find the consecutive ever integers such as the first times the third is 2 greater than 5 times the second.

Question

terenzreignz · Answer

First step to such word problems?
Represent the values :D
If we let
x = the first even integer, what would be your second and third?

anonymous · Answer

its even integers ..  not ever integers.  xD

terenzreignz · Answer

Yes, I assumed that, so... what do you think?

anonymous · Answer

Even consecutive numbers have a difference of 2..
If one number is x, then next number will be : (x + 2)
And the third will be : (x + 2 + 2) = (x + 4)

So, x(x + 4) = 2 + 5(x + 2)

$$x^2 + 4x = 2 + 5x + 10$$

$$x^2 - x - 12 = 0$$

$$(x - 4)(x + 3) = 0$$

x = 4 and x = -3

So, the number are: 4, 6, 8
Or the numbers can be: -3, -1, 1...

anonymous · Answer

no..  i didn't say that you didn't assume that. i just corrected my problem