Question

Solve the following equation.
$4x^2+5y=6+20xy-25y^2+2x$,
$7x-11y=17$.

Answer 1

First equation can be written as
$4x^2-20xy+25y^2+(5y-2x)-6=0$

or $(5y-2x{)}^2+(5y-2x)-6=0$

or $(5y-2x+3)(5y-2x-2)=0$

Hence $5y-2x+3=0$ .$\left(1\right)$

or $5y-2x-2=0$ ..$\left(2\right)$

And the second given equation is
$7x-11y=17$ ..$\left(3\right)$

Solving $\left(1\right)$ and $\left(3\right)$, we get $x=4,y=1$

Again solving $\left(2\right)$ and $\left(3\right)$, we get
$x=\frac{107}{13},y=\frac{48}{13}$

Hence solutions are: $x=4,y=1$

or $x=\frac{107}{13},y=\frac{48}{13}$