Question 741767: find 3 consecutive even integers such that 4 times the sum of the first and the third is 8 less than six times the second
Answer by only4christ(12)   (Show Source):
You can put this solution on YOUR website! Let x = 1st consecutive even integer
Then, x + 2 = 2nd consecutive even integer
And, x + 4 = 3rd consecutive even integer
Now, we build our equation based on the words given to us.
4(x + x + 4) = 6(x + 2) - 8
The part on the left is 4 times the sum of the first and third integer
The part on the right is 8 less than 6 times the sum of the second integer
Now, we just need to remove parenthesis, collect like terms and solve for x.
This "x" will give us the first integer and putting it back into the other expressions given for the 2nd and 3rd integers will give us those.
4(x + x + 4) = 6(x + 2) - 8
4(2x + 4) = 6(x + 2) - 8 Collect like terms on the left side
8x + 16 = 6x + 12 - 8 Remove parenthesis by distributing
8x + 16 = 6x + 4 Collect like terms on the right side
8x - 6x = 4 - 16 Collect like terms: variables on left side and numbers without variables on the right side
2x = -12 Divide both sides by 2
x = -6 This is the first integer
x + 2 = -6 + 2 = -4 This is the second integer
x + 4 = -6 + 4 = -2 This is the third integer
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