if the way is right i have another question:can i skip the last step and leave the answer -1/a-3?
Hallo sabi,
your way is not wrong.
3y+ax=1|x=1a−33y+a(1a−3)=13y+aa−3=1|⋅(a−3)3y⋅(a−3)+a=a−33y⋅(a−3)+a=a−3|−a3y⋅(a−3)+a−a=a−a−33y⋅(a−3)=−3|:3y⋅(a−3)=−1|:(a−3)y=−1a−3y=1−(a−3)y=1−a+3y=13−a
so my way is wrong?
+262 
