SOLUTION: finding three consecutive integers such that the sum of the first, twice the second, and three times the third is 62

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Question 215647: finding three consecutive integers such that the sum of the first, twice the second, and three times the third is 62
Found 2 solutions by drj, checkley77:
Answer by drj(1380)   (Show Source): You can put this solution on YOUR website!
Finding three consecutive integers such that the sum of the first, twice the second, and three times the third is 62


Step 1. Let n be one integer,

Step 2. Let n+1 be the second consecutive number and 2(n+1) is twice the second number

Step 3. Let n+2 is the third consecutive number and 3(n+2) is three times the third

Step 4. Then the sum is n+2(n+1)+3(n+2)=62 as given by the problem statement. The following steps will solve this equation.

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


YOUR ANSWER


  • This is an equation! Solutions: n=9.
  • Graphical form: Equation was fully solved.
  • Text form: n+2*(n+1)+3*(n+2)=62 simplifies to 0=0
  • Cartoon (animation) form:
    For tutors: simplify_cartoon( n+2*(n+1)+3*(n+2)=62 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at .
Moved these terms to the left highlight_green%28+-62+%29
It becomes .
Look at .
Expanded term 2 by using associative property on %28n%2B1%29
It becomes .
Look at .
Multiplied numerator integers
It becomes .
Look at .
Added fractions or integers together
It becomes .
Look at .
Moved -60 to the right of expression
It becomes .
Look at .
Removed extra sign in front of -60
It becomes .
Look at .
Eliminated similar terms highlight_red%28+n+%29,highlight_red%28+2%2An+%29 replacing them with highlight_green%28+%281%2B2%29%2An+%29
It becomes .
Look at .
Added fractions or integers together
It becomes .
Look at .
Remove unneeded parentheses around factor highlight_red%28+3+%29
It becomes .
Look at .
Expanded term 3 by using associative property on %28n%2B2%29
It becomes .
Look at .
Multiplied numerator integers
It becomes .
Look at .
Added fractions or integers together
It becomes .
Look at .
Removed extra sign in front of -54
It becomes .
Look at .
Eliminated similar terms highlight_red%28+3%2An+%29,highlight_red%28+3%2An+%29 replacing them with highlight_green%28+%283%2B3%29%2An+%29
It becomes .
Look at .
Added fractions or integers together
It becomes .
Look at .
Remove unneeded parentheses around factor highlight_red%28+6+%29
It becomes .
Look at .
Solved linear equation highlight_red%28+6%2An-54=0+%29 equivalent to 6*n-54 =0
It becomes .
Result:
This is an equation! Solutions: n=9.

Universal Simplifier and Solver


Done!


With n= 9, then n+1= 10 , and n+2=11

Check sum..9+2*10+3*11=9+20+33=622...which is a true statement.

Step 5. ANSWER: The numbers are 9, 10, and 11.

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



Answer by checkley77(12844)   (Show Source): You can put this solution on YOUR website!
Let x, x+1 & x+2 be the three integers.
x+2(x+1)+3(x+2)=62
x+2x+2+3x+6=62
6x=62-2-6
6x=54
x=54/6
x=9 ans. for the lowest integer.
9+1=10 ans. for the second integer.
9+2=11 ans. for the third integer.
9+2*10+3*11=62
9+20+33=62
62=62
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