Question

How exactly do you know when you need to change a inequality into y=mx+b?

Answer 1

@Loser66

Answer 2

Should I always do it ?

Answer 3

the form is irrelevant to the math of it.

Answer 4

But when graphing it makes a difference

Answer 5

it relates to the previous one

Answer 6

no, its only in the eye of the beholder

Answer 7

wtf...

Answer 8

so a 2 parter eh

Answer 9

To me, ALWAYS

Answer 10

Which inequality will have a shaded area below the boundary line?
y – x > 5
2x – 3y 3
2x – 3y
7x + 2y 2
3x + 4y > 12

Answer 11

So I should change them all to y=mx+b before considering them?

Answer 12

Now, BELOW, right?

Answer 13

Right

Answer 14

not =, but the proper symbol. butill let yall finish up where yall left off

Answer 15

hence, you need y < (lesser) no need to change, just consider which inequality has y < something

Answer 16

Well I have to put them into that form to see which is y<

Answer 17

hahaha, you missed a bunch of symbols from b,c,d

Answer 18

Type the problem carefully, PLEASE

Answer 19

you CAN do it

Answer 20

Give me a second

Answer 21

D?(:

Answer 22

yup

Answer 23

I'm going to make it into that form for every problem.. That couldn't hurt huh?

Answer 24

yes

Answer 25

as long as you remember to flip the symbol when needed

Answer 26

it is easy to do, just practice more

Answer 27

Gotcha! Thanks a million, good teacher

Answer 28

@amistre64 When I graph something such as 3x + 4y > 12 it changes the look of the graph when I put it into the other form...

Answer 29

3x + 4y > 12

let x=0 to investigate how this moves along the y axis

4y > 12

y > 3 the true region is therefore is all the points above 3

Answer 30

7x + 2y <= 2, see how this moves along the y axis, let x=0

2y <= 2

y <= 1 , all the points under y=1 are true, so the shade is under the line

Answer 31

notice that i dont even have to graph them ... i just observe the movement along the y axis

Answer 32

Yeah I see that

Answer 33

Am I missing something.. The top and bottom answer are the same??

Answer 34

an aside ... the intercept form of a line, if you ever want to just graph the intercepts
\[ax+by=c\]
\[\frac{ax}{c}+\frac{by}{c}=\frac cc\]
\[\frac{x}{c/a}+\frac{y}{c/b}=1\]

when x=0, y=c/b
when y=0, x=c/a

Answer 35

the top and bottom graphs do appear to be alike yes

Answer 36

wtf.. They're both the right answer too

Answer 37

Which graph represents the solution set for the system x + 2y 3, x + y 4, and 3x − 2y 4?

Answer 38

ive had some computer programed stuff like that ... it trying to pull a random set

whats the actual question?

Answer 39

thats the question

Answer 40

its missing the <> stuff

Answer 41

lol one second

Answer 42

x + 2y <= 3 ; 3,0 and 0,1.5
2y <= 3
y <= 1.5 we have a shade under the blue line


x + y >= 4 ; 4,0 and 0,4
y >= 4 we have a shade above the red line


3x - 2y >= 4 ; 4/3, 0 and 0,-2
for variety, lets do y=0
3x >= 4
x >= 4/3, its shaded to the right of the green line

Answer 43

WTTTFFFF... How did I not do it right

Answer 44

please curb the profanity ...

Answer 45

What the fudge..

Answer 46

we can always dbl chk the point 4,1

x + 2y <= 3
4 1
4 + 2 = 6 <= 3 is false <---- it doenst work here


x + y >= 4
4 1
4 + 1 = 5 >=4 is true


3x - 2y >= 4
4 1
12 - 2 = 10 >= 4 is true

Answer 47

lol, thats better ... wtf has a specific and profane meaning.

Answer 48

I wish I could show you how I got the answer but It's on a graph

Answer 49

i defined the shaded region by observing the movement along an axis ... then i was able to detemrine which line was which by the intercepts alone; and then i simply applied the shading in the direction

Answer 50

Here let me try to show you what I tried to do and see if you can see why its wrong