```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.