Question 641187
Two consecutive even integers. We'll call them {{{x}}} and {{{x+2}}}. Their reciprocals would be {{{1/x}}} and {{{1/(x+2)}}}. They add to {{{7/12}}}. 

So we have the following equation:

{{{1/x + 1/(x + 2) = 7/12}}}


{{{((x + 2) + x) / (x(x + 2)) = 7/12}}}


{{{(2x + 2) / (x^2 + 2x) = 7/12}}}


{{{12(2x + 2) = 7(x^2 + 2x)}}}


{{{24x + 24 = 7x^2 + 14x}}}


{{{7x^2-24x + 14x-24=0}}}


{{{7x^2-10x-24=0}}}

use quadratic formula to solve for {{{x}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{x = (-(-10) +- sqrt( (-10)^2-4*7*(-24) ))/(2*7) }}}


{{{x = (10 +- sqrt(100+672))/14 }}}


{{{x = (10 +- sqrt(772))/14 }}}


{{{x = (10 +- 27.79)/14 }}}

solutions:

{{{x = (10 + 27.79)/14 }}}

{{{x = 37.79/14 }}}

{{{x = 2.7 }}}


{{{x = (10 - 27.79)/14 }}}

{{{x = 17.79/14 }}}

{{{x = 1.27 }}}



check:

{{{1/x + 1/(x + 2) = 7/12}}}

{{{1/2.7  + 1/(1.27 + 2) = 0.6}}}

{{{0.3  + 1/3.27 = 0.6}}}

{{{0.3  + 0.3 = 0.6}}}

{{{0.6 = 0.6}}}