Question 215647
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.


*[invoke explain_simplification "n+2(n+1)+3(n+2)=62" ] 


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