anonymous
  • anonymous
Solve the following rational inequalities 2x< 4/x+1 Can you show your work please so I can see how to do it. Points will be rewarded Also, if you can can you try 5/x-2>(this one has a line under it, so it's greater than or equal to) 3x/x-2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[2x < \frac{ 4 }{ x+1 }\] That is what first question looks like
anonymous
  • anonymous
\[\frac{ 5 }{ x-2 }\ge \frac{ 3x }{ x-2 }\] That is the second equation
anonymous
  • anonymous
2x^2+2x<4 2x^2+2x-4<0 (x+2)(x-1)

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anonymous
  • anonymous
i multiplied (x+1) and 2x then got them all on the left side and found the zeros
anonymous
  • anonymous
5x-10>3x^2-6x 0>3x^2-11x+10 (x-2)(3x-5)
anonymous
  • anonymous
Thank you!!
anonymous
  • anonymous
you are welcome
anonymous
  • anonymous
oh no
anonymous
  • anonymous
be very careful here you cannot multiply both sides by an expression with a variable because you don't know if it is positive or negative
anonymous
  • anonymous
so if you want to solve \[2x < \frac{ 4 }{ x+1 }\] you have to either solve \[2x-\frac{4}{x+1}<0\] or \[0<\frac{4}{x+1}-2x\]
anonymous
  • anonymous
same for \[\frac{ 5 }{ x-2 }\ge \frac{ 3x }{ x-2 }\] solve \[\frac{5}{x-2}-\frac{3x}{x-2}\geq0\]
anonymous
  • anonymous
is it clear how to continue?
anonymous
  • anonymous
cant u just use plus or minus in case of an uncertainty?
anonymous
  • anonymous
\[2x-\frac{ 4 }{ x+1 }<0,or \frac{ 2x ^{2}+2x-4 }{x+1 }<0\] \[\frac{ 2\left( x ^{2}+x+\left( \frac{ 1 }{ 2 } \right)^{2}-\left( \frac{ 1 }{ 2 } \right)^{2} \right)-4 }{ x+1 }<0\] \[or \frac{ 2\left( x+\frac{ 1 }{ 2 } \right)^{2}-\frac{ 1 }{ 2 } -4}{ x+1 }<0\] \[or \frac{ 2\left( x+\frac{ 1 }{ 2 } \right)^{2}-\frac{ 9 }{2 } }{ x+1 }<0\] Two cases arise ,either numerator is >0 and denominator <0 or viceverse case 1. x+1<0 gives x<-1 \[and 2\left( x+\frac{ 1 }{ 2 } \right)^{2}-\frac{ 9 }{ 2 } >0\] \[\left( x+\frac{ 1 }{2 } \right)^{2}>\frac{ 9 }{4 }\] \[\left| x+\frac{ 1 }{ 2 } \right|>\frac{ 3 }{ 2 }\] \[x+\frac{ 1 }{2 }<-\frac{ 3 }{ 2 },x <-\frac{ 3 }{ 2 }-\frac{ 1 }{ 2 },x<-2\] \[or x+\frac{ 1 }{2 }>\frac{ 3 }{ 2 },x>\frac{ 3 }{2 }-\frac{ 1 }{2 },x>-1\],rejected because x<-1 combining x<-2 you should try to solve case 2 if face any difficulty iwill solve or if you have any quiry,let me know

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