ANSWER
The smaller integer is 4 and the larger integer is 6.
EXPLANATION
Let the first integer be x.
Let the second integer be x + 2 (since they are consecutive even numbers)
The sum of their reciprocals is 5/12. That is
[tex]\frac{1}{x}+\frac{1}{x+2}=\frac{5}{12}[/tex]Simplify the left-hand side:
[tex]\begin{gathered} \frac{x+2+x}{x(x+2)}=\frac{5}{12} \\ \Rightarrow\frac{2x+2}{x^2+2x}=\frac{5}{12} \end{gathered}[/tex]Cross-multiply:
[tex]\begin{gathered} 12(2x+2)=5(x^2+2x) \\ 24x+24=5x^2+10x \\ \Rightarrow5x^2+10x-24x-24=0 \\ 5x^2-14x-24=0 \end{gathered}[/tex]Solve the quadratic equation by quadratic formula:
[tex]\begin{gathered} a=5,b=-14,c=-24\colon \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex][tex]\begin{gathered} x=\frac{14\pm\sqrt[]{(-14)^2-4(5)(-24)}}{2\cdot5} \\ x=\frac{14\pm\sqrt[]{196+480}}{10}=\frac{14\pm\sqrt[]{676}}{10} \\ x=\frac{14\pm26}{10} \\ \Rightarrow x=\frac{14+26}{10},x=\frac{14-26}{10} \\ x=\frac{40}{10},x=\frac{-12}{10} \\ x=4,x=-1.2 \end{gathered}[/tex]Since the number is an even integer, therefore, we have that:
[tex]x=4[/tex]That is the smaller integer.
Therefore, the larger integer is:
[tex]\begin{gathered} x+2 \\ \Rightarrow4+2 \\ \Rightarrow6 \end{gathered}[/tex]Hence, the smaller integer is 4 and the larger integer is 6.