#### Answer

$x=\left\{ -\dfrac{3}{2},4 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given rational equation, $
\dfrac{x+4}{2x}=\dfrac{x-1}{3}
,$ use cross-multiplication and the concepts of quadratic equations.
$\bf{\text{Solution Details:}}$
Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to
\begin{array}{l}\require{cancel}
(x+4)(3)=2x(x-1)
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac$ and the properties of equality, the expression above is equivalent to
\begin{array}{l}\require{cancel}
3x+12=2x^2-2x
\\\\
-2x^2+(3x+2x)+12=0
\\\\
-2x^2+5x+12=0
\\\\
-1(-2x^2+5x+12)=(0)(-1)
\\\\
2x^2-5x-12=0
.\end{array}
In the trinomial expression above the value of $ac$ is $
2(-12)=-24
$ and the value of $b$ is $
-5
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
3,-8
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2x^2+3x-8x-12=0
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(2x^2+3x)-(8x+12)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(2x+3)-4(2x+3)=0
.\end{array}
Factoring the $GCF=
(2x+3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(2x+3)(x-4)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
2x+3=0
\\\\\text{OR}\\\\
x-4=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
2x+3=0
\\\\
2x=-3
\\\\
x=-\dfrac{3}{2}
\\\\\text{OR}\\\\
x-4=0
\\\\
x=4
.\end{array}
Upon checking, none of the solutions above will make the denominator of the original equation equal to zero. Hence, $
x=\left\{ -\dfrac{3}{2},4 \right\}
.$