Four times the first of three consecutive even integers is six more than the product of two and the third integer. Find the integers.
I think there is something missing here you need to have and operator for the first three consecutive even integers. Is it four times the "sum of" three consecutive even integers or is it four times the "multiplication" of the consecutive even integers?
3 consecutive EVEN integers... x, x + 2, x + 4 4(x) = 2(x + 4) + 6 4x = 2x + 8 + 6 4x - 2x = 8 + 6 2x = 14 x = 7 must be doing something wrong because I am getting an odd number
@phi ...help set this up please
Yes, but the integers are x, x+2, x+4
but x has to be an even number...not an odd one
Hm.
@Loser66 ...help
@paki
we WILL figure this out...lol
lol Okay
phi...what did I screw up on ?
x=7 x+2=7+2=9 x+4=7+4=11 ???
but the integers need to be even...not odd
I think the problem is "ill-posed"
You need form the even integer as 2n, 2n+2, 2n+4 , not x
hence "four times the first of them" is 4*2n =8n product of the two and the third is (2n+2)(2n+4) blahblah....
wait....is it the produce of 2 and the 3rd integer....or is it the product of the 2nd and 3rd integer ?
I think its of 2 and the 3rd integer.
yeah it cant be 2nd integer...i tried it already
4(2n) = 2(2n + 4) + 6 8n = 4n + 8 + 6 8n - 4n = 14 4n = 14 nope....I am not understanding this
unless the integers are fractions ?? ....sorry....I do not know this one
nvm..integers cannot be fractions
I think: "No solution" is a solution.
oh.....thats tricky
are you sure it isnt consecutive ODD integers?
now that would work...lol
Suppose it is 2, 4,6 how can 4*2 = (4)(6) + 6??
true ^^
or 12,14,16 4 x 12 = (14 x 16) +6? or 4 x 12= (2x16)+6
Okay.
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