To draw the boundary lines, we will treat the inequalities as equations.
|c|c|
[-1em]
Inequality & Equation -2pt
[-0.7em]
y≤ 2x-2 & y= 2x-2 -2pt [0.5em]
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y<2x-2 & y=2x-2 [0.5em]These equations represent straight lines written in slope-intercept form. Each line has a slope of $m=2$ and a $y\text{-}$intercept of $b=\N2.$ With this information, we can graph their lines. The only difference when graphing them is that we will dash the line of the second inequality. This is because this inequality is strict.
Shading
To complete the graphs we have to shade the correct side of each line. We can do that by testing a point that is not on any of the boundary lines. The easiest point we can choose is the origin.
\begin{array}{|c|c|c|}
\hline \\[-0.8em]
\bm{(x,y)} & \textbf{Inequality} & \textbf{Evaluate} \\[0.2em]
\hline \\[-1em]
(\col{0},\colII{0}) & \colII{0}\stackrel{?}{\leq}2(\col{0})-2 & \ 0\nleq\N2 \, \Large{\colIII{\times}} \\[0.7em]
\hline \\[-1em]
(\col{0},\colII{0}) & \colII{0}\stackrel{?}{<}2(\col{0})-2 & \ 0 \nless\N2 \, \Large{\colIII{\times}} \\[0.7em]
\hline
\end{array}
Since the origin produces a false statement for both inequalities, we should shade the side of the lines that does not contain the origin. We end up with the following diagram.
As we can see, the second inequality excludes the boundary line from the solution set whereas the first inequality does not .
