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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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
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