If the system of inequalities $3x^2+2x-1<0$ and $(3a-2)x-a^2x+2<0$ possesses a solution, find the least natural number $a$ If the system of equations $3x^2+2x-1<0$ and $(3a-2)x-a^2x+2<0$ possesses a solution, find the least natural number $a$
My attempt is as follows:-
$$3x^2+3x-x-1<0$$
$$3x(x+1)-1(x+1)<0$$
$$x\in\left(-1,\dfrac{1}{3}\right)$$
$$(-a^2+3a-2)x<-2$$
$$(a^2-3a+2)x>2$$
Case $1$: $a^2-3a+2<0$
$$a\in(1,2)$$
In this interval there is no natural number, hence no need to proceed further.
Case $2$: $a^2-3a+2\ge0$
$$a\in(-\infty,1]\cup[2,\infty)$$
Checking for $a=1$:
$0>2$, which is not possible 
Checking for $a=2$
$(4-6+2)x>2$
$0>2$, which is not possible
$$x>\dfrac{2}{a^2-3a+2}$$
$$x\in\left(\dfrac{2}{a^2-3a+2},\infty\right)$$
If system of equations possess a solution, then $\dfrac{2}{a^2-3a+2}<\dfrac{1}{3}$
$$6<a^2-3a+2$$
$$a^2-3a-4>0$$
$$a^2-4a+a-4>0$$
$$a(a-4)+(a-4)>0$$
$$(a+1)(a-4)>0$$
$$a\in(-\infty,-1)\cup(4,\infty)$$
So $a=5$ should be the answer, but actual answer is $2$

Inequalities can be sumized in this system:
$x<-1$ and $x>\frac{1}{3}$
$a<=\frac{3\sqrt{x}-\sqrt{x+8}}{2\sqrt{x}}$
$a>=\frac{3\sqrt{x}+\sqrt{x+8}}{2\sqrt{x}}$
The least natural number is $a=4$ for $x=\frac{1}{3}$.
Other values are:
$a=5$ for $x=\frac{1}{6}$;
$a=6$ for $x=\frac{1}{10}$;
$a=7$ for $x=\frac{1}{15}$;
$a=8$ for $x=\frac{1}{21}$;
$a=9$ for $x=\frac{1}{28}$;
$a=10$ for $x=\frac{1}{36}$,
etc…
