Two consecutive positive integers have the property that one integer times three times the order equals 6.What is the sum of the sum of these two integers ?
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The consecutive numbers must both be positive (given), so representing the two numbers as (n) and (n+1), if we multiply n by 3 we have 3*n*(n+1)=6, or if we multiply (n+1) by 3 we get n*3*(n+1)=6 and the left hand side is the same for both equations 3n^2+3n=6 or 3n^2+3n-6=0 Dividing both sides of this equation by 3 we get n^2+n-3=0, this factors as (n+2)(n-1) or n=-2 or n=1. Only n=1, satisfies our initial requirement that n be positive, so the two numbers are 1 and 1+1=2. The sum of 1+2=3.
I hope this helps. John