Question 110886

If the {{{lesser}}} of two consecutive {{{even}}} integers is {{{10}}} {{{more}}} than {{{one-half}}}{{{ the}}}{{{ greater}}}, then we have:


If {{{n}}} is an integer than {{{2n}}} and {{{ 2n+2}}} will be two even consecutive integers where {{{2n}}} is {{{lesser}}}, and {{{2n + 2}}} is {{{greater}}} even integer.

If the {{{lesser}}} of two consecutive {{{even}}} integers is {{{10}}} {{{more}}} than {{{one-half}}}{{{ the}}}{{{ greater}}}, then we have:

{{{2n  + 10 =(1/2)( 2n + 2)}}}……..solve for {{{n}}}

{{{2n  + 10 =(1/2)2n + 2(1/2))}}}……..

{{{2n  + 10 = n + 1)}}}……..move {{{n}}} to the left and {{{10 }}} to the right

{{{2n  -n  = -10 + 1)}}}……..

{{{n  = - 9)}}}……..

Then:
the first even number is {{{2n = 2(-9) = - 18}}}, and second even number is {{{2n + 2 = -18 + 2 = - 16}}}

check if the {{{lesser}}} of two consecutive {{{even}}} integers is {{{10}}} {{{more}}} than {{{one-half}}}{{{ the}}}{{{ greater}}}:

{{{2n  + 10 =(1/2)( 2n + 2)}}}……..plug in values

{{{-18  + 10 =(1/2)( -16)}}}……..

{{{-8   = - 8}}}……..