Let $s=x+y$ and $q=xy$. The identity $s^2=x^2+2xy+y^2$ can connect the first equation to $s$ and $q$.

After solving the quadratic in $s$, use the fact that $x+y$ must be positive to select the appropriate root.

Algebra is faster here because rewriting the system in terms of $s=x+y$ produces one quadratic equation: $s^2+6s-153=0$.

For a quick verification, enter $y=x^2+6x-153$. Click the positive $x$-intercept; it represents the positive value of $s=x+y$.

Enter $9\sqrt{2}-3$ on another line to compare its decimal value with the positive intercept.

Because $x$ and $y$ are positive, $xy\ne 0$. Multiply the first equation by $xy$:

Let $s=x+y$ and $q=xy$. Then

so $q=\frac{2}{9}s^2$.

The second equation becomes

Thus $\frac{3}{2}q+2s=51$. Substitute $q=\frac{2}{9}s^2$:

Multiplying by $3$ gives $s^2+6s-153=0$.

Using the quadratic formula,

Since $s=x+y$ must be positive, $s=9\sqrt{2}-3$. This is choice C.