Thread: I don't understand this problem, rational inequailties!

1. I don't understand this problem, rational inequailties!

Hey guys, what's up? I'm a university student, and well, I suck at math =/
To the owners of this website, thank you, im sure this website has helped many and i hope i can also get help.

my question is about rational inequalities. if you could provide the answer while skimming through the steps, that would be awesome

Solve the inequality. Then graph the solution set on the real number line

x+12
____ >(greater than OR EQUAL TO) 3
x+2

^sorry, if that sounded confusing. I'll write out the equation using words if it would be easier to understand. ' x+12' divided by 'x+2' is greater than or equal to 3

2. Re: I don't understand this problem, rational inequailties!

Originally Posted by captaincrunch383
Hey guys, what's up? I'm a university student, and well, I suck at math =/
To the owners of this website, thank you, im sure this website has helped many and i hope i can also get help.

my question is about rational inequalities. if you could provide the answer while skimming through the steps, that would be awesome

Solve the inequality. Then graph the solution set on the real number line

x+12
____ >(greater than OR EQUAL TO) 3
x+2
If x> -2, x+ 2> 0, so it is positive and you can multiply both sides by it to get x+ 12>= 3(x+ 2). Can you solve that?

If x< -2, x+ 2< 0 so it is negative. Multiplying both sides by it will change the direction of the inequality to get x+ 12<= 3(x+ 2). Can you solve that?

^sorry, if that sounded confusing. I'll write out the equation using words if it would be easier to understand. ' x+12' divided by 'x+2' is greater than or equal to 3[/QUOTE]

3. Re: I don't understand this problem, rational inequailties!

x+12<= 3(x+2) ^^^

would that be .... x+12<= 3x=6

i'm still a bit confused about next steps. sorry =/

my professor has a really strong accent and most of the class doesn't understand a word he says .

4. Re: I don't understand this problem, rational inequailties!

Originally Posted by captaincrunch383
Solve the inequality. Then graph the solution set on the real number line
x+12
____ >(greater than OR EQUAL TO) 3
x+2

I would do it this way:
$\displaystyle \frac{x+12}{x+2}\ge 3\\\frac{x+12}{x+2}-3\ge 0\\\frac{6-2x}{x+2}\ge 0\\\frac{3-x}{x+2}\ge 0\\-2<x\le 3$

5. Re: I don't understand this problem, rational inequailties!

Originally Posted by Plato
I would do it this way:
$\displaystyle \frac{x+12}{x+2}\ge 3\\\frac{x+12}{x+2}-3\ge 0\\\frac{6-2x}{x+2}\ge 0\\\frac{3-x}{x+2}\ge 0\\-2<x\le 3$
thanks. if you don't mind me asking, are you a teacher? or just a really good math student?

the text book said something about critical numbers , would those be -3, and -2?

and then how would i graph -2 < x <= 3 ?

would it be, draw a number line, and at negative 2, have a parentheses like this ( on negative 2, and on 3, have a bracket like this ] on 3 ? and then shade between?

6. Re: I don't understand this problem, rational inequailties!

Originally Posted by captaincrunch383
thanks. if you don't mind me asking, are you a teacher? or just a really good math student?the text book said something about critical numbers , would those be -3, and -2?
and then how would i graph -2 < x <= 3 ?
would it be, draw a number line, and at negative 2, have a parentheses like this ( on negative 2, and on 3, have a bracket like this ] on 3 ? and then shade between?
Look at this webpage for the plot.

The critical numbers are $\displaystyle -2~\&~3$.

I am a retired professor.

7. Re: I don't understand this problem, rational inequailties!

Thank you Plato. So, that graph isn't the type of graph where you would have to test intervals right?