Find the a < 0 for which the inequalities have at least one common solutions Find the $a &lt; 0$ for which the inequalities $\hspace{0.5cm}2\sqrt{ax}&lt;3a—x\hspace{0.5cm}$ and $\hspace{0.5cm}x-\sqrt{\frac{x}{a}}&gt;\frac{6}{a}\hspace{0.5cm}$ have at least one solutions in common.

I don't know how to begin
Please give me a hint
For the first one to make sense, you need $x\lt 3a$. Then square both sides: $$4ax \lt 9a^2 +x^2 -6ax \\ x^2 -10ax +9a^2 \gt 0 \\ (x-9a)(x-a) \gt 0 $$  Since $x$ cannot be greater than $a$, it must be that case that $$x\lt 9a$$ The second one is $$\sqrt{ \frac xa} \lt x-\frac 6 a $$ For this to make sense, you need $x\gt \frac 6a $. Try to do a similar thing now as I did, keeping in mind the restriction. Once you get the two solution sets, you can see what $a$ must be for there to be non-zero intersection between them.
Notice that $x&lt;0$ and remember that multiplying/dividing with negative number changes the inequality sign.
You can rewrite you sytem like this, multiply second by $a$ and you get $$xa-\sqrt{ax} -6&gt;0$$
so $$(\sqrt{ax} -3)(\sqrt{ax} +2)&lt;0 \implies \sqrt{ax} &lt;3$$ First inequality is equivalent to $$( \sqrt{x} +3 \sqrt{a})( \sqrt{x} - \sqrt{a}) \implies  \sqrt{x} &lt; \sqrt{a}$$
So $${9\over a}&lt;x&lt;a \implies a^2&lt;9 \implies a\in (-3,0)$$