Answer
the first even integer, $x,$ is $-10$ and the second consecutive even integer, $x+2,$ is $-8$
Work Step by Step
Let $x$ be the first even integer. Then $x+2$ is the next consecutive even integer.
The conditions of the problem translate to
\begin{array}{l}\require{cancel}
\dfrac{1}{x}+\dfrac{1}{x+2}=-\dfrac{9}{40}
.\end{array}
Using the $LCD=
40x(x+2)
$ and the properties of equality, the equation above is equivalent to
\begin{array}{l}\require{cancel}
40x(x+2)\left(\dfrac{1}{x}+\dfrac{1}{x+2}\right)=\left(-\dfrac{9}{40}\right)40x(x+2)
\\
40(x+2)(1)+40x(1)=x(x+2)(-9)
\\
40x+80+40x=-9x^2-18x
\\
80x+80=-9x^2-18x
\\
9x^2+80x+18x+80=0
\\
9x^2+98x+80=0
\\
(9x+8)(x+10)=0
.\end{array}
Equating each factor to zero (Zero Product Property) and then solving for the variable, then $
x=\left\{ -\dfrac{8}{9},-10 \right\}
.$
Since $x$ is an even integer, then $x=-10$ is the only solution. Hence, the first even integer, $x,$ is $-10$ and the second consecutive even integer, $x+2,$ is $-8.$
