SOLUTION: Find four consecutive intergers such that twice the sum of the two greater intergers exceeds three times the first by 91.

Question 91017: Find four consecutive intergers such that twice the sum of the two greater intergers exceeds three times the first by 91.
Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!
THE 4 CONSECUTIVE NUMBERS ARE X, X+1, X+2 & X+3
2(X+2+X+3)=3(X+X+1)+91
2(2X+5)=3(2X+1)+91
4X+10=6X+3+91
4X-6X=94-10
-2X=84
x=84/-2
x=-42 ANSWER FOR THE SMALLEST NUMBER.
-42+1=-41
-42+2=-40
-42+3=-39
PROOF
2(-40-39)=3(-42-41)+91
2*-79=3*-83+91
-158=-249+91
-158=-158
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