Question

Solve $x+2+y+3+\sqrt{(x+2)(y+3)}=39$.
$(x+2{)}^2+(y+3{)}^2+(x+2)(y+3)=741$.

Answer 1

$x+2+y+3+\sqrt{(x+2)(y+3)}=39$
${(x+2)}^2+{(y+3)}^2+(x+2)(y+3)=741$
Let $x+2=p,y+3=q.$
$p+q+\sqrt{pq}=39⟶I$
$p^2+q^2+pq=741⟶II$
$⟹p+q-39=-\sqrt{pq}$
Squaring on both sides:-
$⟹p^2+q^2+1521+2pq-78p-78q=pq$
$⟹p^2+q^2+pq=78p+78q-1521⟶III$
Subtracting $II$ from $III$
$78p+78q-1521-741=0$
$78p+78q-2262=0$
$p+q=29⟶IV$
From $I$
$\sqrt{pq}=10$
$pq=100⟶V$
$p-q=\sqrt{{(p+q)}^2-4pq}$
$=\sqrt{841-400}$
$=\sqrt{441}=21⟶VI$
Adding $IV$ and $VI$
$2p=50$
$p=25,q=4$
$x+2=25,y+3=4$
$x=23,y=1$