(3) Real numbers $x, y$ satisfy $2 x^{2}+3 y^{2}=6 y$, then the maximum value of $x+y$ is
(3) $1+\frac{\sqrt{10}}{2}$ Hint: Let $x+y=t$, then $x=t-y$, from $2(t-y)^{2}+$ $3 y^{2}=6 y$, we get $5 y^{2}-2(2 t+3) y+2 t^{2}=0$, since $y$ is a real number, then the discriminant $\Delta=$ $4(2 t+3)^{2}-4 \times 5 \times 2 t^{2} \geqslant 0$, we get $\frac{2-\sqrt{10}}{2} \leqslant t \leqslant \frac{2+\sqrt{10}}{2}$.