<s>[INST] solving $|2x+1|-|5x-2|\geq1$ Solve:
$|2x+1|-|5x-2|\geq1$
$\Rightarrow |2x+1|\geq1+|5x-2|$
Then I squared the inequality
$\Rightarrow 4x^2-4x+1\geq1+2|5x-2|+25x^2-20x+4$
$\Rightarrow-21x^2+16x-2|5x-2|-4\geq0$
Then I separated into two cases
The first case
$$\begin{cases}-21x^2+16x-10x+4-4\geq0...(a)\\x\geq0...(b) \end{cases}
$$
From (a)  we get that 
$x\in[0,\frac{2}{7}]$
$(a)\cap(b)\\
\Rightarrow x\in[0,\frac{2}{7}]$
The second case
$$\begin{cases}-21x^2+16x+10-4-4\geq0...(c)\\x<0...(d)\end{cases}$$
From (c) we get that 
$x\in[\frac{4}{7},\frac{2}{3}]$
$(c)\cap(d)\\
\Rightarrow x\in\emptyset$
So the solutions to the initial inequality should be $x\in [0,\frac{2}{7}]$
But wolfram said the solutions were $[\frac{2}{7},\frac{2}{3}]$
Where did I make the mistake???
 [/INST] Your first case is 
a) $-21x^2 + 16x - 10x + 4 - 4 \ge 0$
b) $5x-2\ge 0$
so by a) $(21x-6)x \le 0$ so either $x \le 0$ and $x \ge \frac 27$ (impossible) or $x \ge 0$ and $x \le 27$ so $[0, \frac 27]$.  
From b) $x \ge \frac 25$ and $a\cap b = \emptyset$.
Second case
c) $-21x^2 + 26x -8\ge 0$ and d) $5x - 2< 0$
For c) $(7x -4)(3x -2) \le 0$ and so $\frac 47\le x \le \frac 23$.
And d) yields $x < \frac 25$.  $c\cap d= c$ 
So solution is $[\frac 47, \frac 23]$.
....
Your error was takeing b) and d) as $x\ge 0$ and $x < 0$ and I honestly have no idea why you did that.  The term we had in modulus was $|5x-2|$ so you need to take those as $\ge, < 0$.
</s>