Once we know that $(x,y)$ satisfy $2x + y =  8$ and $x^2 + y^2 = 16$, we can solve for $y$ in the first equation and substitute in the second to obtain $$x^2 + (-2x + 8)^2 = 16,$$ so $5x^2 - 32x + 64 = 16$.

Solving this equation gives $x = 4$ or $x = \dfrac{12}{5}$. Then for each of these we can find $y$, which gives the solutions $(4, 0)$ and $\left(\dfrac{12}{5},\dfrac{16}{5}\right)$.