SOLUTION: Can I ask the following step by step procedures in this Problem so that I could understand the solution.thanks PROBLEM: The sum of twice the 1st, four times the 2nd plus the

Question 629851: Can I ask the following step by step procedures in this Problem so that I could understand the solution.thanks
PROBLEM: The sum of twice the 1st, four times the 2nd plus the 3rd of the three consecutive even integers is equal to 96. find the integers

Found 2 solutions by unlockmath, graphmatics:
Answer by unlockmath(1688) (Show Source): You can put this solution on YOUR website!
Hello,
Setting up these problems is sometimes the toughest.
Let x be the first even integer (x+2) and (x+4) be the next ones.
Now we can write the equation as:
2x+4(x+2)+(x+4)=96
Combine like terms:
7x+12=96
Subtract 10 and divide by 7 to get:
x=12
So therefore the even numbers are:
12,14,16
Make sense?
RJ
www.math-unlock.com
Answer by graphmatics(170) (Show Source): You can put this solution on YOUR website!
The three consecutive even integers are 2n,2n+2 and 2n+4. We have that 2*(2n)+4*(2n+2)+2n+4=96 Do multiplications 4n+8n+8+2n+4=96 Variables on the left and numbers on the right. 4n+8n+2n=96-8-4 14n=84 Divide both sides by 14 14n/14=84/1r n=6
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