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    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.
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    There are three options:
    1. x-1 is negative and x is negative
    2. x-1 is negative and x is non-negative
    3. x-1 is non-negative and x is non-negative

    You need to consider each case individually.

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

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

    So for example in the range x \le 1 we must have |x-1| = -(x-1) = 1-x and |x| = -x (since the inside of the modulus brackets are both negative in this range). So, in the range x \le 1, your inequality becomes:
    1-x+2x \le -x

    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 0 &gt; 1)
    (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 |x| \ge x+3. There are two cases to consider, x \le 0 and x \ge 0. If x \le 0 then |x| = -x, and so we have -x \ge x + 3 \Rightarrow 2x \le -3 \Rightarrow x \le -\frac{3}{2}, which is fine since this is in the range x \le 0. If x \ge 0 then |x|=x, and so we have x \ge x+3. This simplifies to 0 \ge 3 which is always false, so there are no solutions in this range.

    An example of where (d) occurs is the equation |x| \ge x. This is true for all x \ge 0, and false for all x &lt; 0.
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    (Original post by nuodai)
    What we have is three different equations to solve for three different ranges of x.

    So for example in the range x \le 1 we must have |x-1| = -(x-1) = 1-x and |x| = -x (since the inside of the modulus brackets are both negative in this range). So, in the range x \le 1, your inequality becomes:
    1-x+2x \le -x

    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 0 &gt; 1)
    (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 |x| \ge x+3. There are two cases to consider, x \le 0 and x \ge 0. If x \le 0 then |x| = -x, and so we have -x \ge x + 3 \Rightarrow 2x \le -3 \Rightarrow x \le -\frac{3}{2}, which is fine since this is in the range x \le 0. If x \ge 0 then |x|=x, and so we have x \ge x+3. This simplifies to 0 \ge 3 which is always false, so there are no solutions in this range.

    An example of where (d) occurs is the equation |x| \ge x. This is true for all x \ge 0, and false for all x &lt; 0.
    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?
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    (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, [a,b] means the set of numbers x for which a \le x \le b, and (a,b) means a&lt;x&lt;b, and [a,b) means a \le x &lt; b and so on.
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    (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
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    (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 :cool:
 
 
 
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