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.
4 points​

Answer 1

$\huge\mathcal{\fcolorbox{aqua}{azure}{\pink{✯\:Solution}}}$

Assume that three consecutive even integers are x , x + 2 , x + 4

twice the first integer = 2x

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

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

sum = 2x + x + 6 + x - 3   = 4x + 3

Sum = 27

Equate both

4x + 3 = 27

=> 4x = 24

=> x = 6

Hence 6 , 8 and 10 are  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