Answer
$-8 \text{ and } -7$
Work Step by Step
Let $x$ be the integer. Then $x+1$ is the other integer.
The conditions of the problem translate to
\begin{array}{l}\require{cancel}
\dfrac{1}{x}+\dfrac{1}{x+1}=-\dfrac{15}{56}
.\end{array}
Using the properties of equality, the equation above is equivalent to
\begin{array}{l}\require{cancel}
56x(x+1)\left(\dfrac{1}{x}+\dfrac{1}{x+1}\right)=\left(-\dfrac{15}{56}\right)56x(x+1)
\\\\
56(x+1)(1)+56x(1)=x(x+1)(-15)
\\\\
56x+56+56x=-15x^2-15x
\\\\
15x^2+(56x+56x+15x)+56=0
\\\\
15x^2+127x+56=0
\\\\
(x+8)(15x+7)=0
.\end{array}
Equating each factor to $0$ (Zero Product Property) and solving for the variable, then the solutions are $x=\left\{ -8,-\dfrac{7}{15}\right\}.$ Since $x$ is an integer then $x=-8$ is the only acceptable answer. Hence, the integers are $
-8 \text{ and } -7
.$
