Question 1124037

Let the first even integer be {{{n}}}.

Then the second consecutive even integer would be {{{n+2}}}

We must solve for n when:

{{{n+(n+2)=18}}}

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

{{{2n=18-2}}}

{{{2n=16}}}

{{{n=8}}}->the first even integer

{{{n+2=10}}}->the second consecutive even integer