Assume the first even integer is x.
The next even integer will be 2 digits after this digit. So the next even integer will be x+2.
The reciprocals of these two even integers can be written as [tex] \frac{1}{x} [/tex] and [tex] \frac{1}{x+2} [/tex] respectively.
The sum of their reciprocals is 7/24. So we can set up the equation as:
[tex] \frac{1}{x}+ \frac{1}{x+2}= \frac{7}{24} [/tex]
Taking LCM on left hand side:
[tex] \frac{x+2+x}{x(x+2)}= \frac{7}{24} \\ \\
\frac{2x+2}{ x^{2} +2x}= \frac{7}{24} [/tex]
Cross Multiplying the denominators:
[tex]24(2x+2)=7({ x^{2} +2x}) \\ \\
48x+48=7 x^{2} +14x \\ \\
7 x^{2} +14x-48x-48=0 \\ \\
7 x^{2} -34x-48=0 \\ \\
[/tex]
Using the quadratic formula to solve the above equation:
[tex]7 x^{2} -34x-48=0 \\ \\
x= \frac{34+- \sqrt{1156+1344} }{14} \\ \\
x= \frac{34+- \sqrt{2500} }{14} \\ \\
x=\frac{34+-50 }{14} \\ \\
x=6 \\ x=- \frac{8}{7}
[/tex]
Since, the value of x can only be an integer, we discard the fractional value and keep x=6
So the first even integer is 6 and the next even integer is 8. The sum of reciprocals of 6 and 8 is 7/24