find two consecutive odd intergers that twice the larger is fifteen more than three times the smaller

Mathematics

2 answers:

oksano4ka [1.4K]3 years ago

6 0

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

Svetach [21]3 years ago

4 0

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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