Question 1131038


Let the smaller even integer be {{{n}}}, and larger consecutive even integer would be {{{n+2}}}.

if the smaller added to four times the larger gives a sum of {{{48}}}, we have

{{{n+4(n+2)=48}}}....solve for {{{n}}}

{{{n+4n+8=48}}}

{{{5n=48-8}}}

{{{5n=40}}}

{{{n=40/5}}}

{{{n=8}}}


find second number

{{{n+2=8+2=10}}}


so, your numbers are: {{{8}}} and {{{10}}}