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|dw:1372837027496:dw|

is this \[\frac{(x-3)^{2}}{(x+1)(x-2)}\]?

Yes. that is the equation.

ok so the top is >0 for all x does not equal 3 right?

Right.

so we need to make sure the bottom is > 0 right?

Yes.

|dw:1372837247856:dw|

i have forgot 3

the bottom is a parabola that opens up, what are its x intercepts?

-1,2 Right?

right

So far yes.

(-infinity,-1)U(2,3)U(3,infinity)

Okay, I get it now.

good deal:)

I have another quick question

oki

(-infinity,-1)

if it was <=
(-infinity,-1]

The original problem was
\[3(1-\frac{ 3 }{4 }x)>5-\frac{ 1 }{ 4 }x\]

3-(9/4)x>5-(1/4)x
-(8/4)x>5-3
2x<-2
x<-1
correct

you sure it didn't say set builder?

The question said Solve the inequalities and express your answer in Interval notation.

just go talk to your teacher, I make mistakes grading all the time....

Okay, thank you!

np

One more equation?

Inequality*

sure

\[|\frac{x=1 }{ 4}|\ge4\]

is that plus or minus?

I know your supposed to get two answers. I got one which was \[x \ge15\]

Plus. My bad

\[|\frac{ x+1 }{4 }|\]

the very last and should be or

Okay, So if I want the other answer I set the 4 to a negative answer?

so you have abs((x+1)/4)>=4
so
(x+1)/4>=4 gives the solutions you already have

not exactly

now we do
(x+1)/4<=-4
x+1<=-16
x<=-17

set it negative and flip the sign

Ah, that's what confused me.. Didn't know what to do with my sign

you could have done -(x+1)/4<=4
but as a rule I just make the "non absolute value side" negative

Good rule! With interval notation, how would I write this?

I think its
\[(-\infty,-17]U(\infty,15]\]

(-infinity,-17]U[15,infinity)

think of the number line, write it in the order it would fall on the number line

Oh, I see.

<=5

You lost me there..

But, Thank you so much for the help.