Giải thích các bước giải:
Ta có :
$ax-y=2\to y=ax-2$
$\to x+a(ax-2)=3\to x(a^2+1)=2a+3$
$\to x=\dfrac{2a+3}{a^2+1}, a^2+1>0$
$\to y=a.\dfrac{2a+3}{a^2+1}-2=\dfrac{3a-2}{a^2+1}$
b.Để $x+y>0\to \dfrac{2a+3}{a^2+1}+\dfrac{3a-2}{a^2+1}>0$
$\to \dfrac{5a+1}{a^2+1}>0$
$\to 5a+1>0 \to a>-\dfrac 15$
c.Để $x=\sqrt{2}y$
$\to \dfrac{2a+3}{a^2+1}=\sqrt{2}.\dfrac{3a-2}{a^2+1}$
$\to 2a+3=\sqrt{2}.(3a-2)$
$\to \left(2-3\sqrt{2}\right)a=-2\sqrt{2}-3$
$\to a=\dfrac{18+13\sqrt{2}}{14}$