SOLUTION: Find four consecutive even integers such that three times the sum of the first and third numbers is equal to four times the sum of the second and fourth numbers.

Question 1173284: Find four consecutive even integers such that three times the sum of the first and third numbers is equal to four times the sum of the second and fourth numbers.

Found 3 solutions by ikleyn, greenestamps, ewatrrr:
Answer by ikleyn(52834)   (Show Source): You can put this solution on YOUR website!
.

The numbers are n, (n+2), (n+4) and (n+6), in ascending order. The equation is 3*(n + (n+4)) = 4*((n+2) + (n+6)) Simplify 3*(2n+4) = 4*(2n+8) 6n + 12 = 8n + 32 12 - 32 = 8n - 6n -20 = 2n n = = -10. ANSWER. The numbers are -10, -8, -6, -4.

Solved, answered and explained.

Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!

Let the four consecutive even integers be x, x+2, x+4, and x+6.

Three times the sum of the first and third is equal to four times the second and fourth:

ANSWER: -10, -8, -6, and -4

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi four consecutive even integers: n , (n+2), (n+4), (n+6) three times the sum of the first and third numbers is equal to four times the sum of the second and fourth numbers. 3(n + (n+4)) = 4((n+2) + (n+6)) 3(2n+4) = 4(2n+8) 6n + 12 = 8n + 32 -20 = 2n n = -10 four consecutive even integers are: -10, -8, -6, -4 Checking our Answer: 3(-16) = 4(-12) = 48 checks! Wish You the Best in your Studies.

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