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

• RChristenk
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.

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$$x = -y$$

MatinSAR, RChristenk and docnet
RChristenk said:
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 $2x + 1 < 0$?

RChristenk and PeroK
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
RChristenk said:
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) ##.

Gavran said:
## |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) ##

PeroK

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