Question

How do I solve the following equation?

Answer 1

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

`y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{1}{3}}x+7\qquad \impliedby \qquad \begin{array}c \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}`

so we're really looking for the equation of a line whose slope is -1/3 and that it passes through (-2 , 5.5)

`(\stackrel{x_1}{-2}~,~\stackrel{y_1}{5.5})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{1}{3} \\\\\\ \begin{array}c \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5.5}=\stackrel{m}{- \cfrac{1}{3}}(x-\stackrel{x_1}{(-2)}) \implies y -5.5= -\cfrac{1}{3} (x +2)`

`y-5.5=-\cfrac{1}{3}x-\cfrac{2}{3}\implies y=-\cfrac{1}{3}x-\cfrac{2}{3}+5.5\implies y=-\cfrac{1}{3}x-\cfrac{2}{3}+\cfrac{55}{10} \\\\\\ y=-\cfrac{1}{3}x-\cfrac{2}{3}+\cfrac{11}{2}\implies {\Large \begin{array}{llll} y=-\cfrac{1}{3}x+\cfrac{29}{6} \end{array}}`