Answer: \((x,y)=(6,-2)\) or \(\left(-7,\frac35\right)\).

Answer: \((x,y)=(6,-2)\) or \(\left(-7,\frac35\right)\).

From \(x+5y=-4\),

\(x=-4-5y.\)

Substitute this into \(3y-xy=6\):

\(3y-(-4-5y)y=6.\)

So

\(3y+4y+5y^2=6.\)

Hence

\(5y^2+7y-6=0.\)

Factorising,

\((5y-3)(y+2)=0.\)

Therefore

\(y=\frac35\quad\text{or}\quad y=-2.\)

If \(y=\frac35\), then

\(x=-4-5\left(\frac35\right)=-7.\)

If \(y=-2\), then

\(x=-4-5(-2)=6.\)

So the solutions are

\((x,y)=\left(-7,\frac35\right)\quad\text{or}\quad (6,-2).\)