Equation containing fractions

Question

Hi, I don't know where I'm going wrong with this equation but I'm pretty sure I Missed some minor step in solving it. 

 

The equation is : 2x/(x-3) = 10/(x+1) — (7x-27)/(x-3) 

Answer 1

The domain of these rational functions are

all reals EXCEPT x=-1 and x=3

Multiplies both sides by LCD=(x-3)(x+1)

2x(x+1) = 10(x-3) -(7x-27)(x+1)

2x^2 + 2x = 10x - 30 - [7x^2 + 7x - 27x - 27]

2x^2 + 2x = 10x - 30 - [ 7x^2 - 20x - 27]

2x^2 + 2x = 10x - 30 - 7x^2 + 20x + 27

2x^2 + 2x = -7x^2 + 30x - 3

9x^2 - 28x + 3 = 0

(9x - 1)(x - 3) = 0

(x-3) = 0 ---> x = 3 which

is disqualified because it

results in division by zero

9x-1=0 --> x = 1/9

check:

Left side: 2x = 2/9 and x-3 = -2 and 8/9 = -26/9

dividing them : 2/9 * 9/-26 = -1/13

So the left side is -1/13 when x=1/9

Right side:

x-3 = -26/9 is already done...

x+ 1 = 1/9 + 1 = 10/9

7x - 27 = 7/9 - 27 = 7/9 - 243/9 = -236/9

THen 10/(x+1) = 10 divided by 10/9 = 10/1 * 9/10 = 9 <-- first term

the 2nd term is

(7x-27)/(x-3) = (-236/9) divided by (-26/9)

= (-236/9)(9/-26) = 236/26 = 118/13

subtracting them 9 - 118/13 = 117/13 - 118/13 = -1/13

so the right side is -1/13 when x = 1/9

they check !!! the answer is correct !!!

x=1/9