## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=-\dfrac{1}{3}x+\dfrac{23}{3} \\\text{b) Standard Form: } x+3y=23$
$\bf{\text{Solution Outline:}}$ Find the slope of the line defined by the given equation, $3x-y=7 .$ 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} 3x-y=7 \\\\ -y=-3x+7 \\\\ -1(-y)=-1(-3x+7) \\\\ y=3x-7 .\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=3 .\end{array} Since parallel lines have negative reciprocal slopes, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} \text{Through: } (8,5) \\ 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=5 ,\\x_1= 8 ,\\m= -\dfrac{1}{3} ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-5=-\dfrac{1}{3}(x-8) .\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-5=-\dfrac{1}{3}(x)-\dfrac{1}{3}(-8) \\\\ y-5=-\dfrac{1}{3}x+\dfrac{8}{3} \\\\ y=-\dfrac{1}{3}x+\dfrac{8}{3}+5 \\\\ y=-\dfrac{1}{3}x+\dfrac{8}{3}+\dfrac{15}{3} \\\\ y=-\dfrac{1}{3}x+\dfrac{23}{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{23}{3} \right) \\\\ 3y=-x+23 \\\\ x+3y=23 .\end{array} Hence, $\text{a) Slope-Intercept Form: } y=-\dfrac{1}{3}x+\dfrac{23}{3} \\\text{b) Standard Form: } x+3y=23 .$