# Help with a system of inequalities with absolute values

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. 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?

• either plot the graphs of the inequalities or square both the inequalties and solve the resulting quadratic inequalities – vidyarthi Nov 19 '18 at 17:39
• Square $|x-3|<2x$ to $x^2+9-6x<4x^2$ you mean? – Paul Nov 19 '18 at 17:48
• yes, exactly. Now, solve the inequality by factoring the quadratic – vidyarthi Nov 19 '18 at 17:50
• Is this the only way to solve this? I don't really understand how squaring the inequalities takes away the absolute value. – Paul Nov 19 '18 at 17:52
• since square of a real is always positive, so the absolute value is easily removed – vidyarthi Nov 19 '18 at 17:53

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 $$\boxed{x>1}$$.

• so you reduced two squarings to just one! great answer – vidyarthi Nov 19 '18 at 18:10

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

• 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? – Paul Nov 19 '18 at 19:23
• @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 $|x-3|<2x\implies x\ge0$ – vidyarthi Nov 22 '18 at 7:08

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 $$x>1$$.

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 $$x>0$$: $$-2x < x-3 < 2x \Leftrightarrow \begin{cases} 3x > 3 \Leftrightarrow \color{blue}{x> 1} \\ x > -3 \mbox{ does not extend the solution} \end{cases}$$