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25 April, 09:37

6x + 1 = 2 (3x - 2) + 5 A) x = 0 B) x = 12 C) x = 1 12 D) infinitely many solutions

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  1. 25 April, 11:21
    0
    D) infinitely many solutions

    Step-by-step explanation:

    6x + 1 = 2 (3x - 2) + 5

    Distribute

    6x+1 = 6x - 4+5

    Combine like terms

    6x+1 = 6x+1

    Subtract 6x from each side

    6x-6x+1 = 6x-6x+1

    1=1

    This is always true, so x has infinite solutions
  2. 25 April, 13:31
    0
    6x + 1 = 2 (3x - 2) + 5

    6x + 1 = 6x - 4 + 5

    6x + 1 = 6x + 1

    6x = 6x

    x = x

    D.) Infinitely many solutions
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