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Inequalities watch

1. I have to solve the inequality:

|x - 1| +2x <= |x|

I have had practice at solving where there is only one modulus sign, I was told to try two cases (in this question, it would be using -x-1 and x-1 but I didn't even really understand that.

Now I have the modulus on both sides - no idea how to do this.
2. There are three options:
1. is negative and is negative
2. is negative and is non-negative
3. is non-negative and is non-negative

You need to consider each case individually.

Also if anyone says "square both sides", ignore them.
3. Thanks both of you

(Original post by nuodai)
There are three options:
1. is negative and is negative
2. is negative and is non-negative
3. is non-negative and is non-negative

You need to consider each case individually.

Also if anyone says "square both sides", ignore them.
I will try this, in our lecture he did an example that had 2x-1 in the modulus, so there were the two cases - basically he solved the inequality without the modulus using (-2x-1) and (2x+1).

Basically what I mean is, when considering the cases should I just use -x-1 (for negative) and then the answer is where the whole left hand side is positive?

Sorry to sound a bit confusing, I don't really understand it that well.

And yes, I know squaring the sides is stupid
4. (Original post by CJN)
Thanks both of you

I will try this, in our lecture he did an example that had 2x-1 in the modulus, so there were the two cases - basically he solved the inequality without the modulus using (-2x-1) and (2x+1).

Basically what I mean is, when considering the cases should I just use -x-1 (for negative) and then the answer is where the whole left hand side is positive?

Sorry to sound a bit confusing, I don't really understand it that well.

And yes, I know squaring the sides is stupid
What we have is three different equations to solve for three different ranges of .

So for example in the range we must have and (since the inside of the modulus brackets are both negative in this range). So, in the range , your inequality becomes:

You do this with all three possible ranges. There are four possible outcomes:
(a) There is no solution (e.g. if you end up with something like )
(b) There is a solution, but it's not in the given range (in which case it's not actually a solution)
(c) There is a solution and it's in the given range (so it's a solution to the whole thing)
(d) There are an infinite number of solutions (in which case the whole given range is a solution to the equation)

As a simpler example, consider . There are two cases to consider, and . If then , and so we have , which is fine since this is in the range . If then , and so we have . This simplifies to which is always false, so there are no solutions in this range.

An example of where (d) occurs is the equation . This is true for all , and false for all .
5. (Original post by nuodai)
What we have is three different equations to solve for three different ranges of .

So for example in the range we must have and (since the inside of the modulus brackets are both negative in this range). So, in the range , your inequality becomes:

You do this with all three possible ranges. There are four possible outcomes:
(a) There is no solution (e.g. if you end up with something like )
(b) There is a solution, but it's not in the given range (in which case it's not actually a solution)
(c) There is a solution and it's in the given range (so it's a solution to the whole thing)
(d) There are an infinite number of solutions (in which case the whole given range is a solution to the equation)

As a simpler example, consider . There are two cases to consider, and . If then , and so we have , which is fine since this is in the range . If then , and so we have . This simplifies to which is always false, so there are no solutions in this range.

An example of where (d) occurs is the equation . This is true for all , and false for all .
Thanks I think I get it now.

So in case one, where both x-1 and x are <0 the reason that |x-1| becomes -(x-1) is because it has to be positive, so it is given a negative before it to make it positive?

In the end I got x <= -1/2, not idea if that is right.

Our lecturer also makes us write the answers like x ∈ (-∞, -1/2]

In this case the final bracket would be square, since -1/2 is included since its <= ?

Cheers

To update, for case 1 I got the only solution of x <= -1/2

In case 2 the answer was 0 <= -1 so there is no solution

And case 3 I got x <= 1/2 which isn't valid as one of the conditions was x-1 > 0

Does that sound right?
6. (Original post by CJN)
Thanks I think I get it now.

So in case one, where both x-1 and x are <0 the reason that |x-1| becomes -(x-1) is because it has to be positive, so it is given a negative before it to make it positive?

In the end I got x <= -1/2, not idea if that is right.

Our lecturer also makes us write the answers like x ∈ (-∞, -1/2]

In this case the final bracket would be square, since -1/2 is included since its <= ?

Cheers
Yup, that all seems fine to me.

And yes, in general, means the set of numbers x for which , and means , and means and so on.
7. (Original post by CJN)
To update, for case 1 I got the only solution of x <= -1/2

In case 2 the answer was 0 <= -1 so there is no solution

And case 3 I got x <= 1/2 which isn't valid as one of the conditions was x-1 > 0

Does that sound right?
Yup that sounds right! Well done
8. (Original post by nuodai)
Yup that sounds right! Well done
Wow great, thanks for your help! Hopefully I will remember how to do this for my tutorial tomorrow

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