일차함수와 일차방정식

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Answers will vary.

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 $(0,0)$ and see which inequality describes its side of the boundary line.

At $(0,0)$, which inequality is true:

$$\begin{array}{ccccc}y>2x-1 & & \text{or} & & y<2x-1? \\ y>2x-1 & & & & y<2x-1 \\ 0\overset{?}{>}2\cdot 0-1 & & & & 0\overset{?}{<}2\cdot 0-1 \\ 0>-1\,\text{True} & & & & 0<-1\,\text{False}\end{array}$$

Since, $y>2x-1$ is true, the side of the line with $(0,0)$, 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 $(0,0)$? 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 $y<2x-1$.

$y\ge -2x+3$

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

The line $2x+3y=6$ is the boundary line. On one side of the line are the points with $2x+3y>6$ and on the other side of the line are the points with $2x+3y<6$.

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

At $(0,0)$, which inequality is true:

$$\begin{array}{ccccccccccc}2x+3y & > & 6 & & & \text{or} & & & 2x+3y & < & 6? \\ 2x+3y & > & 6 & & & & & & 2x+3y & < & 6 \\ 2(0)+3(0) & \overset{?}{>} & 6 & & & & & & 2(0)+3(0) & \overset{?}{<} & 6 \\ 0 & > & 6\,\text{False} & & & & & & 0 & < & 6\,\text{True}\end{array}$$

So the side with $(0,0)$ is the side where $2x+3y<6$.

(You may want to pick a point on the other side of the boundary line and check that $2x+3y>6$.)

Since the boundary line is graphed as a dashed line, the inequality does not include an equal sign.

The graph shows the solution to the inequality $2x+3y<6$.

$x-4y\le 8$

$3x-y\ge 6$

First we graph the boundary line $x-2y=5$. The inequality is $<$ so we draw a dashed line.

Then we test a point. We’ll use $(0,0)$ again because it is easy to evaluate and it is not on the boundary line.

Is $(0,0)$ a solution of $x-2y<5$?

The point $(0,0)$ is a solution of $x-2y<5$, so we shade in that side of the boundary line.

First we graph the boundary line $y=-4x$. It is in slope–intercept form, with $m=-4\,\text{and}\,b=0$. The inequality is $\le$ so we draw a solid line.

Now, we need a test point. We can see that the point $(1,0)$ is not on the boundary line.

Is $(1,0)$ a solution of $y\le -4x$?

The point $(1,0)$ is not a solution to $y\le -4x$, so we shade in the opposite side of the boundary line. See (the referenced figure).

First we graph the boundary line $y=3$. It is a horizontal line. The inequality is > so we draw a dashed line.

We test the point $(0,0)$.

$$\begin{array}{c} \\ y>3 \\ \\ 0>3\end{array}$$

$(0,0)$ is not a solution to $y>3$.

So we shade the side that does not include (0, 0).