여러 가지 부등식

$x$ is between 3 and 8, not inclusive

ⓐ

$(0,0)$
$\quad$
Simplify.
	So, $(0,0)$ is not a solution to $y>x+4.$

ⓑ

$(1,6)$
$\quad$
Simplify.
	So, $(1,6)$ is a solution to $y>x+4.$

ⓒ

$(2,6)$
$\quad$
Simplify.
	So, $(2,6)$ is not a solution to $y>x+4.$

ⓓ

$(-5,-15)$
Simplify.
	So, $(-5,-15)$ is not a solution to $y>x+4.$

ⓔ

$(-8,12)$
$\,$
Simplify.
	So, $(-8,12)$ is a solution to $y>x+4.$

ⓐ yes ⓑ yes ⓒ yes ⓓ yes ⓔ no

ⓐ yes ⓑ yes ⓒ no ⓓ no ⓔ yes

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 position relative to the boundary line.

At $(0,0),$ which inequality is true: $y>2x-1$ or $y<2x-1?$

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

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: $2x+3y>6$ or $2x+3y<6?$

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

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 shaded region shows the solution to the inequality $2x+3y<6.$

$x-4y\le 8$

$3x-y\ge 6$

All points in the shaded region and on the boundary line, represent the solutions to $y>\frac{5}{2}x-4.$

All points in the shaded region, but not those on the boundary line, represent the solutions to $y<\frac{2}{3}x-5.$

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.

All points in the shaded region, but not those on the boundary line, represent the solutions to $x-2y<5.$

All points in the shaded region, but not those on the boundary line, represent the solutions to $2x-3y<6.$

All points in the shaded region, but not those on the boundary line, represent the solutions to $2x-y>3.$

First, we graph the boundary line $y=-4x.$ It is in slope–intercept form, with $m=-4$ 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.

All points in the shaded region and on the boundary line represent the solutions to $y\le -4x.$

All points in the shaded region, but not those on the boundary line, represent the solutions to $y>-3x.$

All points in the shaded region and on the boundary line, represent the solutions to $y\ge -2x.$

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 $(0,0).$

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

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

So we shade the side that does not include $(0,0)$ as shown in this graph.

All points in the shaded region, but not those on the boundary line, represent the solutions to $y>3.$