Question

Solve the following equation.
$x^2+y(x+1)=17,y^2+x(y+1)=13$.

Answer 1

The given equations are
$x^2+xy+y=17$ $\left(1\right)$
$y^2+xy+x=13$ ..$\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$, we get
$x^2+2xy+y^2+(x+y)=30$
or $(x+y{)}^2+(x+y)=30=0$
or $(x+y+6)(x+y-5)=0$
$\therefore x+y=-6$ $\left(3\right)$
or $x+y=5$. .$\left(4\right)$
We have now to solve equations $\left(1\right)$ and $\left(3\right)$, and $\left(1\right)$ and $\left(4\right)$
Or we may solve $\left(2\right)$ and $\left(3\right)$; and $\left(2\right)$ and $\left(4\right)$
Eliminating y from $\left(1\right)$ and $\left(3\right)$, we get
$-6x-x-6=17$ or $x=-23/7$
Then $y=-x-6=23/7-6=-19/7$
$\therefore$ One solution is:
$x=-23/7,y=-19/7$
Similarly solving $\left(1\right)$ and $\left(4\right)$, we shall get
$x=3$ and $y=2$
Hence the solutions are
$x=-\frac{23}{7},y=-\frac{19}{7}$ or $x=3,y=2$.