## College Algebra (10th Edition)

$\displaystyle \{\frac{1}{3}\}.$
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 $\displaystyle \{\frac{1}{3}\}.$