230. $(x+y)^{2}-(x+y)=150+2 x$.
230. $x+y=\frac{1 \pm \sqrt{8 x+601}}{2}$; therefore, $8 x+601=$ $=t^{2}, t^{2}-1$ is divisible by 8; from this it is clear that $t$ is any odd number, $t=2 t_{1}+1$. Thus, $x+y=$ $=\frac{1 \pm t}{2}, x+y=t_{1}+1$

Answer. $\left\{\begin{array}{l}x=\frac{t_{1}\left(t_{1}+1\right)}{2}-75 \\ y=76-\frac{t_{1}\left(t_{1}-1\right)}{2}\end{array}\right.$ and $\left\{\begin{array}{l}x=\frac{t_{1}\left(t_{1}+1\right)}{2}-75 \\ y=-\frac{t_{1}\left(t_{1}+3\right)}{2}+75 .\end{array}\right.$