Answer
The solution is ($\sqrt{2}, \sqrt{2})$ and ($-\sqrt{2}, -\sqrt{2})$.
See Graph I for the graphing of the solution.
Work Step by Step
$x^2 + 4y^2 = 10 \longrightarrow (i)$
$y = x \longrightarrow (ii)$
Sub. $(ii)$ into $(i)$, we have
$x^2 + 4x^2 = 10$
$5x^2 = 10$
$x^2 = 2$
$x = \sqrt{2}$ or $x = -\sqrt{2}$
From $(ii)$,
when $x = \sqrt{2}$, $y = \sqrt{2}$;
when $x = -\sqrt{2}$, $y = -\sqrt{2}$
The solution is ($\sqrt{2}, \sqrt{2})$ and ($-\sqrt{2}, -\sqrt{2})$.
See Graph I for the graphing of the solution.