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The sq of a nat num is 4 more than 2 times another nat num. The 2nd num is 2 less than 3 times the 1st num. ? the 2 nat num algebraically.?

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  • 2 months ago
    Favorite Answer

    Let x be the first natural number.

    Let y be the second natural number.

    x^2 = 2y + 4

    y = 3x - 2

    Substituting gives:

    x^2 = 2(3x - 2) + 4

    x^2 = 6x - 4 + 4

    x^2 = 6x

    x^2 - 6x = 0

    x(x - 6) = 0

    x = 6 is a solution. (x = 0 is not a natural number, 1,2,3...)

    y = 3x - 2 = 3(6) - 2 = 18 - 2 = 16.

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  • DWRead
    Lv 7
    2 months ago

    "The sq of a nat num is 4 more than 2 times another nat num."

    m is the first number; n is the second number.

    m² = 2n+4

    "The 2nd num is 2 less than 3 times the 1st num."

    n = 3m-2

    m² = 2(3m-2) + 4

    m² = 6m

    m = 6

    n = 3m-2 = 16

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  • 2 months ago

    The square of a natural number is 4 more than 2 times another natural number:

    x² = 2y + 4

    The second number is 2 less than 3 times the first number:

    y = 3x - 2

    we have a system of two equations and two unknowns.  Substitute and solve:

    x² = 2y + 4

    x² = 2(3x - 2) + 4

    x² = 6x - 4 + 4

    x² = 6x

    x² - 6x = 0

    x(x - 6) = 0

    x = 0 and 6

    Since we are told that these are natural numbers, we can throw out the zero, leaving:

    x = 6

    Now we can solve for y to determine the second number:

    y = 3x - 2

    y = 3(6) - 2

    y = 18 - 2

    y = 16

    The second number is 16

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  • 2 months ago

    M^2 = 2N + 4 and

    N + 2 = 3M.

    Multiplying the 2nd eqn by 2, we see that

    M^2 = 6M, so M = 6 and therefore N = 16.

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  • 2 months ago

    x² = 2y + 4

    y = 3x – 2

    x² = 2(3x – 2) + 4

    x² = 6x – 4 + 4

    x² = 6x

    x² – 6x = 0

    x(x – 6) = 0

    x = 0, 6

    y = 3x – 2

    y = 3•0 – 2 = –2 (this is ruled out as negative numbers are not allowed)

    y = 3•6 – 2 = 16

    ans: 6 and 16

    Some definitions, including the standard ISO 80000-2, begin the natural numbers with 0, corresponding to the non-negative integers 0, 1, 2, 3, …, whereas others start with 1, corresponding to the positive integers 1, 2, 3, …, while others acknowledge both definitions.

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