Solve this equation please?

Multiply both sides by x+2 and 2x-3:

6(2x - 3) + x(x+2) = (x+2)(2x - 3)

Expand out:

12x - 18 + x² + 2x = 2x² - 3x + 4x - 6

Combine terms:

x² + 14x - 18 = 2x² + x - 6

Subtract to get everything on one side, I'm going to go to the right side to keep a positive x² term:

0 = x² - 13x + 12

Alternatively, you could go to the left and then reverse the sign (equivalent to multiplying both sides by -1):

-x² + 13x - 12 = 0

x² - 13x + 12 = 0

Now factor:

(x - 1)(x - 12) = 0

x = 1 or x = 12

Double-check:

x = 1:

6/(1+2) + 1/(2 - 3) =? 1

6/3 + 1/-1 =? 1

2 - 1 = 1 (correct)

x = 12:

6/(12+2) + 12/(24 - 3) =? 1

6/14 + 12/21 =? 1

3/7 + 4/7 = 1 (correct)

Answer:

x = 1

or

x = 12

Will assume that question should read as :-

6 / (x + 2) + x / (2x - 3) = 1 ________Note brackets

6 ( 2x - 3 ) + x ( x + 2 ) = ( x + 2 ) ( 2x - 3 )

12x - 18 + x² + 2x = 2x² + x - 6

x² - 13 x + 12 = 0

( x - 12 ) ( x - 1 ) = 0

x = 1 , x = 12

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

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

x^2 + 14x - 18 = 2x^2 + x - 6

x^2 - 13x + 12 = 0

(x - 1)(x - 12) = 0

Solutions:

x = 1

x = 12

Well,

we have :

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

multiplying by (x+2)(2x-3) gives :

6(2x-3) + x(x+2) = (x+2)(2x-3)

12x - 18 + x^2 + 2x = 2x^2 + x - 6

from right to left

x^2 - 13x + 12

1 is obvious root so

(x-1)(x - 12) = 0

finally, the set S of solutions is :

---------------------> S = { 1 , 12 }

hope it will help !!

Do you mean 6/(x+2) + x/(2x-3) = 1 ?

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

6(2x-3) + x(x+2) = (x+2)(2x-3)

12x-18 + x²+2x = 2x² + x - 6

x² - 13x + 12 = 0

(x - 12)(x - 1) = 0

x = 12, 1