## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=3x+7 \\\\\text{b) Standard Form: } 3x-y=-7$
$\bf{\text{Solution Outline:}}$ To find the equation of the line with the following properties: \begin{array}{l}\require{cancel} \text{Slope: } 3 \\\text{Through: } (-1,4) ,\end{array} use the Point-Slope Form of linear equations. Give the equation in the Slope-Intercept Form and in the Standard Form. $\bf{\text{Solution Details:}}$ 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=4 ,\\x_1=-1 ,\\m=3 ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-4=3(x-(-1)) \\\\ y-4=3(x+1) .\end{array} Using the properties of equality, in the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y-4=3(x+1) \\\\ y-4=3(x)+3(1) \\\\ y-4=3x+3 \\\\ y=3x+3+4 \\\\ y=3x+7 .\end{array} Using the properties of equality, in the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y=3x+7 \\\\ -3x+y=7 \\\\ -1(-3x+y)=-1(7) \\\\ 3x-y=-7 .\end{array} The equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=3x+7 \\\\\text{b) Standard Form: } 3x-y=-7 .\end{array}