## College Algebra (11th Edition)

$x=\left\{ -\dfrac{3}{2},4 \right\}$
$\bf{\text{Solution Outline:}}$ 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, $x=\left\{ -\dfrac{3}{2},4 \right\} .$