how do i turn 2y+3=-1/3(x-2) into standard form.

$2y+3=\frac{-1}{3}\times(x-2)$

$2y+3=-\frac{1}{3}\times(x-2)$

$3(2y+3)=3(-\frac{1}{3}\times(x-2))$

$6y+9=3(-\frac{1}{3}\times(x-2))$

$6y+9=-1(x-2)$

$6y+9=-(x-2)$

$6y+9=-x+2$

$6y+9-9=-x+2-9$

$6y+0=-x+2-9$

$6y=-x+2-9$

$6y=-x-7$

$6y+x=-x-7+x$

$6y+x=-7+0$

$6y+x=-7$

$x+6y=-7$