SOLUTION: 3 times the greatest of 3 consecutive even integers exceeds twice the least by 38. I know the answer is 26, 28, 30, but what's the equation?

Algebra ->  Equations -> SOLUTION: 3 times the greatest of 3 consecutive even integers exceeds twice the least by 38. I know the answer is 26, 28, 30, but what's the equation?      Log On


   

Question 226515: 3 times the greatest of 3 consecutive even integers exceeds twice the least by 38. I know the answer is 26, 28, 30, but what's the equation?
Found 2 solutions by user_dude2008, jim_thompson5910:
Answer by user_dude2008(1862) About Me  (Show Source):
You can put this solution on YOUR website!
3(x+4)=2x+38
3x+12=2x+38
3x-2x=38-12
x=26 <--------first number
x+2=28 <--------second number
x+4=30 <--------third number

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=first consecutive even integer, x+2=second consecutive consecutive integer, x+4=third consecutive even integer.

We're also going to assume that all of these integers are positive. So the third is the largest.


Since "three times the greatest of three consecutive even integers exceed twice the least by 38", we get the translation 3%28x%2B4%29=2x%2B38



3%28x%2B4%29=2x%2B38 Start with the given equation.


3x%2B12=2x%2B38 Distribute.


3x=2x%2B38-12 Subtract 12 from both sides.


3x-2x=38-12 Subtract 2x from both sides.


x=38-12 Combine like terms on the left side.


x=26 Combine like terms on the right side.


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Answer:

So the solution is x=26 which means that the first number is 26, the next number is 26+2=28, and the third is 26+4=30