Question

Solve the following systems of equations.
$x^2-xy+y^2=21,y^2-2xy+15=0$

Answer 1

$x^2-xy+y^2=21...\left(i\right)$

$y^2-2xy+15=0....\left(ii\right)$

Simplifying equation (ii)

$y^2-2xy+15=0$

or, $y^2=(-15)+2xy$

or, $y^2=2xy-15...\left(iii\right)$

Putting this value of $y^2$ in equation (i)

$x^2-xy+(2xy-15)=21$

or, $x^2-xy+2xy-15-21=0$

or, $x^2+xy=36$

or, $x(x+y)=36$

or, $(x+y)=\frac{36}{x}...\left(iv\right)$

Hence, solution of the given pair of equations will

be every pair of $x$ & $y$ which satisfies equation (iv).

So, Here will be infinite solutions.