The sum of the reciprocals of two consecutive even integers is 9/40, what are the integers?

May 30, 2015

If the smaller of the two consecutive even integers is $x$
then, we are told,
$\textcolor{red}{\frac{1}{x}} + \textcolor{b l u e}{\frac{1}{x + 2}} = \frac{9}{40}$

So
$\textcolor{w h i t e}{\text{XXXXX}}$generating a common denominator on the left side:
$\left[\textcolor{red}{\frac{1}{x} \cdot \frac{x + 2}{x + 2}}\right] + \left[\textcolor{b l u e}{\frac{1}{x + 2} \cdot \left(\frac{x}{x}\right)}\right] = \frac{9}{40}$

$\left[\textcolor{red}{\frac{x + 2}{{x}^{2} + 2 x}}\right] + \left[\textcolor{b l u e}{\frac{x}{{x}^{2} + 2 x}}\right] = \frac{9}{40}$

$\frac{\textcolor{red}{\left(x + 2\right)} + \textcolor{b l u e}{\left(x\right)}}{{x}^{2} + 2 x} = \frac{9}{40}$

$\frac{2 x + 2}{{x}^{2} + 2 x} = \frac{9}{40}$

$\left(40\right) \left(2\right) \left(x + 1\right) = 9 \left({x}^{2} + 2 x\right)$

$80 x + 80 = 9 {x}^{2} + 18 x$

$9 {x}^{2} - 62 x - 80 = 0$

$\left(9 x + 1\right) \left(x - 8\right) = 0$

Since $x$ is an even integer
the two consecutive even integers are
$8$ and $10$