Algebra 2 Common Core

\begin{array}{l}\require{cancel} \\\text{Slope: } -\dfrac{1}{2} \\\text{$y$-intercept: } \left(0, -\dfrac{5}{2}\right) \end{array}
Using the properties of equality, in the form $y=mx+b,$ the given equation is equivalent to \begin{array}{l}\require{cancel} -3\left( -\dfrac{1}{3}x-\dfrac{2}{3}y \right)=\left( \dfrac{5}{3} \right)(-3) \\\\ -1(-x)-1(-2y)=(5)(-1) \\ x+2y=-5 \\ 2y=-x-5 \\\\ \dfrac{2y}{2}=\dfrac{-x-5}{2} \\\\ y=-\dfrac{1}{2}x-\dfrac{5}{2} .\end{array} Using $y=mx+b$ (where $m$ is the slope and $(0, b)$ is the $y$-intercept) or the Slope-Intercept Form of linear equations, then the equation above has the following properties \begin{array}{l}\require{cancel} \\\text{Slope: } -\dfrac{1}{2} \\\text{$y$-intercept: } \left(0, -\dfrac{5}{2}\right) .\end{array}