Question 222441
Find three consecutive whole numbers such that twice the sum of the two smallest  numbers is 10 more than three times the largest number.


Step 1.  Let n be an integer.


Step 2.  Let n+1 and n+2 be the next two consecutive integers.


Step 3.  Let n+n+1=2n+1 be the sum of the two smallest integers.


Step 4.  Let 2(2n+1) be twice the sum of the two smallest integers.


Step 5.  Let 3(n+2) be three times the largest number


Step 6.  Then using Steps 3 and 4,  2(2n+1)=3(n+2)+10  since twice the sum of the two smallest  numbers is 10 more than three times the largest number.


Step 7.  Solving yields the following steps:


*[invoke explain_simplification "2(2n+1)=3(n+2)+10" ]


n=14, then n+1=15 and n+2=16


Check original equation  {{{2(2n+1)=3(n+2)+10}}} or {{{2(14+15)=3*16+10}}} such that {{{58=58}}} which is a true statement.


Step 8.  ANSWER:  The three consecutive integers are 14, 15, and 16.


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.


Good luck in your studies!


Respectfully,
Dr J



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.


Good luck in your studies!


Respectfully,
Dr J