first off 3) should have been before 1) and 2), so let's first do 3).
now, Check the picture below, the points are really making a line, so let's make a quick table of the material. [tex]\begin{array}{|cc|ll} \cline{1-2} \stackrel{people}{x}&\stackrel{price}{y}\\ \cline{1-2} 2&14\\ 4&20\\ 8&32\\ \cline{1-2} \end{array}[/tex]
now, to get the equation of any straight line, all we need is two points off of it, hmmm for this one let's use hmmm (2 , 14) and (8 , 32)
[tex](\stackrel{x_1}{2}~,~\stackrel{y_1}{14})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{32}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{32}-\stackrel{y1}{14}}}{\underset{run} {\underset{x_2}{8}-\underset{x_1}{2}}}\implies \cfrac{18}{6}\implies 3 \\\\\\ \begin{array}{|c|ll} \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}{14}=\stackrel{m}{3}(x-\stackrel{x_1}{2}) \\\\\\ y - 14 = 3x-6\implies y=3x+8[/tex]
now let's do 1)
[tex]\stackrel{\textit{a bus containing 30 folks}}{y=3(\stackrel{x}{30})+8\implies y=98}[/tex]
now let's do 2)
[tex]\stackrel{\textit{a bus charging 122 bucks}}{\stackrel{y}{122}=3x+8}\implies 114=3x\implies \cfrac{114}{3}=x\implies 38=x[/tex]
