## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=3x+11 \\\text{b) Standard Form: } 3x-y=-11$
$\bf{\text{Solution Outline:}}$ Find the slope of the line defined by the given equation, $3x-y=4 .$ Then use the Point-Slope Form of linear equations with the given point, $(-2,5)$ 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=4 \\\\ -y=-3x+4 \\\\ -1(-y)=-1(-3x+4) \\\\ y=3x-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=3 .\end{array} Since parallel lines have same slopes, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} m=3 \\\text{Through: } (-2,5) .\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= -2 ,\\m= 3 ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-5=3(x-(-2)) \\\\ y-5=3(x+2) .\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=3(x+2) \\\\ y-5=3(x)+3(2) \\\\ y-5=3x+6 \\\\ y=3x+6+5 \\\\ y=3x+11 .\end{array} In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y=3x+11 \\\\ -3x+y=11 \\\\ -1(-3x+y)=-1(11) \\\\ 3x-y=-11 .\end{array} Hence, $\text{a) Slope-Intercept Form: } y=3x+11 \\\text{b) Standard Form: } 3x-y=-11 .$