4.7 Graphs of linear inequalities  (Page 2/10)

Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to indicate whether or not it the line is included in the solution. This is summarized in [link]

Now, let’s take a look at what we found in [link] . We’ll start by graphing the line $y=x+4$ , and then we’ll plot the five points we tested. See [link] .

In [link] we found that some of the points were solutions to the inequality $y>x+4$ and some were not.

Which of the points we plotted are solutions to the inequality $y>x+4$ ? The points $\left(1,6\right)$ and $\left(\mathrm{-8},12\right)$ are solutions to the inequality $y>x+4$ . Notice that they are both on the same side of the boundary line $y=x+4$ .

The two points $\left(0,0\right)$ and $\left(\mathrm{-5},\mathrm{-15}\right)$ are on the other side of the boundary line     $y=x+4$ , and they are not solutions to the inequality $y>x+4$ . For those two points, $y<x+4$ .

What about the point $\left(2,6\right)$ ? Because $6=2+4$ , the point is a solution to the equation $y=x+4$ . So the point $\left(2,6\right)$ is on the boundary line.

Let’s take another point on the left side of the boundary line and test whether or not it is a solution to the inequality $y>x+4$ . The point $\left(0,10\right)$ clearly looks to be to the left of the boundary line, doesn’t it? Is it a solution to the inequality?

Any point you choose on the left side of the boundary line is a solution to the inequality $y>x+4$ . All points on the left are solutions.

Similarly, all points on the right side of the boundary line, the side with $\left(0,0\right)$ and $\left(\mathrm{-5},\mathrm{-15}\right)$ , are not solutions to $y>x+4$ . See [link] .

The graph of the inequality $y>x+4$ is shown in [link] below. The line $y=x+4$ divides the plane into two regions. The shaded side shows the solutions to the inequality $y>x+4$ .

The points on the boundary line, those where $y=x+4$ , are not solutions to the inequality $y>x+4$ , so the line itself is not part of the solution. We show that by making the line dashed, not solid.

The boundary line shown is $y=2x-1$ . Write the inequality shown by the graph.

Solution

The line $y=2x-1$ is the boundary line. On one side of the line are the points with $y>2x-1$ and on the other side of the line are the points with $y<2x-1$ .

Let’s test the point $\left(0,0\right)$ and see which inequality describes its side of the boundary line.

At $\left(0,0\right)$ , which inequality is true:

$\begin{array}{ccccc}\hfill y>2x-1\hfill & & \hfill \text{or}\hfill & & \hfill y<2x-1?\hfill \\ \hfill y>2x-1\hfill & & & & \hfill y<2x-1\hfill \\ \hfill 0\stackrel{?}{>}2·0-1\hfill & & & & \hfill 0\stackrel{?}{<}2·0-1\hfill \\ \hfill 0>\mathrm{-1}\text{True}\hfill & & & & \hfill 0<\mathrm{-1}\text{False}\hfill \end{array}$

Since, $y>2x-1$ is true, the side of the line with $\left(0,0\right)$ , is the solution. The shaded region shows the solution of the inequality $y>2x-1$ .

Since the boundary line is graphed with a solid line, the inequality includes the equal sign.

The graph shows the inequality $y\ge 2x-1$ .

We could use any point as a test point, provided it is not on the line. Why did we choose $\left(0,0\right)$ ? Because it’s the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that $y<2x-1$ .

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Write the inequality shown by the graph with the boundary line $y=\mathrm{-2}x+3$ .

$y\ge \mathrm{-2}x+3$

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Write the inequality shown by the graph with the boundary line $y=\frac{1}{2}x-4$ .

$y<\frac{1}{2}x-4$

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