## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=\dfrac{1}{3}x+\dfrac{10}{3} \\\text{b) Standard Form: } x-3y=-10$
$\bf{\text{Solution Outline:}}$ Find the slope of the line defined by the given equation, $-x+3y=12 .$ Then use the Point-Slope Form of linear equations with the given point, $(-1,3)$ to find the equation of the needed line. Finally, Express the answer in both the Slope-Intercept and Standard forms. $\bf{\text{Solution Details:}}$ In the form $y=mx+b,$ the given equation is equivalent to \begin{array}{l}\require{cancel} -x+3y=12 \\\\ 3y=x+12 \\\\ \dfrac{3y}{3}=\dfrac{x}{3}+\dfrac{12}{3} \\\\ y=\dfrac{1}{3}x+4 .\end{array} Using $y=mx+b$ or the Slope-Intercept Form, where $m$ is the slope, then the slope of the line with the equation above is \begin{array}{l}\require{cancel} m=\dfrac{1}{3} .\end{array} Since parallel lines have the same slope, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} \text{Through: } (-1,3) \\ m=\dfrac{1}{3} .\end{array} Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions, \begin{array}{l}\require{cancel} y_1=3 ,\\x_1= -1 ,\\m= \dfrac{1}{3} ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-3=\dfrac{1}{3}(x-(-1)) \\\\ y-3=\dfrac{1}{3}(x+1) .\end{array} In the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y-3=\dfrac{1}{3}(x)+\dfrac{1}{3}(1) \\\\ y-3=\dfrac{1}{3}x+\dfrac{1}{3} \\\\ y=\dfrac{1}{3}x+\dfrac{1}{3}+3 \\\\ y=\dfrac{1}{3}x+\dfrac{1}{3}+\dfrac{9}{3} \\\\ y=\dfrac{1}{3}x+\dfrac{10}{3} .\end{array} In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} 3(y)=3\left( \dfrac{1}{3}x+\dfrac{10}{3} \right) \\\\ 3y=x+10 \\\\ -x+3y=10 \\\\ -1(-x+3y)=-1(10) \\\\ x-3y=-10 .\end{array} Hence, $\text{a) Slope-Intercept Form: } y=\dfrac{1}{3}x+\dfrac{10}{3} \\\text{b) Standard Form: } x-3y=-10 .$