# SOLUTION: how would I find the product of two consecutive integers with the sum being 71 more than the integers?

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 Question 268563: how would I find the product of two consecutive integers with the sum being 71 more than the integers?Found 2 solutions by solver91311, unlockmath:Answer by solver91311(17077)   (Show Source): You can put this solution on YOUR website! Your question doesn't make sense as posed. What does "71 more than the integers" mean? If your problem were: Find the product of two consecutive integers with the sum being 71. I can do that. Let represent the smaller integer. Then the next consecutive integer must be So, if the sum is 71: Solve for : So the integers are 35 and 36. Multiply 35 times 36 to get the product. WRONG! Got an update on the actual problem. It reads: The product of two consecutive integers is 71 more than their sum. That means that unlockmath (see below) has the correct solution. John Answer by unlockmath(1614)   (Show Source): You can put this solution on YOUR website!Hello, Let x be one integer and x+1 be the next. Now we can set up the following equation: (Note: I'm assuming your problem is "more the the integers added" otherwise it doesn't make sense) x(x+1)=x+(x+1)+71 Rewritten as: x^2+x=2x+72 Subtract 2x and 72 from both sides to get: x^2-x-72=0 Now we can factor this to be: (x-9)(x+8)=0 Solve for x: x=9 x=-8 There we go, the integers would be 9 and 10 or -8 and -7. Make sense? RJ Check out a book I wrote at: www.math-unlock.com