Question

Solve the following systems of equations.
$4x^2+y^2-2xy=7,(2x-y)y=y$

Answer 1

$4x^2+y^2-2xy=\mathrm{7...}\left(i\right)$

$(2x-y)y=y...\left(ii\right)$

Simplifying equation (i)

We know that $a^2+b^2-2ab=(a-b{)}^2...(iii)$

So, $4x^2+y^2-2xy=7$

can be written as

$(2x{)}^2+(y{)}^2-4xy+2xy=7$

or, $(2x{)}^2+y^2-2\times 2x\times y+2xy=7$

or, $(2x-y{)}^2+2xy=7...(iv)$ [from equation (iii)]

Now, simplifying equation (iii)

$(2x-y)y=y$

or, $(2x-y)=\frac{y}{y}=\mathrm{1...}\left(v\right)$

Putting this value in equation (iv)

$1^2+2xy=7$

or, $2xy=7-1=6$

or, $xy=\frac{6}{2}=3$

Hence, the solution of these equations will be all

the values of $(x,y)$ pair which satisfy $xy=3$

So, there will be infinite solutions