How to solve this linear equation? which has an x on each side I have made this equation. 
$5x + 8 = 10x + \dfrac{3}{6}$
And I have achieved this result:
$x = 9$
Is my result correct?
I have already posted two other questions related to this topic, I'm a programmer and am learning Math out of my interest, this is not homework. 
My steps taken:
10x - 5x = 48 - 3
5x = 45
5x/5 = 45/5
x = 9
This is my steps which led to a wrong answer.

A good tip would be to put all of the variables on the left hand side of the equation and all of the numbers on the right side equation.
$5x + 8 = 10x + \dfrac{3}{6}$
First we would reduce the fraction 
$5x + 8 = 10x + \dfrac{1}{2}$
and then we would subtract  $10x $ from both sides, so we will have 
$ 5x-10x + 8 = \dfrac{1}{2}$
Then we would subtract $8$ from both sides and we will have
$ 5x-10x  = \dfrac{1}{2} - 8$
We know that $ 5x - 10x = -5x$ but what about the right hand side of the equation? We need to multiply $8$ by $\frac{2}{2}$
$ 5x-10x  = \dfrac{1}{2} - 8 \times \dfrac{2}{2}$
Now, we must simplify the terms. 
$-5x = \dfrac{1-16}{2}$
$-5x = \dfrac{-15}{2}$
Then, we divide $-5$ from both sides and get 
$ x = \frac{\dfrac{-15}{2}}{-5}$
Flipping the second fraction, we have
$ x = \dfrac{-15}{2}\frac{-1}{5}$
So the final answer is $x = \frac{-15}{10}$ which reduces to $ x = \frac{3}{2}$
Now we shall substitute $ x = \frac{3}{2}$ back into the original equation
$5 \times \dfrac{3}{2} + 8 = 10  \times \dfrac{3}{2} + \dfrac{1}{2}$
$ \dfrac{15}{2}+ 8 = \dfrac{30}{2} + \dfrac{1}{2}$
$ \dfrac{15}{2}+ 8 \times \dfrac{2}{2} = \dfrac{30}{2} + \dfrac{1}{2}$
$ \dfrac{15+16}{2} = \dfrac{30+1}{2}$
$ \dfrac{31}{2}=\dfrac{31}{2}$
Since the left hand side equals the right hand side, we have found our x. 
