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# Graphing Linear Inequalities

## Graphing Linear Inequalities 1.5 - Solution

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

### Boundary Line

To write the equation of the boundary line, replace the inequality sign with an equals sign. \begin{aligned} \text{-}2x+y = \text{-}3 \end{aligned} To draw this line, we will first rewrite the equation in slope-intercept form.
$\text{-}2x+y=\text{-}3$
$y=2x-3$
Now that the equation is in slope-intercept form, we can identify the slope $m$ and the $y\text{-}$intercept $b.$ $\begin{gathered} y=2x-3\quad \Leftrightarrow \quad y=2x+(\text{-}3) \end{gathered}$ We will plot the $y\text{-}$intercept $b=\text{-}3,$ and then use the slope $m=2$ 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 $(0,0)$ as our test point.
$y\lt 2x - 3$
${\color{#009600}{0}} \lt 2\cdot{\color{#0000FF}{0}} - 3$
$0\nless\text{-}3$

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, $y=x-1.$

### 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, $(0,0),$ as a test point to determine which region to shade.
$y
${\color{#009600}{0}} <{\color{#0000FF}{0}} -1$
$0\nless\text{-}1$
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, $y+3x=\frac{1}{2}.$

### Boundary line

First, we need to rewrite the equation in slope-intercept form by isolating $y.$ $y+3x=\frac{1}{2} \quad \Leftrightarrow \quad y=\text{-}3x+\frac{1}{2}$ Thus, the slope is $3$ and the $y$-intercept is $\frac{1}{2}.$ 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.
$y\leq \text{-}3x +\dfrac{1}{2}$
${\color{#009600}{0}} \stackrel{?}{\leq} \text{-}3\cdot{\color{#0000FF}{0}} +\dfrac{1}{2}$
$0 \stackrel{?}{\leq} 0 +\dfrac{1}{2}$
$0 \leq \dfrac{1}{2}$
Since the test point did produce a true statement, we will shade the region contains $(0,0).$