## Elementary Algebra

{$\frac{-1 - \sqrt {73}}{12},\frac{-1 + \sqrt {73}}{12}$}
We will first cross multiply the fractions together in order to create a quadratic equation: $\frac{-1}{3x-1}=\frac{2x+1}{-2}$ $(3x-1)(2x+1)=-1(-2)$ $3x(2x+1)-1(2x+1)=2$ $6x^{2}+3x-2x-1-2=0$ $6x^{2}+x-3=0$ We need to use the quadratic formula to solve this equation: Step 1: Comparing $6x^{2}+x-3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=6$, $b=1$ and $c=-3$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a,b and c in the formula: $x=\frac{-(1) \pm \sqrt {(1)^{2}-4(6)(-3)}}{2(6)}$ Step 4: $x=\frac{-1 \pm \sqrt {1+72}}{12}$ Step 5: $x=\frac{-1 \pm \sqrt {73}}{12}$ Step 6: $x=\frac{-1 - \sqrt {73}}{12}$ or $x=\frac{-1 + \sqrt {73}}{12}$ Step 7: Therefore, the solution set is {$\frac{-1 - \sqrt {73}}{12},\frac{-1 + \sqrt {73}}{12}$}.