Question

Find three consecutive even integers such that the sum of twice the first integer , four more than the second and seven less than the third integer is 27. . Find the integers.
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Answer 1

Answer:

6, 8,and 10

Step-by-step explanation:

let the first even integer be 2x

second even integer will be 2x+2

third even integer will be 2x+4

condition given

the sum of twice the first integer , four more than the second and seven less than the third integer is 27.

so,

twice the first integer =2×2x = 4x

four more than the second integer= 2x+2+4= 2x+6

seven less than the third integer= 2x+4-7= 2x-3

sum of all integer = 27

Therefore,

$(4x) + (2x + 6) + (2x - 3) = 27 \\ \\ 4x + 2x + 6 + 2x - 3 = 27 \\ \\ 8x + 3 = 27 \\ 8x = 27 - 3 \\ 8x = 24 \\ x = 24 \div 8 \\ x = 3$

first even integer be 2x=2×3= 6

second even integer will be 2x+2= 2×3+2= 8

third even integer will be 2x+4= 2×3+4=10