Hello I'm having trouble on what to do first with this fractional 1 varible equation's

Solve for x:
(3 x)/4 + 5/8 = 4 x

Put each term in (3 x)/4 + 5/8 over the common denominator 8: (3 x)/4 + 5/8 = (6 x)/8 + 5/8:
(6 x)/8 + 5/8 = 4 x

(6 x)/8 + 5/8 = (6 x + 5)/8:
(6 x + 5)/8 = 4 x

Multiply both sides by 8: or: cross multiply
6 x + 5 = 32 x

6x - 32x = -5

- 26x = - 5                      multiply both sides by -1

x =5 / 26

It appears there are four "zero-width space" characters in your LaTeX code. Two of them are between the  3/4  and the  x  which is why it doesn't look normal. I found that out by pasting your code into this website: https://www.soscisurvey.de/tools/view-chars.php

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As for your actual question, this is pretty much the method guest used:

$\frac34x+\frac58=4x$

                                And  $\frac34x=\frac{3x}{4}$  so we can rewrite it like this...

$\frac{3x}{4}+\frac58=4x$

                                To get a common denominator, we can multiply  $\frac{3x}4$  by  $\frac22$  which won't change its value.

$\frac22\cdot\frac{3x}4+\frac58=4x$

                                And  $\frac22\cdot\frac{3x}4=\frac{6x}8$     ←That's how it went from  3x  to  6x

$\frac{6x}8+\frac58=4x$

                                Now that the fractions have a common denominator we can combine them.

$\frac{6x+5}{8}=4x$

                                Multiply both sides of the equation by  8

$6x+5=32x$

                                Subtract  6x  from both sides.

$5=32x-6x$

$5=26x$

                                Divide both sides by  26

$\frac5{26}=x$

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Also, we could multiply both sides of the equation by  8  to begin with, like this:

$\frac34x+\frac58\ =\ 4x\\~\\ 8\big(\frac34x+\frac58\big)\ =\ 8\big(4x\big)\\~\\ 8\cdot\frac34\cdot x\ +\ 8\cdot\frac58\ =\ 8\cdot4\cdot x\\~\\ 6x\ +\ 5\ =\ 32x\\~\\ 5\ =\ 26x\\~\\ \frac{5}{26}\ =\ x$_

$\dfrac{3}4 x + \dfrac{5}8 = 4x\\ 6x + 5 = 32x\\ 26x = 5\\ x = \dfrac{5}{26}$