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???
 A: What about separating it by cases? 
Let us consider that $x\geq 2/5$. Then we have
\begin{align*}
|2x + 1| - |5x - 2| = (2x + 1) - (5x - 2) = 3 - 3x \geq 1 \Longleftrightarrow 3x \leq 2 \Longleftrightarrow x\leq 2/3
\end{align*}
Hence the first solution set is given by $S_{1} = [2/5,2/3]$.
We shall then solve the inequation when $-1/2\leq x \leq 2/5$. In this case, we get
\begin{align*}
|2x + 1| - |5x - 2| = (2x + 1) + (5x - 2) = 7x - 1 \geq 1 \Longleftrightarrow 7x \geq 2 \Longleftrightarrow x\geq 2/7
\end{align*}
Consequently, the second solution set is given by $S_{2} = [2/7,2/5]$.
Finally, we have the case when $x\leq -1/2$, from whence we obtain
\begin{align*}
|2x + 1| - |5x - 2| = -(2x + 1) + (5x - 2) = 3x - 3 \geq 1 \Longleftrightarrow 3x \geq 4 \Longleftrightarrow x \geq 4/3
\end{align*}
Therefore the third solution set is given by $S_{3} = \varnothing$.
Gathering all the solution sets, it results that $S = S_{1}\cup S_{2}\cup S_{3} = [2/7,2/3]$, which coincides with the solution proposed by wolfram alpha.
Hopefully it helps.
A: 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 $\ge, < 0$.
