Question

Solve the following systems of equations.
$2y^2-4xy+3x^2=17,y^2-x^2=16$

Answer 1

$2y^2-4xy+3x^2=17....\left(i\right)$

$y^2-x^2=\mathrm{16...}\left(ii\right)$

Simplifying equation (ii)

$y^2-x^2=16$

$y^2=16+x^2...\left(iii\right)$

Putting this value in equation (i)

$2(16+x^2)-4xy+3x^2=17$

or, $32+2x^2-4xy+3x^2-17=0$

or, $5x^2-4xy+15=0$

or, $x(5x-4y)+15=0$

or, $5x-4y=\frac{-15}{x}$

or, $4y-5x=\frac{15}{x}$

Hence, every pair of $x$ & $y$ satisfying this equation

will be a solution of given pair of equations.

So, there will be infinite solutions.