Solve ##\left| x+3 \right|= \left| 2x+1\right|##

[RChristenk]

Homework Statement

Solve $\left| x+3 \right|= \left| 2x+1\right|$

Relevant Equations

Absolute values

Both sides are in absolute values, i.e. positive, so the solution is straightforward: $\left| x+3 \right|= \left| 2x+1\right| \Rightarrow x+3=2x+1 \Rightarrow x=2$.

But the solution presents another case: $x+3 = -(2x+1)$. How is this possible if both sides are in absolute values, i.e. positive?

I understand $\left| x\right|= \pm c$, but when there are absolute values on both sides, I don't understand why you can remove the absolute values by setting one side to negative. Thanks for the help.

 

[PeroK]

If $|x| = |y|$ then there are four possible solutions:
$$x = y$$or$$x = -y$$or$$-x = y$$or$$-x = -y$$This, however, simplifies to two solutions.
$$x = y$$or$$x = -y$$

 

[pasmith]

Homework Statement: Solve $\left| x+3 \right|= \left| 2x+1\right|$
Relevant Equations: Absolute values

Both sides are in absolute values, i.e. positive, so the solution is straightforward: $\left| x+3 \right|= \left| 2x+1\right| \Rightarrow x+3=2x+1 \Rightarrow x=2$.

But the solution presents another case: $x+3 = -(2x+1)$. How is this possible if both sides are in absolute values, i.e. positive?

If you were asked to solve $|x| = |3|$, would you exclude $x = -3$ as a solution?

Here, is there any reason to exlude the case $-3 < x < -\frac12$, where $x + 3 > 0$ but $2x + 1 < 0$?

 

[FactChecker]

When you remove the absolute values, you do not know if the inside is positive or negative and you have to account for either possibility.

 

[Gavran]

I understand $\left| x\right|= \pm c$, but when there are absolute values on both sides, I don't understand why you can remove the absolute values by setting one side to negative. Thanks for the help.

$|x| = c \Rightarrow x = \pm c$ where $c \ge 0$. By using this, there will be $|x+3| = |2x+1| \Rightarrow x+3 = \pm |2x+1|$ and finally $x+3 = \pm (2x+1)$ because $\pm |2x+1|$ can be rewritten as $\pm (2x+1)$.

 

[Chestermiller]

What about squaring both sides and solving the resulting quadratic equation?

 

[Mark44]

$|x| = c \Rightarrow x = \pm c$ where $c \ge 0$. By using this, there will be $|x+3| = |2x+1| \Rightarrow x+3 = \pm |2x+1|$ and finally $x+3 = \pm (2x+1)$ because $\pm |2x+1|$ can be rewritten as $\pm (2x+1)$.

You can skip a step in the above. $|x+3| = |2x+1| \Rightarrow x+3 = \pm (2x+1)$