Inequalities & Graphing

7/30/2019 Inequalities & Graphing

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Systems of Linear Equalities

By L.D.


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Problem 1

y > - 2x + 1


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Problem 1

y > - 2x + 1Now lots of people may not like graphing

inequalities like this, so an alternate way to

see this is y = -2x + 1, basically we are just

graphing an inequality in the y = mx + bmethod, m is the slope and b is the y

intercept. So lets graph it using the alternate

way first on the other page. The slope is red

and the y intercept is blue.


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Easy Right


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To make this an

inequality wemust look at our

original problem

y > - 2x + 1


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dMini Lesson

So, to tell the difference between > and >

when doing and inequality what you do is

you think that since > has more ink underit, it gets a full line like ______ and since

> has less ink under it, it can only afford a

dotted line _ _ _ _.


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After that short

lesson, we nowknow that the

line shouldnt be

solid since the

problem is

y > - 2x + 1. And

that inequality

sign doesnt have

enough ink.


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After that short

lesson, we nowknow that the

line shouldnt be

solid since the

problem is

y > - 2x + 1. And

that inequality

sign doesnt have

enough ink.


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Now we need to

bring in the

inequality part.

Since we dont know

what y is iny > - 2x + 1 we need

to fill up the empty

space on one side.

We choose the sideto shade based on if

the y is more or less

than the problem, if

it is more then we

shade the top, if it is

less, than we shade

the bottom. In this

case we shade


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THE TOP, THATS

ALL YOU NEEDSINCE THIS IS THE

FINAL ANSWER

(for problem 1)


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Problem 2

x 4y > -8

y 2x < -1


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Problem 2

x 4y > -8

y 2x < -1

So the first thing we

need to do is reorder

these so they fit the

y = mx + b format.


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Problem 2

x 4y > -8

y 2x < -1

(x 4y > -8) 4

x/4 y > -2

+y +yx/4 > -2 + y

+2 +2

x/4 + 2 > y


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Problem 2

x/4 + 2 > y

y 2x < -1

y 2x < -1

+2x +2x

y < -1 2x or -1 2x > y


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Problem 2

x/4 + 2 > y

-1 2x > y

Next we need to graph

them, just graph them,

dont shade yet. I will

use green for the topand purple for the

bottom.


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Now try shading


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The answer to

our problem is

the area theyboth shade over.

To check the

problem you can

choose a pointthey both shade

over, I choose (-

3,0), and then

you cansubstitute it into

both problems

individually.


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Problem 2

(-3)/4 + 2 > (0)

1 > 0

-1 2(-3) > (0)

-1 + 6 > 0

5 > 0


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Problem 2

(-3)/4 + 2 > (0)

1 > 0

-1 2(-3) > (0)

-1 + 6 > 0

5 > 0

Since both formulas

turn out correct, we cannow assume my

graphing was correct.


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Problem 3

x > 0

y > 0

y > 2x + 3


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Problem 3

If you are currently wondering how to deal with

the 0s, wonder no longer. Treat the zero like if

you had seen something like 5 > y or 6 < x. For

the former you would draw a horizontal lineacross the 5 on the y axis and for the latter you

would draw a vertical line across the 6 on the x

axis. Its the same with the 0s, you draw ahorizontal or vertical line for them. So now try to

graph the lines with shading.


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x > 0 (red dots)

y > 0 (green dots)y > 2x + 3 (blue dots)


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Problem 4

y > -5

y < 3


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Problem 4

This is the last one, so give it a try graphing-

wise. Remember to shade and check!


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Inequalities & Graphing