SOLUTION: One-sixth of the smallest of three consecutive even integers is three less than one-tenth the sum of the other even integers. Find the integers. 1/6x +3 = 1/10 (x+2)+(x+4) 1/6x

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Question 214750: One-sixth of the smallest of three consecutive even integers is three less than one-tenth the sum of the other even integers. Find the integers.
1/6x +3 = 1/10 (x+2)+(x+4)
1/6x +3 = 1/10 (2x+6)
1/6x +3 -3 = 1/10 (2x+3)
here is where I lose it, do not understand what to do with the fractions.
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
One-sixth of the smallest of three consecutive even integers is three less than one-tenth the sum of the other even integers. Find the integers.

Step 1. Let x be an even integer, x+2 and x+4 be the next two consecutive integers.

Step 2. x+2+x+4 is the sum of the two even integers or 2x+6
Step 3. (2x+6)/10 is one-tenth of the sum of the other even integers.

Step 4. x/6 is one-sixth of the smallest even itneger

Step 5. x%2F6=%282x%2B6%29%2F10-3 as given by the problem statement.

Step 6 Multiply 60 to both sides of the equation to get rid of the fractions.

60%2Ax%2F6=60%2A%28%282x%2B6%29%2F10-3%29

10x=6%2A%282x%2B6%29-180

10x=12x%2B36-180

10x=12x-144

Add 144-10x to both sides

10x%2B144-10x=12x-144%2B144-10x

2x=144

x=72

Step 7. The numbers are 72, 74 and 76.

I hope the above steps were helpful.

For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J