Answer
$\{5\}.$
Work Step by Step
The denominator on the RHS can be factored to $(x-2)(x+1)$.
We must exclude $x=-1 $and $x=2$ from the solutions,
as the equation would not be defined for these values.
Multiply both sides with the LCD, $(x-2)(x+1)$
$x(x+1)+2(x-2)=7x+1$
Distribute and simplify (write in standard fom).
$x^{2}+x+2x-4-7x-1=0$
$x^{2}-4x-5=0$
we can factor the LHS, as -5 and +1 are factors of -5,
and their sum is -4.
$(x-5)(x+1)=0$
By the zero product rule, x is either 5 or -1.
BUT, we excluded -1 in the first step, so
the solution set is $\{5\}.$
