SOLUTION: find four consecutive odd intergers such that twice the sum of the first two numbers is eight more than the sum of the last two numbers

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Question 53020: find four consecutive odd intergers such that twice the sum of the first two numbers is eight more than the sum of the last two numbers
Answer by checkley71(8403) About Me  (Show Source):
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2(X+X+2)=(X+4)+(X+6)+8 OR 2X+2X+4=2X+10+8 OR 4X-2X=18-4 OR 2X=14 THEN X=7
SO THE NUMBERS ARE 7,9,11 & 13
PROOF 2(7+9)=11+13+8 OR 2*16=32 OR 32=32