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: 
It is very easy to see if your result is NOT correct. If $x=9$, then the left side of your equation is
$$5 \cdot 9  + 8 = 45+8=53$$
and the right is
$$10\cdot 9 + \frac36 = 90 + \frac12$$
which is not the same.

Hint: 
When solving the equation, first move all the values containing $x$ on one side, so you get
$$5x-10x = -8+\frac36$$ and solve from here on.
A: $5x + 8 = 10x + 3/6$
Let's gather all the x's on one side and the other stuff on the other side as well as reduce the fraction to lowest terms first:
$10x-5x = 8-\frac{1}{2}$
$5x = 8-.5$
$5x = 7.5$
$x = 1.5$ or $x=\frac{3}{2}$ or $x=1+\frac{1}{2}$  
Now, to compute each side as a verfication:
$LHS= 5x+8 = 5*\frac{3}{2}+8 = 7.5+8 = 15.5$
$RHS = 10x+\frac{3}{6} = 10*\frac{3}{2}+\frac{1}{2} = 15+0.5 = 15.5$
A: $$5x+8 = 10x + \frac{3}{6} \overset{(1)}{\iff} 8-\frac{3}{6}=10x-5x \overset{(2)}{\iff} \frac{45}{6} = 5x  \overset{(3)}\iff x=\frac{9}{6}\overset{(4)}{=} \frac{3}{2} $$
Steps:
(1) subtract $5x+\frac{3}{6}$ on both sides of the equation
(2) Simplify, note that $8-\frac{3}{6} = \frac{48}{6}-\frac{3}{6} = \frac{45}{6}$
(3) Divide by $5$ (or, equivalently, multiply by $\frac{1}{5}$), note that $\frac{45}{6}\cdot\frac{1}{5} = \frac{9\cdot 5}{6}\cdot\frac{1}{5} = \frac{9}{6}$
(4) Simplify, note that $\frac{9}{6}=\frac{3\cdot 3}{2\cdot 3} = \frac{3}{2}$.         
A: It's very simple to see whether your solution is correct or not. You can simply take your $x$ and put back into the original equation. Remember, the equal sign ($=$) means that both sides have to be the same and therefore, if both sides are not equivalent then your $x$ value is wrong and you have gone wrong somewhere. 
$$5 * (9) + 8 \not= 10 * (9) + \frac{3}{6}$$
If you to solve the equation then follow these steps:


*

*Simplify both sides as much as possible. For example, get rid of all the fractions

*Bring all your $x$'s to one side

*Isolate $x$ by following order of operations BEDMAS/PEDMAS

*Plug $x$ back into the original equation and see if you are correct


Good luck!
A: 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. 
A: It looks like your first attempt to solve this went wrong at the first step. By the usual method you would get $10x-5x$ on one side of the equation and $8-\frac{3}{6}$ on the other. You got the $10x-5x$ on one side all right, but for some reason you decided to multiply the other side by a factor of $6$. 
A step that multiplies things by $6$ is a reasonable step for this problem, but you have to do it on both sides of the equation at the same time, not just on one side.
