Asked by NEED HELP NOW!!
                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.
            
        also, my teacher told me to use an equation.
Answers
                    Answered by
            JJ
            
    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?
    
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?
                    Answered by
            bobpursley
            
    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.
    
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.
                    Answered by
            bobpursley
            
    No JJ, 2,6,8 are not consecutive even numbers.  Your equation is wrong.
    
                    Answered by
            JJ
            
    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.
    
                    Answered by
            bobpursley
            
    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.
    
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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