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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.

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    $\begingroup$ But you haven't actually asked a question. $\endgroup$ – Gerry Myerson Sep 4 '13 at 12:03
  • $\begingroup$ Well, you see, clearly I've done something wrong in this exercise, and maybe someone could perhaps indicate what was it precisely. $\endgroup$ – Bak1139 Sep 4 '13 at 12:10
  • $\begingroup$ Ah, so you want to solve the equation and the inequality! Why didn't you say so? $\endgroup$ – Gerry Myerson Sep 4 '13 at 12:12
  • $\begingroup$ basically, yeah, that's my general intent.. $\endgroup$ – Bak1139 Sep 4 '13 at 12:16
  • $\begingroup$ In first problem, there are two cases, (i) $5x+2\ge 0$ and (ii) $5x+2\lt 0$. In Case 1, you are solving $5x+2=x-1$. That gives $x=-\frac{3}{4}$. But at that $x$ we have $5x+2\lt 0$! So Case (i) has no solution. In Case (ii), we are solving $-(5x+2)=x+1$. There is a solution $x=-\frac{1}{6}$, but it does not satisfy the condition $5x+2\lt 0$. Thus there are no solutions. $\endgroup$ – André Nicolas Sep 4 '13 at 12:23
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$|5x+2| = x - 1$ 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 ;)

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  • $\begingroup$ Very detailed and concise! $\endgroup$ – Bak1139 Sep 4 '13 at 12:28
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    $\begingroup$ I got that only 2 has to be excluded on the last step, when i was considering the case $x\ge 2$, and when i formed the respective quadratic inequality to solve, the polinomial had two roots, one of them was $-\frac{74}6$, the other was 2, so, since the inequality was strict, this number 2 fell out. The other two points didn't. $\endgroup$ – W_D Sep 4 '13 at 16:47
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    $\begingroup$ In the other cases it turned out that these segments which represent conditions under which we open the modules ($x_1\le x\le x_2$ etc.) where contained in the sets of solutions of the respective inequalities i obtained after "opening" the modules, so nothing had to be excluded... Perhaps you were considering the conditions of the form $x_1 < x < x_2$ indeed? $\endgroup$ – W_D Sep 4 '13 at 16:51
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    $\begingroup$ it will not be a mistake in case you include each end in at least one condition, because the idea is to check all points of course. For example, if there was inequality of the kind, say, $|x-1|>3$ and you would choose this method (though here it is possible to simply use the definition of the absolute value), you'd first note that $x-1$ turns into 0 at point 1. Then you'd have 2 cases to consider: 1) $x\le 1$ --> $-(x-1)>3$ --> $x<-2$ ---> together with our condition we get the same result - $x<-2$; 2) $x\ge 1$ --> $x-1>3$ --> $x>4$ ---> $x>4$, and, taking the union of these two, $\endgroup$ – W_D Sep 4 '13 at 19:54
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    $\begingroup$ you obtain $x\in (-\infty,-2)\cup(4,\infty)$. Well, yes, the conditions seem a kind of superflous here, but it's just to illustrate... You could include this point 1 only in one condition - e.g. 1) $x\le 1$ 2) $x>1$ or 1) $x<1$ 2) $x\ge 1$, but to form these conditions like 1) $x<1$ 2) $x>1$ would be a mistake, since you exclude point 1 from both of them and hence, even in case it satisfies your inequality, it will not appear in your answer (well, yes, here it doesn't satisfy, but in general this could be the case) $\endgroup$ – W_D Sep 4 '13 at 19:56

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