Problema Solution
Find three consecutive odd integers such that 8 more than twice the first integer is equal to 19 less than three times the third integer.
Answer provided by our tutors
every odd integer can be written as 2k + 1, where k is integer
the 3 consecutive odd integers can be written as:
2k - 1, 2k + 1, 2k + 3
8 more than twice the first integer is equal to 19 less than three times the third integer
8 + 2(2k + 1) = 3(2k + 3) - 19
by solving we find
k = 10
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the numbers are
2*10 - 1 = 19
2*10 + 1 = 21
2*10 + 3 = 23
the numbers are 19, 21 and 23.