Question 1049905
Lets call the first integer {{{x}}}.
For the second integer to be a consecutive even interger it must be {{{x+2}}}}

 so far we have:
first interger = {{{x}}}
second interger = {{{x+2}}}

if given that the sum of {{{twice}}} the {{{lesser}}} and {{{three}}} times the {{{greater}}} is {{{66}}}, we have

{{{2x+3(x+2)=66}}}...............solve for {{{x}}}

{{{2x+3x+6=66}}}

{{{5x=66-6}}}

{{{5x=60}}}

{{{x=60/5}}}

{{{x=12}}}

and, second one is {{{x+2=14}}}