+262 
if the way is right i have another question:can i skip the last step and leave the answer -1/a-3?
+26387 $$\begin{array}{rcl}
3y+ax &=& 1 \qquad | \qquad x=\dfrac{1}{a-3}\\\\
3y + a\left( \dfrac{1}{a-3} \right) &=& 1\\\\
3y + \dfrac{a}{a-3} &=& 1 \qquad | \qquad \cdot(a-3)\\\\
3y \cdot(a-3) + a &=& a-3 \\\\
3y \cdot(a-3) + a &=& a-3 \qquad | \qquad -a\\\\
3y \cdot(a-3) + a-a &=& a-a-3 \\\\
3y \cdot(a-3) &=& -3 \qquad | \qquad :3\\\\
y \cdot(a-3) &=& -1 \qquad | \qquad : (a-3)\\\\
y &=& \dfrac{-1}{ a-3 }\\\\
y &=& \dfrac{1}{ -(a-3) }\\\\
y &=& \dfrac{1}{ -a+3 }\\\\
\mathbf{y} & \mathbf{=} & \mathbf{\dfrac{1}{ 3-a } }
\end{array}$$
