Question 1086610
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Find 3 consecutive odd integers such that twice the product of the first two is 7 more than the product of the last two. 
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Let the middle integer be n.
{{{2(n-2)n=7+n(n+2)}}}
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{{{2n^2-4n=n^2+2n+7}}}
{{{n^2-6n-7=0}}}
{{{(n-7)(n+1)=0}}}
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n=-1, or n=7.


Do they both work?
{{{2(-1-2)*(-1)=7+(-1)(-1+2)}}}
{{{-6(-1)=7-(1)}}}
{{{6=6}}}
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{{{2(7-2)*7=7+7(7+2)}}}
{{{70=7+7*9}}}
{{{70=7+63}}}
{{{70=70}}}
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Both solutions for n are good.


The three consecutive odd numbers:
-3, -1, 1
OR
5, 7, 9




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Find 2 consecutive integers whose product is 1 less than their sum.
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You try this one.