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Find three consecutive integers such that 5 times the first decreased by three times the
second is 5 more than the third.


Sagot :

[tex]5n-3(n+1)=n+2+5\\ 5n-3n-3=n+7\\ 5n-3n-n=7+3\\ n=10\\ n+1=11\\ n+2=12[/tex]

10,11,12