SOLUTION: Find two consecutive odd integers such that three times the second is twelve more than the first.

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Question 116381: Find two consecutive odd integers such that three times the second is twelve more than the first.
Answer by MathLover1(20849) About Me  (Show Source):
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Find two consecutive odd integers such that three times the second is twelve more than the first.
let two consecutive odd be:
x
x%2B+2
three times the second is twelve more than the first, then
3%28x%2B+2%29+=+x+%2B+12
3x%2B+6+=+x+%2B+12
3x+-+x+=+12+-+6
2x+=+6
x+=+3……….first odd number

x%2B+2=+5……second odd number
Check if
three times the second is twelve more than the first
3%2A5+=+3%2B12
15+=+15