Solve the simultaneous equations
$4x^{2}+y^{2}=10, 2x-y=4$

:

$2x-y=4 \\ \therefore y=2x-4$

Substituting the value of y in $4x^{2}+y^{2}=10$, we get

$4x^{2}+\left( 2x-4\right) ^{2}=10 \\ 4x^{2}+\left( 2x\right) ^{2}+2\times \left( 2x\right) \times \left( -4\right) +\left( -4\right) ^{2}=10 \\ 4x^{2}+4x^{2}-16x+16=10 \\ 8x^{2}-16x+16-10=0 \\ 8x^{2}-16x+6=0$

Dividing both sides by 2, we get

$4x^{2}-8x+3=0$ which can be factorised as follows;

$4x^{2}-6x-2x+3=0 \\ 2x\left( 2x-3\right) -1\times \left( 2x-3\right) =0 \\ \left( 2x-3\right) \left( 2x-1\right) =0$

$\therefore 2x-3=0$ or $2x-1=0$

or, $2x=3$ or $2x=1$

thus, $x=\dfrac {3}{2}$ or $x=\dfrac {1}{2}$

Going back to the simultaneous equations, $y=2x-4$

Thus for $x=\dfrac {3}{2}$, $\space y=2\times \dfrac {3}{2}-4=3-4=-1$

For $x=\dfrac {1}{2}$, $\space y=2\times \dfrac {1}{2}-4=1-4=-3$