SOLUTION: fine three consecutive even integers such that the sum of the smallest and twice the second is 20 more than the third

SOLUTION: fine three consecutive even integers such that the sum of the smallest and twice the second is 20 more than the third

Algebra.Com

Question 120932: fine three consecutive even integers such that the sum of the smallest and twice the second is 20 more than the third
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
X, X+1 & X+2 ARE THE 3 CONSECUTIVE INTEGERS.
X+2(X+1)=(X+2)+20
X+2X+2=X+2+20
X+2X-X=2+20-2
2X=20
X=20/1
X=10 ANSWER FORT THE SMALLEST INTEGER.
10+1=11 FOR THE NEXT INTEGER.
10+2=12 FOR THE LARGEST INTEGER.
PROOF
10+2*11=12+20
10+22+32
32=32

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