i need help with this problem

janetfizz

New member
Joined
Mar 14, 2006
Messages
2
1/3x < 21 :?
 

emmaiskool242

Junior Member
Joined
Mar 29, 2006
Messages
66
i'm not sure how to do this one, what is the section called in your book, then i might be able to help you more. Because I think I know how to do it but i'm not quite sure
 
G

Guest

Guest
janetfizz said:
1/3x < 21 :?

1/3x < 21

Multiply both sides by three to get rid of the fraction.

x < 63.

:)


Sorry, I made a really stupid mistake. I wasn't thinking.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,829
You have not given a very clear statement of your problem!
Depending of which it means:
\(\displaystyle \begin{array}{l}
\frac{1}{{3x}} < 21\quad \Rightarrow \quad \frac{1}{{63}} < x \\
\frac{1}{3}x < 21\quad \Rightarrow \quad x < 63 \\
\end{array}\)
Please learn to format the question you mean to ask!


EDIT: Aly, before you answer please be sure you know what you are doing!
 
G

Guest

Guest
Re: thanks

janetfizz said:
thank you very much how do you do 5p-7=18 and y-14.4< 18.39
Well, I don't want to do your homework for you, but the help I can give you is so straightforward that it'll give you the answer. I'll help you with the second one, and then I'll give you a hit about the first one.

y-14.4 < 18.39

Add 14.4 to both sides.

y < 22.79


For the first one, try adding the seven. Then finish out the problem. See what you get. Always make sure you have your numbers on one side and your variable on the other when working problems like these out.
 
G

Guest

Guest
pka said:
You have not given a very clear statement of your problem!
Depending of which it means:
\(\displaystyle \begin{array}{l}
\frac{1}{{3x}} < 21\quad \Rightarrow \quad \frac{1}{{63}} < x \\
\frac{1}{3}x < 21\quad \Rightarrow \quad x < 63 \\
\end{array}\)
Please learn to format the question you mean to ask!


EDIT: Aly, before you answer please be sure you know what you are doing!

Gack, you're right. I meant to ask anyway but I automatically assumed, since it appears to be just a basic inequality. I'll bear that in mind next time. Thanks, pka.
 
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