Can this method be also right?
\(\displaystyle \L \;[1]5x\,-\,y\,=\,2\)Yes
\(\displaystyle \L \;[2]-y\,=\,2\,-\,5x\)Yes
\(\displaystyle \L \;[3]y\,=\,-(2\,-\,5x)\,=\,5x\,-\,2\)No because you are distributing the negative but the 2 has no negative. If you wanted to make the right side negative you would have to multiply both sides of the equation by a negative number.
So on step [2] lets divide both sides by negative 1:\(\displaystyle \L \;[4]y\,=\,5x\,-\,2\)
That is what you got but your steps weren't correct.
\(\displaystyle \L \;y\,=\,5x\,-\,2\) or \(\displaystyle x\,=\,2\,+\,y\,/5\,\)
.....
No. Once we got this equation we switch the x and y... NOT solve for x. When we switch the
x and y we then solve for y.
\(\displaystyle \L
\begin{array}{l}
[4]\underbrace y_{} = 5\underbrace x_{} - 2 \\
[5]\underbrace x_{} = 5\underbrace y_{} - 2 \\
\end{array}\)
Now solve [5] for y.