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# Solving Systems of Linear Inequalities Graphically

## Solving Systems of Linear Inequalities Graphically 1.8 - Solution

Let's graph each of the inequalities separately and then place the graphs on the same coordinate plane.

### Inequality (I)

In order to graph an inequality we need the boundary line. A boundary line can be written by replacing the inequality symbol with an equals sign. \begin{aligned} \underline{\textbf{Inequality}} \quad & \quad \underline{\textbf{Boundary Line}}\\ y\textcolor{magenta}{\geq}2 \qquad \ & \quad\qquad \ y\textcolor{magenta}{=}2 \end{aligned} The boundary line of the inequality is $y=2,$ which is a horizontal line whose $y\text{-}$intercept is $2.$ Notice that the line is solid because the inequality is not strict.

Since all $y$ values greater than or equal to $2$ are in the solution set, we will shade above the line.

### Inequality (II)

To start, let's determine the boundary line. \begin{aligned} \underline{\textbf{Inequality}} \quad & \quad \underline{\textbf{Boundary Line}}\\ y\textcolor{magenta}{\leq} x+5 \ \quad & \ \qquad y\textcolor{magenta}{=}x+5 \end{aligned} Notice that the boundary line is in slope-intercept form. In order to draw the line, we will plot the $y\text{-}$intercept $5$ and use the slope $1$ to determine a second point. Also, the boundary line will be solid because our inequality is non-strict.

Next, we will test a point. If the point satisfies the inequality, we shade the region that contains the test point. Otherwise, we will shade the other region. Let's test $(0,0)$ for simplicity.
$y\leq x+5$
${\color{#009600}{0}}\stackrel{?}{\leq} {\color{#0000FF}{0}}+5$
$0\leq 5 \ {\color{#009600}{\huge{\checkmark}}}$
Since the test point satisfies the inequality, we will shade the region that contains the test point. Let's add this graph to the same coordinate plane with Inequality (I).

### Inequality (III)

In the same manner as before, let's determine the boundary line for Inequality (III). \begin{aligned} \underline{\textbf{Inequality}} \quad & \quad \underline{\textbf{Boundary Line}}\\ y\textcolor{magenta}{\leq} \text{-} x+5 \quad & \ \qquad y\textcolor{magenta}{=}\text{-} x+5 \end{aligned} Once again, let's test the point $(0,0).$
$y\leq \text{-} x+5$
${\color{#009600}{0}}\leq \text{-}{\color{#0000FF}{0}}+5$
$0\leq 5 \ {\color{#009600}{\huge{\checkmark}}}$
Since the point satisfies the inequality, we will shade the region that contains the point. This boundary will be solid as well because the inequality is not strict. Let's add it to the same coordinate plane as the first two inequalities.

### Viewing the Solution

Next, let's remove all of the parts of the shading that aren't including in the overlapping sections.

### Finding the Shape and its Area

The overlapping section is a triangle. To calculate its area, we need to determine the base and height. Then we can use the formula for the area of a triangle.

As we can see from the graph, $b=6$ and $h=3.$ Let's find the area.
$A=\dfrac{b\cdot h}{2}$
$A=\dfrac{{\color{#0000FF}{6}}\cdot {\color{#009600}{3}}}{2}$
$A=\dfrac{18}{2}$
$A=9$
As a result, the area is $9 \ \text{units}^2.$