Posted by NEED HELP NOW!! on Friday, February 22, 2013 at 5:12pm.
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  JJ, Friday, February 22, 2013 at 5:14pm
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  bobpursley, Friday, February 22, 2013 at 5:15pm
let n be first number
n+2 be second number
n+4 be third number.
3n2(n+2)=1/3 * (n+4)
solve for n, then the other numbers.

Algebra 1  bobpursley, Friday, February 22, 2013 at 5:16pm
No JJ, 2,6,8 are not consecutive even numbers. Your equation is wrong.

Algebra 1  JJ, Friday, February 22, 2013 at 5:18pm
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  bobpursley, Friday, February 22, 2013 at 5:20pm
3(n)  2(2n+2) = 1/3(n+4)
is not the same as
3n2(n+2)=1/3 * (n+4)
again, 2,6, 8 are not consecutive even numbers.
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