"#