algebra homework

Question

if the sum of four consecutive even integers is 268, what is the smallest of the four integers?

Answer 1

Hi, so the formula for the 4 consecutive EVEN integers goes:

n+(n+2)+(n+3)+(n+4)

n would be the smallest integer, (n+2) would be the next EVEN integer.....and so on

You would then set this equation equal to 268 to yield:

n+(n+2)+(n+4)+(n+6) = 268

Since the whole equation is additive in nature, we can remove the parenthesis without any consequence, and we can condense the equation into:

(n+n+n+n) + (2+4+6) = 268

4n+12=268

4n=256

n=64

Since the question asks for the smallest number, the answer is 64, however, if the equation were to ask the second, third, or fourth number in the sequence the answers would be 66,68,70 respectively.

Answer 2

a + b + c + d = 268

If they're consecutive and even, then we'll just use (a) as a base starting point.

a=(b-2) : (a + 2) = b

a=(c-4) : (a + 4) = c

a=(d-6) : (a + 6) = d

Plugging everything into the original formula, you get:

a + (a+2) + (a+4) + (a+6) = 268,

= 4a + 12 = 268

= 4a = 256

= a = 64

a=64, b=66, c=68, d=70

64 + 66 + 68 + 70 = 268