How to solve an equation where both sides are rational functions of $x$?

Question

How would I solve

$$\frac{x+3}{3x+1}=\frac{2x-1}{3-2x} $$

I attempted multiplying $3x+1$, then multiplying $3-2x$

Then factorising although when I get to solving the quadratic equation it is incorrect and I'm not sure where I have gone wrong with solving it

Answer 1

\begin{align} \frac{x+3}{3x+1}=\frac{2x-1}{3-2x}\,\,\,&\overset{\text{your}}{\underset{\text{approach}}{\iff}} (x+3)(3-2x)=(2x-1)(3x+1) \\ &\iff -2 x^2-3 x+9 = 6 x^2-x-1 \\ &\iff -8 x^2-2 x+10 = 0\tag{1} \\\,\\\ &\overset{\text{quadratic}}{\underset{\text{formula}}\iff} x=\dfrac{2\pm\sqrt{4-4\cdot(-8)\cdot10}}{2(-8)} \\ \end{align}

You could also factor out $2$ in $(1)$ to get: $-4x^2-x+5=0$.

\begin{align}\iff x=-\dfrac54\text{ or }x=1\end{align}

Answer 2

Hint: You should end up with $4x^2 + x - 5 = 0$, which factors as $(4x + 5)(x - 1) = 0$.