## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=4x-29 \\\\\text{b) Standard Form: } 4x-y=29$
$\bf{\text{Solution Outline:}}$ Find the slope of the given equation, $4x-y=3 .$ Then use this slope and the given point , $( 7,-1 )$ to find the equation of the needed line. Give the equation in the Slope-Intercept Form and in the Standard Form. $\bf{\text{Solution Details:}}$ In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the given equation is equivalent to \begin{array}{l}\require{cancel} 4x-y=3 \\\\ -y=-4x+3 \\\\ -1(-y)=-1(-4x+3) \\\\ y=4x-3 .\end{array} Hence, the slope of the given line is $m=4 .$ Since parallel lines have the same slope, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} m=4 \\\text{Through: } (7,-1) .\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=-1 ,\\x_1=7 ,\\m=4 ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-(-1)=4(x-7) \\\\ y+1=4(x-7) .\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+1=4(x-7) \\\\ y+1=4(x)+4(-7) \\\\ y+1=4x-28 \\\\ y=4x-28-1 \\\\ y=4x-29 .\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=4x-29 \\\\ -4x+y=-29 \\\\ -1(-4x+y)=-1(-29) \\\\ 4x-y=29 .\end{array} The equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=4x-29 \\\\\text{b) Standard Form: } 4x-y=29 .\end{array}