$$x-y+3=0, { x }^{ 3 } - { y }^{ 3 } +49=$$
$x=y-3\text{, }y=\frac{\sqrt{169-4z}}{6}+\frac{3}{2}\text{, }z\in \mathrm{C}$
$x=y-3\text{, }y=-\frac{\sqrt{169-4z}}{6}+\frac{3}{2}\text{, }z\in \mathrm{C}$
$x=y-3\text{, }y=\frac{\sqrt{169-4z}}{6}+\frac{3}{2}\text{, }z\leq \frac{169}{4}$
$x=y-3\text{, }y=-\frac{\sqrt{169-4z}}{6}+\frac{3}{2}\text{, }z\leq \frac{169}{4}$
