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Mathematics 21 Online
OpenStudy (anonymous):

Which of the following ordered pairs is not a solution to the inequality y < -x + 4? (1, 2) (2, 1) (4, 0) (3, -2)

OpenStudy (anonymous):

just sub in all your x and y into the equations until you get something that doesn't make sense.

OpenStudy (anonymous):

how do i do that

OpenStudy (anonymous):

your points are int he form of (x,y)

OpenStudy (anonymous):

(1,2) for example x=1 and y=2

OpenStudy (anonymous):

y+x<4 just plugin the values of x&y to satisfy the above condition.

OpenStudy (anonymous):

y<-x+4 so for the first one you would end up with 2<-1+4 which equals 2<3

OpenStudy (anonymous):

is it D

OpenStudy (anonymous):

b

OpenStudy (anonymous):

no what did you get for b as you equation?

OpenStudy (anonymous):

2 < -1 +4

OpenStudy (anonymous):

solve the right hand side

OpenStudy (anonymous):

-1 + 4 = 3

OpenStudy (anonymous):

yup so your equation is 2<3 is this right or rwrong? this equation says that 2 is less than 3

OpenStudy (anonymous):

its right

OpenStudy (anonymous):

wait is it c because its 4 < 4

OpenStudy (anonymous):

yup and your wuestion is asking what point makes this pint wrong.

OpenStudy (anonymous):

yup :)

OpenStudy (anonymous):

you got it

OpenStudy (anonymous):

can you help with one more?

OpenStudy (anonymous):

yea no problem

OpenStudy (anonymous):

Part 1: Solve the compound inequality 3x + 6 > 36 and 3x + 6 54 and show all of your work. You can use <= to represent the less than or equal to symbol. Part 2: Use complete sentences to describe the graph of the solution from Part 1.

OpenStudy (anonymous):

okay lets look at your first compound inequality. 3x+6>36 attempt to solve this , what do you get?

OpenStudy (anonymous):

well do i - 6 from both sides? or the 3?

OpenStudy (anonymous):

subtract the 6

OpenStudy (anonymous):

if you subtracted 3 you would get 3x+3>33 which is useless if your trying to get x all by itself on a side

OpenStudy (anonymous):

from 36 and 3 right

OpenStudy (anonymous):

or from 6 because you wanna cancel it out?

OpenStudy (anonymous):

from the left side and the right side. so 3x + 6 -6>36-6

OpenStudy (anonymous):

3x>42

OpenStudy (anonymous):

you sure?

OpenStudy (anonymous):

3x>30?

OpenStudy (anonymous):

yup. now what would be your next step if you wanted to get one side to have just x ?

OpenStudy (anonymous):

divide by 3 on both sides

OpenStudy (anonymous):

x=10

OpenStudy (anonymous):

yup and now whats your next equation?

OpenStudy (anonymous):

oh wait x doest EQUAL to 10 but is rather ...

OpenStudy (anonymous):

so now i do the other side the same

OpenStudy (anonymous):

your equation was 3x>30 so x>10 not equal to 10

OpenStudy (anonymous):

x=16

OpenStudy (anonymous):

> yes

OpenStudy (anonymous):

sorry

OpenStudy (anonymous):

that's okay and I cant see your second equation what was it again?

OpenStudy (anonymous):

x<16 (3x + 6 > 36 and 3x + 6 54)

OpenStudy (anonymous):

okay for your second equation all I can see is 3x + 6 54 which makes no sense to me whats your sign > < or =

OpenStudy (anonymous):

less than or equal to sorry didnt see that

OpenStudy (anonymous):

now that i have that do i just put x<10 and x is less than or equal to 16 and thats the answer ?

OpenStudy (anonymous):

x>10

OpenStudy (anonymous):

now you have to graph it

OpenStudy (anonymous):

and in you graph your x values hve to be less than ten or greater than o equal to 16

OpenStudy (anonymous):

6 WOULD BE CLOSED CIRCLE TO THE RIGHT?

OpenStudy (anonymous):

...where di you get 6 from?

OpenStudy (anonymous):

16

OpenStudy (anonymous):

yup then

OpenStudy (anonymous):

and 10 would closed circle to the left

OpenStudy (anonymous):

thats how i graph it right?

OpenStudy (anonymous):

I think since your value cant be ten you would have to have an open acircle

OpenStudy (anonymous):

and yes I think so

OpenStudy (anonymous):

thank you so much!

OpenStudy (anonymous):

no problem im glad to help :)

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