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Math Help - Basic Inequality questions

  1. #1
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    Basic Inequality questions

    Hi,

    I want to check if my solution of the following question is correct :

    1) Solve :  |x+1|=5

    A: x \leq 4


    2) a) |x-1| \leq 3
    b) |x+1| \leq 3

    A: a) x \leq 4  ,
    b)  x \leq 4


    3) x < \frac{1}{x}

    I don't know how to solve the last one


    thx in advance
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  2. #2
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    Quote Originally Posted by Musab View Post
    Hi,

    I want to check if my solution of the following question is correct :

    1) Solve :  |x+1|=5

    A: x \leq 4
    Hi

    Solve :  |x+1|=5

    Case 1 : x \leq -1
    The equation becomes -(x+1)=5 or x=-6 and -6 < -1

    Case 2 : x \geq -1
    The equation becomes x+1=5 or x=4 and 4 > -1

    2 solutions : -6 and 4
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  3. #3
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    Quote Originally Posted by Musab View Post
    Hi,

    I want to check if my solution of the following question is correct :

    1) Solve :  |x+1|=5

    A: x \leq 4


    2) a) |x-1| \leq 3
    b) |x+1| \leq 3

    A: a) x \leq 4  ,
    b)  x \leq 4


    3) x < \frac{1}{x}

    I don't know how to solve the last one


    thx in advance
    1) |x+1| = 5

    x+1 = 5 , x + 1 = -5

    x = 4 , x = -6


    2) |x-1| \le 3

    -3 \le x - 1 \le 3

    -2 \le x \le 4


    3) x < \frac{1}{x}

    x - \frac{1}{x} < 0

    \frac{x^2-1}{x} < 0

    critical values are x = -1 , x = 0 , and x = 1

    for x < -1 ... original inequality is true

    for -1 < x < 0 ... original inequality is false

    for 0 < x < 1 ... original inequality is true

    for x > 1 ... original inequality is false

    solution set is x < -1 or 0 < x < 1
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  4. #4
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    Quote Originally Posted by skeeter View Post


    3) x < \frac{1}{x}

    x - \frac{1}{x} < 0

    \frac{x^2-1}{x} < 0

    critical values are x = -1 , x = 0 , and x = 1

    for x < -1 ... original inequality is true

    for -1 < x < 0 ... original inequality is false

    for 0 < x < 1 ... original inequality is true

    for x > 1 ... original inequality is false

    solution set is x < -1 or 0 < x < 1
    Can you explain how you solved this question please.

    First I have to find the critical points, and then check the values between these points against the original inequality ?

    Is that right ?
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  5. #5
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    Quote Originally Posted by Musab View Post
    Can you explain how you solved this question please.

    First I have to find the critical points, and then check the values between these points against the original inequality ?

    Is that right ?
    yes
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Musab View Post
    Can you explain how you solved this question please.

    First I have to find the critical points, and then check the values between these points against the original inequality ?

    Is that right ?
    There is an easier way.

    Assume x>0, then if we multiply both sides by x, the sign of the inequality is preserved.

    x^2<1. Since we know that x is positive, the solution is 0<x<1

    Now assume x is negative.... You try!
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