The equation 3x+6=2x+10+x-4 is true for all real numbers. The equation 6x+2-2x=4x+1 has no solution. When do you think an equation has all r

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The equation 3x+6=2x+10+x-4 is true for all real numbers. The equation 6x+2-2x=4x+1 has no solution. When do you think an equation has all real numbers as its solutions?​

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Minh Khuê 3 years 2021-08-09T14:23:12+00:00 1 Answers 19 views 0

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  1. King
    0
    2021-08-09T14:25:07+00:00
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    Answer:

      when it simplifies to 0=0

    Step-by-step explanation:

    An equation has “all real numbers” as its solution when it can be simplified to …

      0 = 0 . . . has “all real numbers” as a solution set

    __

    Example:

      3x +6 = 2x +10 +x -4

    Subtract (3x+6) from both sides:

      (3x +6) -(3x +6) = (2x +10 +x -4) -(3x +6)

      0 = x(2 +1 -3) +(10 -4 -6)

      0 = 0x +0

      0 = 0 . . . . has “all real numbers” as a solution set (True for every value of x.)

    Example 2:

      6x +2 -2x = 4x +1

    Subtract (4x+1) from both sides:

      (6x +2 -2x) -(4x +1) = (4x +1) -(4x +1)

      x(6 -2 -4) +(2 -1) = 0

      0x +1 = 0

      1 = 0 . . . . . does not have “all real numbers” as a solution set. There is no solution. (No value of x will make this true.)

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )