\((x+3) \propto y\\ ∴x + 3 = ky\hspace{1mm}when\hspace{1mm}x = 3, y = 12\\ 3+3 = 12k\\ ∴ k = \frac{1}{2}\\ \Longrightarrow x + 3 = \frac{1}{2}y\) to find x when y = 8 \(x + 3 = \frac{1}{2}\times 8\\ x=4-3\\ x = 1\)