Basic Algebra questions The following questions regard basic algebra;

$|5x+2| = x-1$
$|6x^2 + 7x + 2 |  + 55 |x-2| > 40$   .

I have tackled them on my self several times, and now asking for advice.
Solving the second inequality, the answer I got was, all x ( all real x) except $-\frac{2}{3}$ , $-\frac{1}{2}$, $2$.
The first inequality yields x= $-\frac{1}{6}$ , $-\frac{3}{4}$ .
According to the book, there is no solutions for the equality, and the answer to the above inequality is wrong.
 A: $|5x+2| = x - 1$ means that 1) $x-1 \ge 0$ and 2) $5x+2 = x-1$ or $5x+2 = -(x-1)$. The latter two equations should be solved:
$5x+2 = x-1$
$4x = -3$
$x = -\frac3{4}$ - but this root doesn't satisfy 1), hence it's not the root of the original equation.
$5x+2 = -x+1$
$6x = -1$
$x = -\frac1{6}$ - also doesn't satisfy 1), hence not a root.
As for the inequality, you first have to determine when the expressions under modules turn to 0 and change sign (ths is necessary to get rid of the modules and solve ordinary inequality. Since when you know that a sign of an expression is, say, -, on a given interval, you can "open" the module in the respective way. So you have to divide the real line into several intervals, on each of which all your expressions under modules have constant signs).
So first solve $6x^2+7x+2 = 0$. Determine the roots - $x_1 = -\frac2{3},x_2 = -\frac1{2}$. Then $x-2$ turns to 0 at $2$.
Obviously, $6x^2+7x+2 \ge 0$ when $x\le x_1$,  $6x^2+7x+2 \le 0$ when $x_1\le x \le x_2$ and $6x^2+7x+2 ge 0$ when $x\ge x_2$.
So there are several cases to consider. 
First, assume that $x\le x_1$. The inequality will be rewritten as $6x^2+7x+2+55(-x+2) > 40$, or $6x^2-48x+72 >0$ ... solve this and then "intersect" the solution with the condition $x\le x_1$. Thus you'll obtain one part of the whole solution. The other possibilities to consider separately are $x_1\le x \le x_2$, $x_2\le x \le 2$ and $2\le x$   ... Hope this helps... Ihope there are no mistakes ;)
