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

Conner Baxter

Mathematics

30 December, 14:36

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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Jaylin Blackwell · 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