Solve x³+3xy²=-49 and x²-8xy+y²=8y-17x

Solve for real solutions for the following system of equations:

$x^3+3xy^2=-49\\x^2-8xy+y^2=8y-17x$

My solution:

From $x^2-8xy+y^2=8y-17x$, we do some algebraic manipulation to get

\[ (y-4x)^2-16x^2+x^2=8y-17x\\(y-4x)^2=8y-17x+15x^2-(1) \]

\[ x^2-8xy+y^2=8y-17x\\8xy=x^2+y^2-8y+17x-(2) \]

From $x^3+3xy^2=-49$, in trying to relate the $y$ term to $y-4x$, we get

\[ x^3+3xy^2=-49\\x^3+3x(y-4x+4x)^2=-49\\x^3+3x[(y-4x)^2+8x(y-4x)+16x^2]=-49\\x^3+3x(y-4x)^2+24x^2(y-4x)+48x^2=-49\\49(x^3+1)+3x[(y-4x)^2+8x(y-4x)]=0\\49(x^3+1)+3x(8y-17x+15x^2+8xy-32x^2)=0\text{   from (1)}\\49(x+1)(x^2-x+1)+3x[8y(x+1)-17x(x+1)]=0\\(x+1)[49(x^2-x+1)+3x(8y-17x)]=0\\(x+1)[49x^2-49x+49+24xy-51x^2]=0\\(x+1)[-2x^2-49x+49+3(8xy)]=0\\(x+1)[-2x^2-49x+49+3x^2+3y^2-24y+51x]=0\text{   from (2)}\\(x+1)(x^2+2x+1+3y^2-24y+48)=0\\(x+1)[(x+1)^2+3(y-4)^2]=0 \]

This is possible if and only if $x=-1$.

When $x=-1$, we get

\[ -1-3y^2=-49\\y^2=16\\y=\pm 4 \]