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How do I solve the following absolute value inequality and inequality problems properly?

1) $\newcommand\abs[1]{|#1|}\abs{2x+9}>x$

Solving this problem algebraically, I get

When $x > 0, x > -9$

Steps:

$2x + 9 > x$

$x > -9$

When $x < 0, x < -3$

Steps:

$2x + 9 < -x$

$x < -3$

However, the complete answer is all real numbers. What are the steps that I need to take to get the complete answer?

2) $15x^2 - 2 > x$

My solution: $15x^2 - 2 > x$

$15x^2 - 2 - x > 0$

$(3x+1)(5x-2) > 0$

$(3x+1) > 0$ and $(5x-2) > 0$

or

$(3x+1) < 0$ and $(5x-2) < 0$

Case: $(3x+1) > 0$ and $(5x-2) > 0$

Solution: $x > -1/3$ and $x > 2/5$

Case: $(3x+1) < 0$ and $(5x-2) < 0$

Solution: $x < -1/3$ and $x < 2/5$

Final Solution: $x > -1/3$ and $x > 2/5$ or $x < -1/3$ and $x < 2/5$

However, the answer to this problem is: $x > 2/5$ or $x < -1/3$

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  • $\begingroup$ For the second one, your Final Solution is technically correct. However, saying that $x\gt -1/3$ and $x\gt 2/5$ is a little strange, since $x\gt 2/5$ implies $x\gt -1/3$. So the condition $x\gt -1/3$ and $x\gt 2/5$ is equivalent to $x\gt 2/5$. For the first problem, the algebraic solution went awry. Note that if $x\lt 0$ the inequality automatically holds. $\endgroup$ – André Nicolas Dec 9 '13 at 1:35
  • $\begingroup$ I just added my steps for the first problem. Where did I go wrong algebraically? $\endgroup$ – FearlessFuture Dec 9 '13 at 1:47
  • $\begingroup$ There is no absolute value around the $x$. We have (i) $|2x+9|=2x+9$ if $x\ge -9/2$, and (ii) $|2x+9|=-(2x+9)$ if $x\lt -9/2$. Case (i): We want $2x+9\gt x$, that is, $x\gt -9$. We also need $x\ge -9/2$. This combines to simply $x\ge -9/2$. Case (ii) We want $-2x-9\gt x$, so $x\lt -3$. We also want $x\lt -9/2$, so all we need is $x\lt -9/2$. We have seen the inequality holds if (i) $x\ge -9/2$ and (ii) for $x\lt -9/2$, in other words, always. $\endgroup$ – André Nicolas Dec 9 '13 at 1:55
  • $\begingroup$ The above solution used the definition of absolute value mechanically. It is better to note that the inequality automatically holds if $x\lt 0$. And if $x\ge 0$, $|2x+9|=2x+9$. But $2x+9$ is clearly $\gt x$ if $x\ge 0$. $\endgroup$ – André Nicolas Dec 9 '13 at 2:05
  • $\begingroup$ @AndréNicolas When I apply the "mechanical" method to $\newcommand\abs[1]{|#1|}\abs{3x+5}<14$, I get the following solutions: $x < 3$ and $x > -5/3$ and $x > -19/3$ and $x < -5/3$ The true solution for this is $-19/3 < x < 3$ In this case, would I just throw out $x > -5/3$ and $x < -5/3$ since they both can't be true? $\endgroup$ – FearlessFuture Dec 9 '13 at 2:28

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