algebra help please?

Question

the product of two consecutive even integers is the same six more than three times the sum of the integers. find the pair of integers.show work and give explanations for each step.

how do i slove this problem???
thanks

Answer 1

Let the smaller integer be 2n (this makes sure it is even), then the larger one will be 2n+2. So the product is 2n (2n+2) = 4n^2 + 4n.
Six more than three times their sum is 3((2n) + (2n+2)) + 6
= 3(4n+2) + 6
= 12n + 12.
So we have
4n^2 + 4n = 12n + 12
<=> n^2 - 2n - 3 = 0
<=> (n-3)(n+1) = 0
<=> n = -1 or n = 3.
So the integers are -2 and 0, or else 6 and 8. You can check that both combinations work.

Answer 2

(2k)(2k+2)=3(2k+2k+2)+6
expand and divide by 2:
2k^2 + 2k = 6k + 6
2k^2 - 4k -6=0 => k^2 - 2k -3 =0
k = -1, k = 3
so the integers are 6 and 8 (if they should be positive)
and -2,-4 if negative

Answer 3

n(n+1) = 6 + 3[(n) + (n+1)]
n² + n = 6 + 3(2n + 1)
n² + n - 6 + 6n + 3 = 0
n² + 7n - 3 = 0
d = 49 - 4.1.-3
d = 49 + 12
...

Answer 4

x (x+2) = 6+3(x+x+2)
x^2 + 2x = 6+6x+6
x^2 - 4x - 12 = 0
(x+2)(x-6) = 0
x= -2, 6

Answer 5

First, solve your lack of proper English. What do you mean?