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# Question on Absolute Values Watch

1. Why is the solution of |1+3x|<6x only x>1/3?
After applying the properties of modulus, I get -6x<1+3x<6x
And after solving each inequality, I get x>-1/9 and x>1/3, but why is x>-1/9 rejected?
2. (Original post by javainstaller)
Why is the solution of |1+3x|<6x only x>1/3?
After applying the properties of modulus, I get -6x<1+3x<6x
And after solving each inequality, I get x>-1/9 and x>1/3, but why is x>-1/9 rejected?
Have you looked at the graph?

Anyways, for x < -1/3 x you get x > -1/9. But the intersetion of these is the empty set, hence rejected.

i.e: to get |1+3x| = -(1 + 3x) you are assuming that x < -1/3.

So your solution to x > -1/9 corresponds to x < -1/3. You need to combine these two solutions together and that's nothing. There's no number bigger than -1/9 and simultaneously smaller than -1/3.
3. (Original post by Zacken)
Have you looked at the graph?

Anyways, for x < -1/3 x you get x > -1/9. But the intersetion of these is the empty set, hence rejected.

i.e: to get |1+3x| = -(1 + 3x) you are assuming that x < -1/3.

So your solution to x > -1/9 corresponds to x < -1/3. You need to combine these two solutions together and that's nothing. There's no number bigger than -1/9 and simultaneously smaller than -1/3.
Hey Zacken, thank you so much for your reply!!! I finally understand the solution, thank you so much!!!
4. (Original post by javainstaller)
Hey Zacken, thank you so much for your reply!!! I finally understand the solution, thank you so much!!!

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