Question 1152325
{{{x/(x+1)>3x}}}

1. Work the problem

{{{x/(x+1)>3x}}}

{{{x>3x(x+1)}}}

{{{x>3x^2+3x}}}...add {{{-x}}} to both sides

{{{x-x>3x^2+3x-x}}}

{{{0>3x^2+3x-x}}}......switch sides

{{{3x^2+3x-x<0}}}

{{{3x^2+2x<0}}}



2. Set to zero and factor.

{{{3x^2+2x=0}}}

{{{(3x+2)x=0}}}

3. Find zeros.
{{{(3x+2)x=0}}}...this will be equal to zero if {{{(3x+2)=0}}} or {{{x=0}}}

so, zeros are:
{{{x=0}}}
{{{(3x+2)=0}}}=>{{{3x=-2}}}=>{{{x=-2/3}}}

4. Find intervals.

{{{-2/3<x<0}}}

5. Test intervals.

let say {{{x=-1/3}}}
{{{3(-1/3)^2+2(-1/3)<0}}}
{{{3(1/9)-2/3<0}}}
{{{cross(3)(1/cross(9)3)-2/3<0}}}
{{{1/3-2/3<0}}}
{{{-1/3<0}}}-> which is true

6. Write answer in interval notation and graph.
({{{-2/3}}},{{{ 0}}})



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