How do i solve this equation?

please answer i have test in 2 days

Solve for x:
(5 x)/(6)-3/8 = (5 x)/12+11/24

Put each term in (5 x)/(6)-3/8 over the common denominator 24: (5 x)/(6)-3/8 = (20 x)/24-(9)/24:
(20 x)/24-(9)/24 = (5 x)/12+11/24

(20 x)/24-(9)/24 = (20 x-9)/24:
(20 x-9)/24 = (5 x)/12+11/24

Put each term in (5 x)/12+11/24 over the common denominator 24: (5 x)/12+11/24 = (10 x)/24+11/24:
(20 x-9)/24 = (10 x)/24+11/24

(10 x)/24+11/24 = (10 x+11)/24:
(20 x-9)/24 = (10 x+11)/24

Multiply both sides by 24:
20 x-9 = 10 x+11

Subtract 10 x from both sides:
(20 x-10 x)-9 = (10 x-10 x)+11

20 x-10 x = 10 x:
10 x-9 = (10 x-10 x)+11

10 x-10 x = 0:
10 x-9 = 11

Add 9 to both sides:
10 x+(9-9) = 11+9

9-9 = 0:
10 x = 11+9

11+9 = 20:
10 x = 20

Divide both sides of 10 x = 20 by 10:
(10 x)/10 = 20/10

10/10 = 1:
x = 20/10

The gcd of 20 and 10 is 10, so 20/10 = (10×2)/(10×1) = 10/10×2 = 2:
Answer: |x = 2

How do i solve this equation?

5x/6 - 3/8 = 11/24 + 5x/12

$\begin{array}{|rcll|} \hline x\cdot\frac56 - \frac38 &=& \frac{11}{24} + x\cdot\frac{5}{12} \qquad &| \qquad \cdot 12\\\\ x\cdot\frac56 \cdot 12 - \frac38 \cdot 12 &=& \frac{11}{24} \cdot 12 + x\cdot\frac{5}{12} \cdot 12\\\\ x\cdot 5 \cdot 2 - 3\cdot \frac{12}{8} &=& \frac{11}{2} + x\cdot 5\\\\ 10x- 3\cdot \frac{3}{2} &=& \frac{11}{2} + 5x \\\\ 10x - \frac{9}{2} &=& \frac{11}{2} + 5x \qquad &| \qquad -5x\\\\ 10x -5x -\frac{9}{2} &=& \frac{11}{2} \qquad &| \qquad + \frac{9}{2}\\\\ 10x -5x &=& \frac{11}{2} + \frac{9}{2}\\\\ 5x &=& \frac{11+9}{2} \\\\ 5x &=& \frac{20}{2} \\\\ 5x &=& 10 \qquad &| \qquad : 5 \\\\ x &=& \frac{10}{5} \\\\ \mathbf{ x } & \mathbf{=} & \mathbf{2} \\ \hline \end{array}$