Absolute Value Inequalities Analytical Approach

Question

For $|x-3|-|2x+1|<0$, I considered adding $|2x+1|$ on both sides and solving it. $$|x-3|<|2x+1|$$ $$x-3< |2x+1|$$ $$x-3 < 2x+1$$ However, I keep getting $x>-4$ and $x<2/3$ instead of $x<-4$ and $x>2/3$. I know I can do a graphical approach and find the values, but why do I keep getting the wrong answer?

Answer 1

You have $|x-3| < |2x+1|$. Now there are 4 possible cases:

1) $x-3\ge 0$ and $2x + 1 \ge 0$.

This means:

a) $x - 3 \ge 0$ so $x \ge 3$.

b) $2x + 1 \ge 0$ so $x \ge -\frac 12$

c) $x - 3 < 2x + 1$ so $-4 < x$ or $x > -4$.

Combining we get $x \ge 3$.

2) $x-3 < 0$ and $2x +1 \ge 0$.

This means

a) $x - 3 < 0$ so $x < 3$

b) $2x + 1 \ge 0$ so $x \ge -\frac 12$

c) $3-x < 2x + 1$ so $2 < 3x$ and $x > \frac 23$

Combining we get $\frac 23 < x < 3$.

Case 3) $x-3 \ge 0$ and $2x + 1 < 0$ then

a) $x -3 \ge 0$ and $x > 3$

b) $2x + 1 < 0$ and $x < -\frac 12$

c) $x-3 < -1-2x$ and $3x < 2$ and $x < \frac 23$.

Combining we get contradictions. This is not possible.

Case 4) $x-3 < 0$ and $2x + 1 < 0$ then

a) $x - 3 < 0 $ so x < 3$

b) $2x +1< 0$ so x < -\frac 12$.

c) $3-x < -2x - 1$ so $x < -4$

Combining we get $x < -4$.

So we have 3 possible results $x \ge 3$ or $\frac 23 < x < 3$ or $x < -4$. Combining we get either $x < -4$ of $x > \frac 23$.

Answer 2

Analytically you may also proceed as follows:

\begin{eqnarray*} |x-3| < |2x+1| & \Leftrightarrow & (x-3)^2 < (2x+1)^2 \\ & \Leftrightarrow & 0 < (2x+1)^2 -(x-3)^2 \stackrel{a^2-b^2 =(a-b)(a+b)}{=} (2x+1 - (x-3))(2x+1 + (x-3))\\ & \Leftrightarrow & 0 < (x+4)(3x-2) \\ & \Leftrightarrow & \boxed{x< -4 \mbox{ or } x>\frac{2}{3}}\\ \end{eqnarray*}