5x+2/2=2x-1

I did consider this.

$$\frac{5x+2}{2}=2x-1$$

$$5x+2=2{(2x-1)}$$

$$5x+2=4x-2$$

$$x=-4$$

$$5x+\frac{2}{2}=2x-1$$

$$5x+1=2x-1$$

$$5x=2x-2$$

$$3x=-2$$

$$x=\frac{-2}{3}$$

Nice work Zacismyname!!!

But is it possible that  there is a couple of brackets missing?

$${\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}{{\mathtt{2}}}} = {\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}$$        multiplie all terms with 2 in order to remove the denominator on the left side.

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}$$      get all x to the left side and all terms without x on the other side.

$$\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right){\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}} = \left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right){\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}$$

Good luck