Question 221309
Find the three consecutive integers such that the sum of the first and second is 9 more than half of the third


Step 1.  Let n be the first 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 first and second integers.


Step 4.  Let {{{(n+2)/2+9}}} since 9 more than half of the third integer.


Step 5.  Then {{{2n+1=(n+2)/2+9}}} since the sum of the first and second is 9 more than half of the third


Step 6.  Solving yields the following steps


Multiply 2 to both sides of the equation


{{{2(2n+1)=2(n+2)/2+2*9}}} 


{{{4n+2=n+2+18}}} 


{{{4n+2=n+20}}}


Subtract n+2 from both sides of the equation


{{{4n+2-n-2=n+20-n-2}}}


{{{3n=18}}}


Divide 3 to both sides of the equation


{{{3n/3=18/3}}}


{{{n=6}}}  {{{n+1=7}}} and {{{n+2=8}}}


Check  {{{2n+1=(n+2)/2+9}}} in Step 5  2*6+1=8/2+9 or 12+1=4+9 which is a true statement.


Step 7.  ANSWER:  The three consecutive and even integers are 6, 7 and 8.


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