How to solve equation with unknown in the denominator eg 2x+3/x-4 - 2x-8/2x+1=1
+129849 (2x+3) / (x-4) - (2x-8) / (2x+1) =1 multiply both sides by the common denominator of (x -4)(2x + 1)
So we have....
(2x + 3)(2x+ 1) - [(2x - 8)(x -4)] = (x -4)(2x + 1) simplify
4x^2 + 8x + 3 - [ 2x^2 - 16x + 32] = 2x^2 - 7x - 4
2x^2 + 24x -29 = 2x^2 - 7x - 4
Subtract 2x^2 from both sides........add 7x, 29 to both sides
31x = 25 divide both sides by 31
x = 25 / 31
+154 What you do is that you time the denominator with the number diagonally opposite.
You will get (2x+3)(2x+1)-(x-4)(2x-8)=1
It is now an wasy quadratic equation which can now be solved easily :)
+154 By the way im back!!!
Sorry for the absence, hopefully it wont happen again until June (When i have my very important maths exam)
+129849 (2x+3) / (x-4) - (2x-8) / (2x+1) =1 multiply both sides by the common denominator of (x -4)(2x + 1)
So we have....
(2x + 3)(2x+ 1) - [(2x - 8)(x -4)] = (x -4)(2x + 1) simplify
4x^2 + 8x + 3 - [ 2x^2 - 16x + 32] = 2x^2 - 7x - 4
2x^2 + 24x -29 = 2x^2 - 7x - 4
Subtract 2x^2 from both sides........add 7x, 29 to both sides
31x = 25 divide both sides by 31
x = 25 / 31
+154 ohhhh yes.
I completely forgot to times 1 by the two equations...
Im an idiot.
Anyway good working Cpill, hope you appreciate the 5 stars :)
