Question

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Answer 1

Did I do this right? \[\frac{ x }{ x+1 }>3 = \frac{ x }{ x+1 }-3>0=\frac{ x }{ x+1 }-\frac{ 3x+1 }{ x+1 }=\frac{ x-3x-1 }{ x+1 }>0=\frac{ -2x-1 }{ x+1 }>0\]

Answer 2

\[\frac{ -2x-1 }{ x+1 } > ??\] i can't see the last step
can you type it again..

Answer 3

just a >0 there

Answer 4

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

Answer 5

hmm there is a mistake \[\huge\rm \frac{ x }{ x+1 } - \frac{3\color{Red}{(x+1)}}{x+1}\] don't forget the parentheses there becuase to find the common denominator you're multiplying 3 by (x+1) not just x

Answer 6

Ah gotcha. Thank you! You are the best!

Answer 7

but hey wait let me tell you the easy way \[\frac{ x }{ x+1 } >3\]
why not multiply both sides by `x+1`
\[ (x+1)\cdot \frac{ x }{ x+1 } >3 \cdot (x+1)\]
\[ \cancel{ (x+1)}\cdot \frac{ x }{\cancel{ x+1} } >3 \cdot (x+1)\]\[x>3x+3\]
now get the x terms on one side and constant on the other

Answer 8

hmm.... I am supposed to end up with a 0 on one side so that I can graph it on a number line and them express the solution in interval notation...

Answer 9

hmm okay so \[\frac{ -2x-3 }{ x+1 } >0\]
what would you do after this ??
how would you solve for x

Answer 10

What is the question?

Answer 11

fine the x=0's ... -3/2 and -1

Answer 12

jeez.. can't type