System of parametric inequalities I'm still struggling with the parametric inequalities.I'm doing some systems of parametric inequalities, but I can not understand how to proceed.This is the system:
$x-2a<1+a$
$\frac x2 -2a<2x-(a-3)$
$\frac 13 -x<a$ per $a>0$
These are the solutions I've found:
$x<3a+1$
$x>{-2a-6\over 3}$
$x>{-3a+1\over 3}$
But now I do not know how to proceed. I thought I could compare the solutions like:
$3a+1>{-2a-6\over 3}>{-3a+1\over 3}$
If you could help me, I'd be really grateful. 
And if it's not too much trouble, if you could recommend some books or a detailed explanation on the literal inequalities and elementary algebra.
 A: The strategy when you have two inequalities, like $x>\frac{6-2a}3$ and $x>\frac{1-3a}3$, where the signs are the same (i.e. the both say $x>\text{expression}$) is to compare the things $x$ is compared to. So we will determine whether $\frac{-6-2a}3>\frac{1-3a}3$ or not.
For that inequality, one should solve the corresponding equation $\frac{-6-2a}3=\frac{1-3a}3$. Its solution is $a=7$. Now, one can solve the inequality. Pick a test value $a=0$, and plug it in both sides. On the left hand side it is $-2$ and on the right, $1/3$. Thus $\frac{-6-2a}3<\frac{1-3a}3$ holds for all $a<7$, while $\frac{-6-2a}3>\frac{1-3a}3$ holds for $a>7$.
Therefore, the assertion that $x$ is greater than both $\frac{-6-2a}3$ and $\frac{1-3a}3$ is the same as saying $x>\frac{1-3a}3$ when $a\le 7$ and $x>\frac{-6-2a}3$ when $a\ge7$. 
Since we only have one inequality of the form $\text{expression}>x$, we do not have to do anything else, as the system of inequalities now splits into two pieces: $$1+3a > x >\frac{1-3a}3\text{ if }a\le7\qquad1+3a>x>\frac{-6-2a}3 \text{ if }a\ge7$$ which, without any further information, counts as a solution.
