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Solving inequalities dilemma

"Solve the following inequality":



My working:



Contradictory answers? :/
I get x>12x > \frac{1}{2} in one instance & x<12x < \frac{1}{2} in another instance ;/ Whats going wrong?

The right answer is x<12 x < \frac{1}{2}
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Original post by Sidhant Shivram
"Solve the following inequality":



My working:



Contradictory answers? :/
I get x>12x > \frac{1}{2} in one instance & x<12x < \frac{1}{2} in another instance ;/ Whats going wrong?

The right answer is x<12 x < \frac{1}{2}


You've written lots of implication signs, when they should be "is implied by" signs. ∣x+1∣<∣x+2∣  ⟹  x>12 or x<12|x+1| < |x+2| \implies x > \frac{1}{2} \ \text{or} \ x < \frac{1}{2} is what you've shown. You've given a list of possibilities such that the correct answer is some subset of the list. You need to check that every one of your outcomes is in fact true (and here, one is not).
Original post by Smaug123
You've written lots of implication signs, when they should be "is implied by" signs. ∣x+1∣<∣x+2∣  ⟹  x>12 or x<12|x+1| < |x+2| \implies x > \frac{1}{2} \ \text{or} \ x < \frac{1}{2} is what you've shown. You've given a list of possibilities such that the correct answer is some subset of the list. You need to check that every one of your outcomes is in fact true (and here, one is not).


Ah okay. So you mean, I have worked out a list of possible solutions for the inequality and now, I have to manually check (by plugging in these possible solutions back into the inequality) which one(s) is/are the actual, right solutions?

Thank you
Reply 3
Avatar for TeeEm
TeeEm
19
Original post by Sidhant Shivram
Ah okay. So you mean, I have worked out a list of possible solutions for the inequality and now, I have to manually check (by plugging in these possible solutions back into the inequality) which one(s) is/are the actual, right solutions?

Thank you


to add to smaug123 comment, from my experience modulus inequalities are best solved as equations followed by a graphical aproach
Original post by Sidhant Shivram
Ah okay. So you mean, I have worked out a list of possible solutions for the inequality and now, I have to manually check (by plugging in these possible solutions back into the inequality) which one(s) is/are the actual, right solutions?

Thank you

Yep. (Graphical approach is a bit more intuitive, as TeeEm says.)
Reply 5
Avatar for Hasufel
Hasufel
16
square both sides and simplify? - there only is one result!
Original post by Sidhant Shivram
"Solve the following inequality":



My working:



Contradictory answers? :/
I get x>12x > \frac{1}{2} in one instance & x<12x < \frac{1}{2} in another instance ;/ Whats going wrong?

The right answer is x<12 x < \frac{1}{2}


You have not established the conditions for each section
Section 1 you make no change you you are saying both sides are positive
In section 2 you change the LHS so you are saying that it is negative, it is not possible for the LHS to be negative if the RHS is positive so that section should not exist

Etc
Sidhant Shivram
...


If both sides of the equation/inequality are inside modulus signs the standard/easiest technique is to square both sides. You just need to solve a quadratic inequality.

Edit: Sorry didn't see Hasufel had already said this.
(edited 9 years ago)

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