## Algebra 2 Common Core

$x=\left\{ -\dfrac{3}{2}, 2 \right\}$
Using the properties of equality, the given equation, $\dfrac{2}{2x-1}=\dfrac{x}{3} ,$ is equivalent to \begin{align*}\require{cancel} 2(3)&=(2x-1)x &\text{(cross multiply)} \\ 2(3)&=2x(x)-1(x) &\text{(use Distributive Property)} \\ 6&=2x^2-x \\ 0&=2x^2-x-6 .\end{align*} Using the factoring of trinomials in the form $ax^2+bx+c,$ the expression \begin{align*} 2x^2-x-6 \end{align*} has $ac= 2(-6)=-12$ and $b= -1 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -4,3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{align*} 2x^2-4x+3x-6=0 .\end{align*} Grouping the first and second terms and the third and fourth terms, the expression above is equivalent to \begin{align*} (2x^2-4x)+(3x-6)=0 .\end{align*} Factoring the $GCF$ in each group results to \begin{align*} 2x(x-2)+3(x-2)=0 .\end{align*} Factoring the $GCF= (x-2)$ of the entire expression above results to \begin{align*} (x-2)(2x+3)=0 .\end{align*} The factored form of the given equation is $(x-2)(2x+3)=0.$ Equating each factor to zero (Zero Product Property) and then solving for the variable, then \begin{array}{lcl} x-2=0 &\text{ OR }& 2x+3=0 \\ x=2 && 2x=-3 \\ && x=-\dfrac{3}{2} \end{array} Hence, the solutions are $x=\left\{ -\dfrac{3}{2}, 2 \right\} .$