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I'm trying to solve this system on inequalities

$$ \left\{ \begin{array}{c} |x-3|<2x \\ |2x+5|>3 \end{array} \right. $$

The steps I'm taking are:

Finding the absolute values sings, so for $x-3 \geq 0$ we have $x \geq 3$ therefore $$|x-3| = \left\{ \begin{array}{c} x-3 & \text{for $x \geq 3$} \\ -x+3 & \text{for $x < 3$} \end{array} \right. $$

and

$2x+5 \geq 0$ we have $x \geq \frac{-2}{5}$ therefore $$|2x+5| = \left\{ \begin{array}{c} 2x+5 & \text{for $x \geq \frac{-2}{5}$} \\ -2x-5 & \text{for $x<\frac{-2}{5}$} \\ \end{array} \right. $$

So I build a few systems with the complete inequalities, for the first one we have:

$$ \left\{ \begin{array}{c} x \geq 3 \\ x-3<2x = x>-3 \end{array} \right. $$

So the solution here would be $x>3$, then:

$$ \left\{ \begin{array}{c} x<3 \\ -x+3<2x = x>1 \end{array} \right. $$

The solution would be $1<x<3$. Then

$$ \left\{ \begin{array}{c} x \geq \frac{-2}{5} \\ 2x+5>3 = x>-1 \end{array} \right. $$

So the solution of the system is $x>-1$, then

$$ \left\{ \begin{array}{c} x< \frac{-2}{5} \\ -2x-5>3 = x<-4 \end{array} \right. $$

And the solution is $x<-4$

Now the solution my book gives is x>1 for the initial system. But I can't find that one. I can't get a solution at all. I tried finding a common point between the 4 solutions I found (as if it was a 4-inequalities system), but there isn't one really. What am I doing wrong?

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  • $\begingroup$ either plot the graphs of the inequalities or square both the inequalties and solve the resulting quadratic inequalities $\endgroup$ – vidyarthi Nov 19 '18 at 17:39
  • $\begingroup$ Square $|x-3|<2x$ to $x^2+9-6x<4x^2$ you mean? $\endgroup$ – Paul Nov 19 '18 at 17:48
  • $\begingroup$ yes, exactly. Now, solve the inequality by factoring the quadratic $\endgroup$ – vidyarthi Nov 19 '18 at 17:50
  • $\begingroup$ Is this the only way to solve this? I don't really understand how squaring the inequalities takes away the absolute value. $\endgroup$ – Paul Nov 19 '18 at 17:52
  • $\begingroup$ since square of a real is always positive, so the absolute value is easily removed $\endgroup$ – vidyarthi Nov 19 '18 at 17:53
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Since $2x>|x-3|$ we get $x>0$ so $2x+5>0$ and so $2x+5>3$ so $x>-2$ which is nothing new. So $x>0$ and $|x-3|<2x$ so after squaring we get $$ x^2-6x+9<4x^2\implies 3x^2+6x-9>0$$

or $$(x+3)(x-1)>0\implies x-1>0$$ so $\boxed{x>1}$.

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  • $\begingroup$ so you reduced two squarings to just one! great answer $\endgroup$ – vidyarthi Nov 19 '18 at 18:10
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We have $$|x-3|<2x\implies 3x^2+6x-9>0\implies (x+3)(x-1)>0$$ and $$|2x+5|>3\implies 4x^2+20x+16>0\implies (x+4)(x+1)>0$$. I think you could proceed now? For a quick method on how to solve further, see here

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  • $\begingroup$ If I try to graph this I get x<-4 and x>1 as solutions. It should only be x>1. Where am I wrong? $\endgroup$ – Paul Nov 19 '18 at 19:23
  • $\begingroup$ @Paul yes, by graphing technique, you get the answer as $x<-4$ or $x>1$ as answers. Here you have to use the fact that $|x-3|<2x\implies x\ge0$ $\endgroup$ – vidyarthi Nov 22 '18 at 7:08
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Since $0\le\left|x-3\right|$ and $\left|x-3\right|\le2x$ we conclude $2x\ge0$ and then $x\ge0$. From $\left|x-3\right|<2x$ and since $x\ge0$, we get $-2x< x-3<2x$ and therefore we have $\left\{\begin{array}{c}x-3<2x \\x-3>-2x\end{array}\right.$ and therefore $\left\{\begin{array}{c}x>-3 \\3x>3\end{array}\right.$ and then $\left\{\begin{array}{c}x>-3 \\x>1\end{array}\right.$, so $x>1$ which satisfies the initial statement $x\ge0$. So far we see $x$ has to be greater than $1$ to satisfy $\left|x-3\right|<2x$. Let us consider the next inequality $|2x+5|>3$ which leads to $2x+5>3$ or $2x+5<-3$. Then we get $x>-1$ or $x<-4$. And finally if we combine this with the result of first inequality, $x>1$ we get $x>1$.

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There is actually only the first inequality to be solved:

$$ |2x+5| \stackrel{\color{blue}{0}\leq|x-3| \color{blue}{< 2x}}{>} |x-3|+5 > 3$$ So, you only need to solve $|x-3| < 2x$ while $x>0$: $$ -2x < x-3 < 2x \Leftrightarrow \begin{cases} 3x > 3 \Leftrightarrow \color{blue}{x> 1} \\ x > -3 \mbox{ does not extend the solution} \end{cases} $$

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