Question

HELP PLEASE I DONT GET THIS

Answer 1

so the idea being, we have a system of equations of two variables and 4 equations, each one rendering a line, for this case these aren't equations per se, they're INEquations, so pretty much the function will be the same for an equation but we'll use > or < instead of =, but fairly the function is basically the same, the behaviour differs a bit.

we have a line passing through (-6,0) and (0,8), side one

we have a line passing through the x-axis and -6, namely (-6,0) and the y-axis and -4, namely (0,-4), side two

we have a line passing through (0,-4) and (6,4), side three

now, side four is simply the line connecting one and three.

the intersection of all four lines looks like the one in the picture below, so what are those lines with their shading producing that quadrilateral?

well, we have two points for all four, and that's all we need to get the equation of a line, once we get the equation, with its shading like that in the picture, we'll make it an inequality.

`(\stackrel{x_1}{-6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{8}-\stackrel{y1}{0}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{(-6)}}} \implies \cfrac{8 -0}{0 +6} \implies \cfrac{ 8 }{ 6 } \implies \cfrac{4}{3}`

`\begin{array} \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}{0}=\stackrel{m}{ \cfrac{4}{3}}(x-\stackrel{x_1}{(-6)}) \implies y -0 = \cfrac{4}{3} ( x +6) \\\\\\ y=\cfrac{4}{3}x+8\hspace{5em}\stackrel{\textit{side one} }{\boxed{y < \cfrac{4}{3}x+8}}`

`\rule{34em}{0.25pt}\\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{-4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-4}-\stackrel{y1}{0}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{(-6)}}} \implies \cfrac{-4 -0}{0 +6} \implies \cfrac{ -4 }{ 6 } \implies - \cfrac{2}{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}{0}=\stackrel{m}{- \cfrac{2}{3}}(x-\stackrel{x_1}{(-6)}) \implies y -0 = - \cfrac{2}{3} ( x +6) \\\\\\ y=-\cfrac{2}{3}x-4\hspace{5em}\stackrel{\textit{side two} }{\boxed{y > -\cfrac{2}{3}x-4}} \\\\[-0.35em] \rule{34em}{0.25pt}`

`\stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{(-4)}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{0}}} \implies \cfrac{4 +4}{6 -0} \implies \cfrac{ 8 }{ 6 } \implies \cfrac{4}{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}{(-4)}=\stackrel{m}{ \cfrac{4}{3}}(x-\stackrel{x_1}{0}) \implies y +4 = \cfrac{4}{3} ( x -0) \\\\\\ y=\cfrac{4}{3}x-4\hspace{5em}\stackrel{ \textit{side three} }{\boxed{y > \cfrac{4}{3}x-4}} \\\\[-0.35em] \rule{34em}{0.25pt}`

`(\stackrel{x_1}{6}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{8}) ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{8}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{6}}} \implies \cfrac{ 4 }{ -6 } \implies - \cfrac{2}{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}{4}=\stackrel{m}{- \cfrac{2}{3}}(x-\stackrel{x_1}{6}) \\\\\\ y=-\cfrac{2}{3}x+8\hspace{5em}\stackrel{ \textit{side four} }{\boxed{y < -\cfrac{2}{3}x+8}}`

now, we can make that quadrilateral a trapezoid by simply moving one point for "side four", say we change the point (0 , 8) and in essence slide it down over the line to (-3 , 4). Notice, all we did was slide it down the line of side one, that means the equation for side one never changed and thus its inequality is the same function.

now, with the new points for side for of (-3,4) and (6,4), let's rewrite its inequality

`(\stackrel{x_1}{-3}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{(-3)}}} \implies \cfrac{4 -4}{6 +3} \implies \cfrac{ 0 }{ 9 } \implies 0`

`\begin{array}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}{4}=\stackrel{m}{ 0}(x-\stackrel{x_1}{(-3)}) \implies y -4 = 0 ( x +3) \\\\\\ y=4\hspace{5em}\stackrel{ \textit{side four changed} }{\boxed{y < 4}}`