Question

what are the solution of the inequality?check the solutions 4x+6<-6

Answer 1

no

Answer 2

Can you at least guess how you can solve for the variable 'x'?

Answer 3

6

Answer 4

What do you do with the number 6's?

Answer 5

You want to try to get 4x by it's self, to do so.. What do you have to do to both sides?

Answer 6

multiply

Answer 7

not mulitiply divide sorry

Answer 8

Wrong. Give you a hint, you will take something from one side, and do it to the other side. 6<-6

To get that 6 away from the 4x side, you have to do what? Keyword: Away

Answer 9

subtract

Answer 10

Correct, so the problem should look something like this..

4x+6<-6
-6 -6

This cancels out the 6 over by 4x and leaves you with that?

Answer 11

With what* sorry.

Answer 12

the answer i come up with is x<+6

Answer 13

You forgot the 4 infront of that X. Now you want to X by it's self to find out what X is right?

Answer 14

Sorry, it would be 4x<-12

Answer 15

the option of the answers none of them have the 4 infront the x

Answer 16

I understand this, what do you have to do to get x by it's self so it shows x<(number here)?

Answer 17

First off, what are your options anyhow? :)

Answer 18

A.x<-3 Bx>-3 C.x>-6 D.x<+6

Answer 19

Okay, so how would you solve 4x<-12 ? Do you know?

Answer 20

divid 12 by 2

Answer 21

No, sorry.

4x<-12

4x divided by 4.

That cancels the 4 beside the X so we can get just "X" like we wanted.

Then what you do to one side, you must to to the other side so let's take -12 divided by 4. Tell me what you get.

Answer 22

-3

Answer 23

Correct, so now that you divided by a negative, you must do what to the "<" sign?

Answer 24

turn the sign to a greater than sign

Answer 25

Correct! Great job Devonhinds =]! So your final answer would be x>-3

Answer 26

thnks

Answer 27

Anytime :) Glad I could help!

Answer 28

what number is a solution of the inequality?10.6<b

Answer 29

Give the Best Response button for this question, then close this question, and make another. I'll help you ;). Be sure to put within the question @skiller8860 so you can tag me in it :).