Remember to subtract you change the sign and add. Now the problem is 1/(x-3) +  -2/(3-x) = 4 (x^2-9).

Next put the minus sign into the denominator of the second term for 2/-(3-x).  Multiply the minus sign through the parenthesis to have -3 + x and rearrange x - 3.  Now the problem is 1/(x-3) + 2/(x-3) = 4 (x^2 - 9).

Factor the denominator in the answer to have x^2 - 9 = (X-3) times ( x+ 3). 

 

Multiply both sides of the equation by x-3 to have 1 + 2 = 4/ (x+3)

Multiply both sides by x + 3 to have 3x + 9 = 4

Subtract 9 form both sides to have 3 x = 5

Divide both sides by 3 to have x = 5/3

 

1/(x-3)-2/(3-x)=4/(x^2-9)

1/(x-3)+2/(x-3)=4/(x^2-9)

(1+2)/(x-3)=4/(x-3)(x+3) .....(x-3) cancels out from both sides

3=4/(x+3)  ......cross multiply

3(x+3)=4

3x+9=4

3x=-5

x=-5/3

First recall the formulas -(a - b) = (b - a) and x^2 + a^2 = (x + a)(x - a). If you rearrange the equation with these formulas in mind you get: 1/(x-3) + 2/(x-3) = 4/[(x + 3)(x - 3)] Now you can add the terms on the left hand side and multiply through by [(x + 3)(x - 3)]. That should give you a good start. Hint: the answer is a really simple number.