Find three consecutive odd integers such that eight more than the sum of the first two is equal to eleven less then three times the third

Question

Mathematics

Boston Golden

26 February, 00:00

Find three consecutive odd integers such that eight more than the sum of the first two is equal to eleven less then three times the third

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Answers (1)

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Colt Barrett · Answer 1

A+b+8 = 11-3c

a+b+3c = 11-8

a+b+3c = 3

a = b-1

c = b+1

b-1+b+3 * (b+1) = 3

2b-b+3b+3 = 3

4b = 3-3

4b = 0

b = 0

a = b-1

a = 0-1

a = - 1

c = b+1

c = 0+1

c = 1

a = - 1, b = 0, c = 1