Question

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

Answer 1

\[2x < \frac{ 4 }{ x+1 }\]

That is what first question looks like

Answer 2

\[\frac{ 5 }{ x-2 }\ge \frac{ 3x }{ x-2 }\]
That is the second equation

Answer 3

2x^2+2x<4
2x^2+2x-4<0
(x+2)(x-1)

Answer 4

i multiplied (x+1) and 2x then got them all on the left side and found the zeros

Answer 5

5x-10>3x^2-6x
0>3x^2-11x+10
(x-2)(3x-5)

Answer 6

Thank you!!

Answer 7

you are welcome

Answer 8

oh no

Answer 9

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

Answer 10

so if you want to solve
\[2x < \frac{ 4 }{ x+1 }\] you have to either solve
\[2x-\frac{4}{x+1}<0\] or
\[0<\frac{4}{x+1}-2x\]

Answer 11

same for
\[\frac{ 5 }{ x-2 }\ge \frac{ 3x }{ x-2 }\]
solve
\[\frac{5}{x-2}-\frac{3x}{x-2}\geq0\]

Answer 12

is it clear how to continue?

Answer 13

cant u just use plus or minus in case of an uncertainty?

Answer 14

\[2x-\frac{ 4 }{ x+1 }<0,or \frac{ 2x ^{2}+2x-4 }{x+1 }<0\]
\[\frac{ 2\left( x ^{2}+x+\left( \frac{ 1 }{ 2 } \right)^{2}-\left( \frac{ 1 }{ 2 } \right)^{2} \right)-4 }{ x+1 }<0\]
\[or \frac{ 2\left( x+\frac{ 1 }{ 2 } \right)^{2}-\frac{ 1 }{ 2 } -4}{ x+1 }<0\]
\[or \frac{ 2\left( x+\frac{ 1 }{ 2 } \right)^{2}-\frac{ 9 }{2 } }{ x+1 }<0\]
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