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

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

\(6y+0=-x+2-9\)

 

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

 

\(6y=-x-7\)

 

\(6y+x=-x-7+x\)

 

\(6y+x=-7+0\)

 

\(6y+x=-7\)

 

\(x+6y=-7\)

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