solve |2x-1| less than or egual to 3?

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  • 3 months ago

    |2x - 1| ≤ 3

    so, either 2x - 1 ≤ 3 or -(2x - 1) ≤ 3

    i.e. either 2x - 1 ≤ 3 or 2x - 1 ≥ -3

    so, 2x ≤ 4 or 2x ≥ -2

    i.e. x ≤ 2 or x ≥ -1

    Combining, we have:

    -1 ≤ x ≤ 2....or [-1, 2] using interval notation

    :)>

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  • Pope
    Lv 7
    3 months ago

    Both sides are non-negative, so there is no problem with squaring both.

    |2x - 1| ≤ 3

    (2x - 1)² ≤ 9

    (2x - 1)² - 9 ≤ 0

    [(2x - 1) + 3][(2x - 1) - 3] ≤ 0

    (2x + 2)(2x - 4) ≤ 0

    (x + 1)(x - 2) ≤ 0

    -1 ≤ x ≤ 2

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  • ted s
    Lv 7
    3 months ago

    you have 2 | x - 1 / 2|  | ≤ 3 ===> | x - 1 / 2 | ≤ 3/2 ===> all reals within 3/2 of 1 / 2 =====> [ -1 , 2 ]

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  • 3 months ago

    x is less than or equal to 2

    add the one, then divide the 2 by the 4 

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