SOLUTION: find three consecutive even integers such that the sum of the first interger is three times the thrid integer is 64 more that the second integer

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Question 225514: find three consecutive even integers such that the sum of the first interger is three times the thrid integer is 64 more that the second integer
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Find three consecutive even integers such that the sum of the first integer and three times the third integer is 64 more than the second integer.

Step 1. Let n be the first even integer.

Step 2. Let n+2 and n+4 be the next two consecutive even integers.

Step 3. Let 3(n+4) be three times the third integer.

Step 4. Let n+3(n+4) be the sum of the first integer and 3 times the third integer

Step 5. Let n+2+64=n+66 be 64 more than the second integer.

Step 6. Then, n+3(n+4)=n+66 since the sum of the first integer and three times the third integer is 64 more than the second integer

Step 7. Solving yields the following steps

Solved by pluggable solver: EXPLAIN simplification of an expression
Your Result:


YOUR ANSWER


  • This is an equation! Solutions: n=18.
  • Graphical form: Equation n%2B3%2A%28n%2B4%29=n%2B66 was fully solved.
  • Text form: n+3*(n+4)=n+66 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+n%2B3%2A%28n%2B4%29=n%2B66+%29
    For tutors: simplify_cartoon( n+3*(n+4)=n+66 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at n%2B3%2A%28n%2B4%29=highlight_red%28+n%2B66+%29.
Moved these terms to the left highlight_green%28+-n+%29,highlight_green%28+-66+%29
It becomes n%2B3%2A%28n%2B4%29-highlight_green%28+n+%29-highlight_green%28+66+%29=0.
Look at highlight_red%28+n+%29%2B3%2A%28n%2B4%29-highlight_red%28+n+%29-66=0.
Eliminated similar terms highlight_red%28+n+%29,highlight_red%28+-n+%29 replacing them with highlight_green%28+%281-1%29%2An+%29
It becomes highlight_green%28+%281-1%29%2An+%29%2B3%2A%28n%2B4%29-66=0.
Look at highlight_red%28+%281-1%29%2An+%29%2B3%2A%28n%2B4%29-66=0.
Since highlight_red%28+%281-1%29%2An+%29 has zero as a factor, it should be replaced with a zero
Look at 0%2B3%2A%28n%2B4%29-66=0.
Added fractions or integers together
It becomes 0%2B3%2A%28n%2B4%29-66=0.
Look at highlight_red%28+0+%29%2B3%2A%28n%2B4%29-66=0.
Remove extraneous zero highlight_red%28+0+%29
It becomes 3%2A%28n%2B4%29-66=0.
Look at highlight_red%28+3%2A%28n%2B4%29+%29-66=0.
Expanded term 3 by using associative property on %28n%2B4%29
It becomes highlight_green%28+3%2An+%29%2Bhighlight_green%28+3%2A4+%29-66=0.
Look at 3%2An%2Bhighlight_red%28+3+%29%2Ahighlight_red%28+4+%29-66=0.
Multiplied numerator integers
It becomes 3%2An%2Bhighlight_green%28+12+%29-66=0.
Look at 3%2An%2Bhighlight_red%28+12+%29-highlight_red%28+66+%29=0.
Added fractions or integers together
It becomes 3%2An%2Bhighlight_green%28+-54+%29=0.
Look at 3%2An%2Bhighlight_red%28+-54+%29=0.
Removed extra sign in front of -54
It becomes 3%2An-highlight_green%28+54+%29=0.
Look at highlight_red%28+3%2An-54+%29=0.
Solved linear equation highlight_red%28+3%2An-54=0+%29 equivalent to 3*n-54 =0
It becomes highlight_green%28+0+%29=0.
Result: 0=0
This is an equation! Solutions: n=18.

Universal Simplifier and Solver


Done!



With n=18, then n+2=20 and n+4=22

Check equation in Step 6: n+3(n+4)=n+66...18+3*22=18+66 or 84=84...which is a true statement.

Step 8 The three consecutive even integers are 18, 20, and 22.

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J
http://www.FreedomUniversity.TV