Answer
$\displaystyle \{\frac{1}{3}\}.$
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, $(x+2)(x-1)$
$3x(x-1)+1(x+2)=4-7x$
Distribute and simplify (write in standard fom).
$3x^{2}-3x+x+2-4+7x=0$
$3x^{2}+5x-2=0$
Two factors of $ac=-6$ whose sum is $b=5$
are +6 and -1. Rewrite the LHS:
$3x^{2}+6x-x-2=0$
$3x(x+2)-(x+2)=0$
$(3x-1)(x+2)=0$
By the zero product rule, x can be either -2 or $\displaystyle \frac{1}{3}.$
But, we excluded -2 as a possible solution at the beginning, so
the solution set is $\displaystyle \{\frac{1}{3}\}.$
