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

  • #1
RChristenk
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7
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.
 
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  • #2
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$$x = -y$$
 
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  • #3
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 [itex]|x| = |3|[/itex], would you exclude [itex]x = -3[/itex] as a solution?

Here, is there any reason to exlude the case [itex]-3 < x < -\frac12[/itex], where [itex]x + 3 > 0[/itex] but [itex]2x + 1 < 0[/itex]?
 
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  • #4
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.
 
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  • #5
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) ##.
 
  • #6
What about squaring both sides and solving the resulting quadratic equation?
 
  • #7
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) ##
 
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