Fanduekisses
  • Fanduekisses
**CHECK MY WORK PWEASE?? **Am I doing this right? SOLVING INEQUALITY.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Fanduekisses
  • Fanduekisses
\[\frac{ 3 }{ x-1 }+\frac{ 2x}{ x+1 }>-1\]
Fanduekisses
  • Fanduekisses
\[\frac{ 3(x+1)+2x(x-1)+(x-1)(x+1) }{ (x-1)(x+1) } >0\]
Fanduekisses
  • Fanduekisses
\[\frac{ 3x^2+x-2 }{ (x-1)(x+1) }\]

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Fanduekisses
  • Fanduekisses
I did a sign chart but I get all positive, I know I must've done something wrong because the answer is \[(-\infty, -1) U (1,\infty)\]
Fanduekisses
  • Fanduekisses
I did a sign chart but I get all positive, I know I must've done something wrong because the answer is \[(-\infty, -1) U (1,\infty)\]
Fanduekisses
  • Fanduekisses
@countonme123 @dan815 @Data_LG2
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
there is some mistake here, lets see if we can find it
Fanduekisses
  • Fanduekisses
:)
misty1212
  • misty1212
first lets check the algebra
Fanduekisses
  • Fanduekisses
ok
misty1212
  • misty1212
ooh i see it!!
misty1212
  • misty1212
your numerator is wrong you wrote \[\frac{ 3x^2+x-2 }{ (x-1)(x+1) }\] but it should be \[\frac{ 3x^2+x+2 }{ (x-1)(x+1) }\]
Fanduekisses
  • Fanduekisses
Ohhhhhh I see it now too lol
misty1212
  • misty1212
now the numerator in this case is always positive so you can ignore it
misty1212
  • misty1212
and \[(x+1)(x-1)>0\] outside the zeros, on \(x<-1\) and \(x>1\)
Fanduekisses
  • Fanduekisses
So I only look at x=+1 then
misty1212
  • misty1212
zeros are at \(1\) and \(-1\)
Fanduekisses
  • Fanduekisses
Thank you! :D
Fanduekisses
  • Fanduekisses
I have another question, how do you find the domain of a function that has a fraction inside a square root? As in:\[f(x)=\sqrt{\frac{ x }{ x^2-2x-35 }}\]
Fanduekisses
  • Fanduekisses
I factored the denominator: (x-7)(x+5)
Fanduekisses
  • Fanduekisses
so x cannot equal 7 or -5

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