find two consecutive odd intergers that twice the larger is fifteen more than three times the smaller
2 answers:
Answer:
11, 13
Step-by-step explanation:
Let n and n+2 be the consecutive odd integers
2(n+2) = 3n + 15
2n + 4 = 3n + 15
n = 11
n + 2 = 13
Answer:
11, 13
Step-by-step explanation:
an odd number can be represented by 2n+1
since they are consecutive, the larger odd number will be 2n + 1 + 2
now,
2(2n+3) = 2n + 1 +15
solving this eqn, we get n = 5
so the two numbers are (2n+1) = 11 and 11+2= 13
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