0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ How do you justify simply erasing the absolute-value signs? $\endgroup$ – Henning Makholm Dec 19 '18 at 1:19
1
$\begingroup$

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*}

$\endgroup$
1
$\begingroup$

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 $x > \frac 23$.

$\endgroup$
  • $\begingroup$ If $3 - x < 2x + 1$, then $2 < 3x \implies x > \frac{2}{3}$. $\endgroup$ – N. F. Taussig Dec 19 '18 at 10:24
  • $\begingroup$ Dang. Arithmetic in your head is hard ..... $\endgroup$ – fleablood Dec 19 '18 at 17:18
  • $\begingroup$ Yee gods! I made at least 3 errors and a totally wrong answer! Still, I think cases are the most straight forward way to learn it. $\endgroup$ – fleablood Dec 19 '18 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.