SOLUTION: please solve the following inequalities as i am having some difficulties: 4x + 12 >= 36 + 4 8z + 4z < 6z - 12 thank you :)

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Question 13471: please solve the following inequalities as i am having some difficulties:
4x + 12 >= 36 + 4
8z + 4z < 6z - 12
thank you :)

Answer by elima(1433) About Me  (Show Source):
You can put this solution on YOUR website!
Do not get confused with that inequality sign(<,>), you solve the problem just as if it was an = sign.
4x + 12 >= 36 + 4
We first add the 36+4=40;
4x%2B12%3E=40
Now we subtract 12 from both sides;
4x%2B12-12%3E=40-12
4x%3E=28
Now divide both sides by 4;
4x%2F4%3E=28%2F4
x%3E=7
8z + 4z < 6z - 12
First we want to add the 8z+4z, which we are able to do because they are like terms.8z+4z=12z
12z<6z-12
Now we want to get the variable z on one side. So we will subtract 6z from both sides.
12z-6z<6z-12-6z
6z<-12
Divide both sides by 6;
6z%2F6%3C-12%2F6
z<-2
Don't forget to check the answer.
1. 4%287%29%2B12%3E=40
28%2B12%3E=40
40%3E=40
2. 12(-2)<6(-2)-12
-24<-12-12
-24<-24
Hope this helps
=)