Which Inequality is Represented by the Graph

We tin graph inequalities with ane variable on a number line. We apply a closed dot,

$$

∙
,

\bullet,

to correspond

$$

≤

\leq

and

$$

≥
.

\geq.

Nosotros use an open up dot,

$$

∘
,

\circ,

to represent

$$

<

<

and

$$

>
.

>.

If

$$

10
≥
−
1
,

x \geq -one,

the we know that

$$

x

x

could be any number that is greater than or equal to -ane. We bear witness this solution on a number line by placing a closed dot at -1, to indicate that -1 is a solution, and shade all values on the number line that are greater than -i.

If

$$

2
≤
y
<
viii
,

2 \leq y <8,

then we know that

$$

y

y

can be any number greater than or equal to ii and less than 8. We evidence this solution on a number line past placing a airtight dot at 2 to indicate that 2 is a solution, placing an open up dot at 8 to signal that eight is not a solution, and shading all values on the number line between two and 8.

Write the inequality that is represented by the number line below.

---

The graph shows all numbers between -five and 1, but not including -v or 1. Therefore, the inequality represented by the graph is

$$

−
5
<
x
<
1.

-5<x<1.

The solution set to a two-variable linear inequality is shown as a shaded graph on the coordinate aeroplane. Shaded regions show the areas that comprise points in the solution. If a line is solid, then the points on the line are contained in the solution. If a line is dashed, then the points on the line are non independent in the solution.

To graph a ii-variable linear inequality:

• Put the inequality into slope-intercept class.

• Graph the line that bounds the inequality. Apply a solid line for
  
  

  $$
  
  ≤
  
  \leq
  
  

  
  
  or
  
  

  $$
  
  ≥
  
  \geq
  
  

  
  
  and a dotted line for
  
  

  $$
  
  <
  
  <
  
  

  
  
  or
  
  

  $$
  
  >
  .
  
  >.
  
  

  
  

• Shade above the line for
  
  

  $$
  
  >
  
  >
  
  

  
  
  or
  
  

  $$
  
  ≥
  .
  
  \geq.
  
  

  
  
  Shade below the line for
  
  

  $$
  
  <
  
  <
  
  

  
  
  or
  
  

  $$
  
  ≤
  .
  
  \leq.
  
  

  
  

Write an inequality to describe the graph.

The dotted line has a gradient of

$$

−

1
ii


-\frac{1}{2}

and and a

$$

y

y

-intercept of 1, so the equation of the line is

$$

y
=
−

1
two

10
+
ane.

y=-\frac{1}{2}x+1.

The line is dotted, then the solution set does not include the values on on the line.

In addition, the region below the line is shaded, indicating all of the

$$

y

y

values below the line. Therefore,

$$

y
<
−

1
2

x
+
ane.

y < -\frac{1}{2}x+1.

When we graph systems of linear inequalities, we graph ane inequality at a time. The solution is the shaded region that is the intersection of the inequalities.

$$

{





y
≤
2
x
+
iii








y
>
−

1
3

x
−
one






\brainstorm{cases} y \le 2x+three \\ y > -\frac{1}{3}ten-ane \end{cases}

Which shaded region represents the solutions to both of these inequalities?

$$

y
>
x

y
<
−
3
10
+
iv

y > x \\ y < – 3x + 4

---

Notice how the two lines split the graph into four regions. The prepare of solutions that satisfy both inequalities will always be ane of these four regions.

For this problem, the solutions to

$$

y
>
x

y > 10

will be above the line, in regions 1 or 2. The solutions to

$$

y
<
−
iii
x
+
4

y < -3x + four

volition be below the line, in regions 2 or 3.

Therefore, the regions that represents the solution set for both inequalities is 2.

For this gear up of inequalities, where on the coordinate plane exercise we find points that are solutions to both inequalities?

$$

y
≤
ten
+
2

y
≥
10
+
2

y\leq ten + two \\ y \geq x + ii

---

Information technology is not possible for a

$$

y

y

value to be both greater than

$$

ten
+
2

x+two

and less than

$$

x
+
2.

x+2.

Therefore, the only possible solutions are on the line of

$$

y
=
10
+
2.

y=ten+two.