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Find 3 consecutive integers such that the smallest is ten less than twice the greatest

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Let the integers be x, x+1, x+2
Now the smallest is ten less than twice the greatest
So (x+2)2-10=x
2x+4-10=x
2x-x-6=0
x-6=0
x= 6
The integers then are;
6, 7, 8
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