Question

I've been absent from school and doing assignments from home. I don't know how to find the inequalities.

Answer 1

Check the picture below.

so to get the EQUATion of each lines, we'll use those points you see in the picture for each, now we're only getting their "equation" only just yet, then we'll do the inequality part.

for the blue line

`(\stackrel{x_1}{0}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{0}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{0}-\stackrel{y1}{(-5)}}}{\underset{\textit{\large run}} {\underset{x_2}{5}-\underset{x_1}{0}}} \implies \cfrac{0 +5}{5} \implies \cfrac{ 5 }{ 5 } \implies 1`

`\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}{(-5)}=\stackrel{m}{ 1}(x-\stackrel{x_1}{0}) \implies y +5 = 1 ( x -0) \\\\\\ y+5=x\implies \boxed{y=x-5}`

for the red line

`(\stackrel{x_1}{-4}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-2}-\stackrel{y1}{3}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{(-4)}}} \implies \cfrac{-5}{6 +4} \implies \cfrac{ -5 }{ 10 } \implies - \cfrac{ 1 }{ 2 }`

`\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}{3}=\stackrel{m}{- \cfrac{ 1 }{ 2 }}(x-\stackrel{x_1}{(-4)}) \implies y -3 = - \cfrac{ 1 }{ 2 } ( x +4) \\\\\\ y-3=- \cfrac{ 1 }{ 2 }x-2\implies \boxed{y=- \cfrac{ 1 }{ 2 }x+1}`

now, the dashed line for the red one, means the borderline is not included so whatever "y" is, is either > or <.

the solid line for the blue line means, the borderline is included, so whatever "y" is, is ⩾ or ⩽.

now, what area do we shaded, let's deal with the red one first.

well, do usually a true/false region check, so we pick a point on either side of the line, hmmmm for simplicity let's pick the origin, (0,0), which is below the red line, that means x = 0 and y = 0

`y ~~ \square- \cfrac{ 1 }{ 2 }x+1\implies 0~~ \square-\cfrac{1}{2}(0)+1\implies 0~~ \square ~~ 1\implies 0 < 1`

so the sign that will make that statement true can only possible by "<", meaning that "0 is less than 1", that means that equation is

`{\Large \begin{array}{llll} y ~~ < - \cfrac{ 1 }{ 2 }x+1 \end{array}}`

now let's deal with the blue line.

same gig, we'll do a true/false region check, hmm let's pick for the sake of simplicity and slacking the same point, (0,0) which is above the blue line

`y ~~ \square ~~ x-5\implies 0~~ \square ~~0-5\implies 0~~ \square ~~-5\implies 0\geqslant -5`

so only sign that makes that true is "⩾" for the blue line, because "0 is indeed greater or equal than -5", so we get

`{\Large \begin{array}{llll} y \geqslant x-5 \end{array}}`

now, bear in mind that we could have pick some other point, on either side, and the issue is, to make it a true or false statement by using either inequality, for example if we end up with two values such as -9 [ ] -1, well, -9 is lesser than -1, so it can only be -9 < -1 or -9 ⩽ -1 to make it true, now if we want to make that statement false, we simply do -9 > -1 or such, and that part is "no shaded", because is false, -9 is not greater than -1.