Question 128459:
Find three consecutive even integers such that the sum of the first and third exceeds one half the second by 15 Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! LET X, X+2 & X+4 BE THE THREE CONSECUTIVE EVEN INTEGERS.
X+(X+4)=(X+2)/2+15
2X+4=(X+2+30)/2
2X+4=(X+32)/2 NOW CROSS MULTIPLY
2(2X+4)=X+32
4X+8=X+32
4X-X=32-8
3X=24
X=24/3
X=8 ANSWER FOR THE FIRST INTEGER.
8=2=10 THE SECOND INTEGER.
8=4=12 THE THIRD INTEGER.
PROOF:
8+12=10/2+15
20=5+15
20=20