Algebra 1

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Find three consecutive even numbers such that the difference between 3 times the first number and twice the second number is 1/3 the third number.

also, my teacher told me to use an equation.

  • Algebra 1 -

    Even numbers have a differnce of two.

    2, 6, and 8 are consecutive even numbers.

    Let n = 1st
    Let n + 2 = 2nd
    Let n + 4 = 3rd

    3(n) - 2(2n+2) = 1/3(n+4)

    Here is the equation. Can you solve it?

  • Algebra 1 -

    let n be first number
    n+2 be second number
    n+4 be third number.

    3n-2(n+2)=1/3 * (n+4)
    solve for n, then the other numbers.

  • Algebra 1 -

    No JJ, 2,6,8 are not consecutive even numbers. Your equation is wrong.

  • Algebra 1 -

    That isn't the answer. That was just an example of what consecutive even numbers mean to show why we are using n, n+2 and n+4. My equation is the same as yours.

  • Algebra 1 -

    3(n) - 2(2n+2) = 1/3(n+4)
    is not the same as
    3n-2(n+2)=1/3 * (n+4)

    again, 2,6, 8 are not consecutive even numbers.

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