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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]
$Ax+By<C$ | $Ax+By\le C$ |
$Ax+By>C$ | $Ax+By\ge C$ |
Boundary line is not included in solution. | Boundary line is included in solution. |
Boundary line is dashed. | Boundary line is solid. |
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.
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:
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 $y<2x-1$ .
Write the inequality shown by the graph with the boundary line $y=\mathrm{-2}x+3$ .
$y\ge \mathrm{-2}x+3$
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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