Given $\tan(2x+y)=2$ and $\tan 2x=\dfrac32$, the value of $\tan y$ is:

1. Write $y$ as a difference: $y=(2x+y)-2x$. 2. Apply $\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$ with $A=2x+y$ and $B=2x$. 3. Numerator: $2-\dfrac32=\dfrac12$. 4. Denominator: $1+2\times\dfrac32=1+3=4$. 5. So $\tan y=\dfrac{1/2}{4}=\dfrac18$. Subtracting the tangents directly, without the denominator, gives $\tfrac12$. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 5: Trigonometry_