Question 224489
Find three consecutive even integers such that twice largest integer exceeds the sum of the sum of the first two by 6.


Step 1.  Let n be the first even integer.


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


Step 3.  Let 2(n+4)=2n+8 be twice the largest integer.


Step 4.  Let n+n+2=2n+2 be the sum of the first two.


Step 5.  Then,  2n+8=2n+2+6 since twice largest integer exceeds the sum of the sum of the first two by 6.


Step 6.  Solving the equation in Step 5 yields the following steps.


*[invoke explain_simplification "2n+8=2n+2+6" ]


Step 7.  ANSWER:  The three consecutive integers can be anything that the satisfies equation in Step 5. 


For example 2, 4, 6 will yield 2(6)=2+4+6 which is a true statement


Another example,  20, 22, 24 will yield 2(24)=20+22+6  which is another true statement


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