SOLUTION: find three consecutive integers such that the sum of the first and twice the second is 98 minus three times the third

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Question 656350: find three consecutive integers such that the sum of the first and twice the second is 98 minus three times the third
Answer by checkley79(3341)   (Show Source): You can put this solution on YOUR website!
Let X= THE FIRST INTEGER
Let (X+1)= THE SECOND INTEGER.
Let (X+2)=THE THIRD INTEGER.
X+2(X+1)=98-3(X+2)
X+2X+2=98-3X-6
3X+3X=98-6-2
6X=90
X=90/6
X=15 ANS FOR THE FIRST INTEGER.
PROOF:
15+2(15+1)=98-3(15+2)
15+2*16=98-3(17)
15+32=98-51
47=47

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