Find all the values of the parameter p for which the following system: $$\begin{cases}px + 3y = -p \\ 3x+py \; \;= 8 \end{cases}$$ has $x \ge 0, y\ge 0$ solutions.
I got $(-3;8]$ but in answers there is $[-3;0]$. Just added the two equations and got $$x+y=(8-p)/(p+3).$$ Where is the mistake? Please help me without using calculus. Thanks.
Hint: From the first equation we get $$y=-\frac{1}{3}(p+px)$$ plugging this in the second equation we obtain $$3x-\frac{p^2}{3}-\frac{p^2}{3}x=8$$ so $$x(9-p^2)=24+p^2$$ Can you finish? From the solution for $$x$$ we know that $$9>p^2$$ or $$3>|p|$$ for $y$ we get $$y=-\frac{11p}{9-p^2}$$ Since we have $$3>|p|$$ we get $$-11p<0$$ so $$p<0$$ and we get for $p$: $$-3<p<0$$