5. Find the maximum and minimum of $x+y$ such that
$$
x+y=\sqrt{2 x-1}+\sqrt{4 y+3}
$$
5. We have
$$
(x+y) / 3=(\sqrt{2 x-1}+\sqrt{y+3 / 4}+\sqrt{y+3 / 4}) / 3 \leq \sqrt{(2 x+2 y+1 / 2) / 3} .
$$
and
$$
(x+y)=\sqrt{2 x-1}+\sqrt{y+3 / 4}+\sqrt{y+3 / 4} \geq \sqrt{2 x+2 y+1 / 2} .
$$

Let $t=x+y$. Then we have
$$
\begin{array}{l} 
2 t^{2}-12 t-3 \leq 0 \Rightarrow 3-\sqrt{21 / 2} \leq t \leq 3+\sqrt{21 / 2} \\
\text { and } \quad 2 t^{2}-4 t-1 \geq 0 \quad \Rightarrow \quad t \geq 1+\sqrt{3 / 2}, \quad t \leq 1-\sqrt{3 / 2}
\end{array}
$$

Since $t \geq 0$, the maximum value is $3+\sqrt{21 / 2}$ and the minimum value is $1+\sqrt{3 / 2}$. Note that the maximum value is attained by the solution of
$$
x+y=3+\sqrt{21 / 2} \text { and } 2 x-1=\sqrt{y+3 / 4}
$$
while the minimum is attained by the solution of
$$
x+y=1+\sqrt{3 / 2} \text { and } y+3 / 4=0 .
$$