The sum of three integers equal 45, The sum of the first and second integers is 9 more than the third, The sum of the first and third integers is two times the second integer
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Let the integers be A, B, C. A+B+C=45, so A+C=45-B

A+B=C+9, A+C=2B, so 45-B=2B, 3B=45 and B=15.

A-C=9-B and A+C=2B so add these two equations: 2A=9+B. Plug in B=15: A=12 and C=45-(A+B)=45-27=18.

SOLUTION: A=12, B=15, C=18
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