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Graphing Linear Inequalities

Graphing Linear Inequalities 1.5 - Solution

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a
Graphing an inequality involves two main steps.
  1. Plotting the boundary line.
  2. Shading the solution.

Boundary Line

To write the equation of the boundary line, replace the inequality sign with an equals sign. -2x+y=-3\begin{aligned} \text{-}2x+y = \text{-}3 \end{aligned} To draw this line, we will first rewrite the equation in slope-intercept form.
-2x+y=-3\text{-}2x+y=\text{-}3
y=2x3y=2x-3
Now that the equation is in slope-intercept form, we can identify the slope mm and the y-y\text{-}intercept b.b. y=2x3y=2x+(-3)\begin{gathered} y=2x-3\quad \Leftrightarrow \quad y=2x+(\text{-}3) \end{gathered} We will plot the y-y\text{-}intercept b=-3,b=\text{-}3, and then use the slope m=2m=2 to plot another point on the line. We will then connect these points with a straight line. This is our boundary line. Since we have a strict inequality, the boundary line will be dashed.

Test a point

To decide which side of the boundary line to shade, we will substitute a test point that is not on this line. If the substitution creates a true statement, we shade the region that includes the test point. Otherwise, we shade the opposite region. Let's use (0,0)(0,0) as our test point.
y<2x3y\lt 2x - 3
0<203{\color{#009600}{0}} \lt 2\cdot{\color{#0000FF}{0}} - 3
0-30\nless\text{-}3

Shade the correct region

Since the substitution of the test point did not create a true statement, we will shade the region that does not contain the point.

b
To graph the given inequality, we will begin by graphing the boundary line, y=x1.y=x-1.

Boundary line

Note that, since the inequality is less than, the boundary will not be included and the line should be dashed.

Test a point

Let's now use the origin, (0,0),(0,0), as a test point to determine which region to shade.
y<x1y<x-1
0<01{\color{#009600}{0}} <{\color{#0000FF}{0}} -1
0-10\nless\text{-}1
Thus, the point is not a solution to the inequality. This means that we should shade the region below the graph.
c
To graph the given inequality, we will begin by graphing the boundary line, y+3x=12.y+3x=\frac{1}{2}.

Boundary line

First, we need to rewrite the equation in slope-intercept form by isolating y.y. y+3x=12y=-3x+12 y+3x=\frac{1}{2} \quad \Leftrightarrow \quad y=\text{-}3x+\frac{1}{2} Thus, the slope is 33 and the yy-intercept is 12.\frac{1}{2}. We'll use this information to graph the boundary.

Test a point

Let's us the origin as a test point to determine what region to should shade.
y-3x+12y\leq \text{-}3x +\dfrac{1}{2}
0?-30+12{\color{#009600}{0}} \stackrel{?}{\leq} \text{-}3\cdot{\color{#0000FF}{0}} +\dfrac{1}{2}
0?0+120 \stackrel{?}{\leq} 0 +\dfrac{1}{2}
0120 \leq \dfrac{1}{2}
Since the test point did produce a true statement, we will shade the region contains (0,0).(0,0).