Question

**CHECK MY WORK PWEASE?? **Am I doing this right? SOLVING INEQUALITY.

Answer 1

\[\frac{ 3 }{ x-1 }+\frac{ 2x}{ x+1 }>-1\]

Answer 2

\[\frac{ 3(x+1)+2x(x-1)+(x-1)(x+1) }{ (x-1)(x+1) } >0\]

Answer 3

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

Answer 4

I did a sign chart but I get all positive, I know I must've done something wrong because the answer is \[(-\infty, -1) U (1,\infty)\]

Answer 5

I did a sign chart but I get all positive, I know I must've done something wrong because the answer is \[(-\infty, -1) U (1,\infty)\]

Answer 6

@countonme123 @dan815 @Data_LG2

Answer 7

HI!!

Answer 8

there is some mistake here, lets see if we can find it

Answer 9

:)

Answer 10

first lets check the algebra

Answer 11

ok

Answer 12

ooh i see it!!

Answer 13

your numerator is wrong you wrote \[\frac{ 3x^2+x-2 }{ (x-1)(x+1) }\] but it should be \[\frac{ 3x^2+x+2 }{ (x-1)(x+1) }\]

Answer 14

Ohhhhhh I see it now too lol

Answer 15

now the numerator in this case is always positive so you can ignore it

Answer 16

and \[(x+1)(x-1)>0\] outside the zeros, on \(x<-1\) and \(x>1\)

Answer 17

So I only look at x=+1 then

Answer 18

zeros are at \(1\) and \(-1\)

Answer 19

Thank you! :D

Answer 20

I have another question, how do you find the domain of a function that has a fraction inside a square root? As in:\[f(x)=\sqrt{\frac{ x }{ x^2-2x-35 }}\]

Answer 21

I factored the denominator: (x-7)(x+5)

Answer 22

so x cannot equal 7 or -5