Answer
{$\frac{-1 - \sqrt {73}}{12},\frac{-1 + \sqrt {73}}{12}$}
Work Step by Step
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 {$\frac{-1 - \sqrt {73}}{12},\frac{-1 + \sqrt {73}}{12}$}.
