Question

Solve the following equation.
$x+y+xy=11,x^{2}y+x{y}^2=30$.

Answer 1

Put $x+y=u$ and $xy=v$. Then the given equations can be written as
$u+v=11,uv=30$
$\therefore$ u and v are the roots of $t^2=11t+30=0$
Hence $u=6,v=5$ or $u=5,v=6$.
Thus $x+y=6,xy=5$ or $x+y=5,xy=6$
First we take $x+y=6$ ..$\left(1\right)$
and $xy=5$
$\therefore$ x and y are the roots of $t^{-}6t+5=0$
or $(t-5)(t-1)=0$,
$\therefore t=5,1$
Hence $x=5,y=1$ or $x=1,y=5$.
We now take $x+y=5,xy=6$.
Solving these, we shall get $t^2-5t+6=0$
$x=3,y=2$ or $x=2,y=3$
Hence the solution set is $(5,1),(1,5).(3,2),(2,3)$.