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.

3n-2(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

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

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

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