document.write( "Question 1152325: It is not a system, I can not find anything, anywhere that is just solving one nonlinear equation. \r
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Algebra.Com's Answer #774504 by ikleyn(52780)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "    \"graph%28+330%2C+330%2C+-5%2C+5%2C+-5%2C+20%2C%0D%0A++++++++++x%2F%28x%2B1%29%2C+3x%0D%0A%29\"\r\n" );
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document.write( "    Plot y = \"x%2F%28x%2B1%29\"  (red)  and  y = 3x (green).\r\n" );
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\n" ); document.write( "\n" ); document.write( "From the plot, it is clear that the solution is the union of TWO intervals:\r
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\n" ); document.write( "\n" ); document.write( "        one interval is semi-infinite (\"-infinity\",\"-1\")   and the other interval is somewhere between -1 and 0.\r
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\n" ); document.write( "\n" ); document.write( "The solution by @MathLover1 gives only the second interval and entirely misses the first semi-infinite interval.\r
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\n" ); document.write( "\n" ); document.write( "So,  her solution is  INCORRECT.   Therefore,  I came to bring the correct solution.\r
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\n" ); document.write( "\n" ); document.write( "After completing my solution,  I will point the error in the MathLover1' solution.\r
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document.write( "So, we start from the given inequality\r\n" );
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document.write( "    \"x%2F%28x%2B1%29\" > 3x.       (1)\r\n" );
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document.write( "Move 3x to the left side\r\n" );
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document.write( "    \"x%2F%28x%2B1%29\" - 3x > 0\r\n" );
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document.write( "Write with the common denominator\r\n" );
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document.write( "    \"x%2F%28x%2B1%29\" - \"%28%283x%29%2A%28x%2B1%29%29%2F%28x%2B1%29\" > 0\r\n" );
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document.write( "Make equivalent transformations\r\n" );
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document.write( "    \"%28x+-+3x%2A%28x%2B1%29%29%2F%28x%2B1%29\" > 0\r\n" );
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document.write( "    \"%28-3x%5E2+%2B+x+-+3x%29%2F%28x%2B1%29\" > 0\r\n" );
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document.write( "    \"%28-3x%5E2+-+2x%29%2F%28x%2B1%29\" > 0\r\n" );
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document.write( "    \"%28-x%2A%283x%2B2%29%29%2F%28x%2B1%29\" > 0\r\n" );
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document.write( "Multiply both sides by (-1) and change the inequality sign\r\n" );
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document.write( "    \"%28x%2A%283x%2B2%29%29%2F%28x%2B1%29\" < 0.\r\n" );
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document.write( "Divide by 3 both sides\r\n" );
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document.write( "    \"%28x%2A%28x%2B2%2F3%29%29%2F%28x%2B1%29\" < 0.      (2)\r\n" );
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document.write( "So, our original inequality is equivalent to the inequality (2).\r\n" );
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document.write( "Now I will solve the inequality (2).\r\n" );
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document.write( "The factors in inequality (2) have three critical points, where linear terms become equal to zero and change the sign.\r\n" );
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document.write( "These points are  -1,  \"-2%2F3\"  and  0.\r\n" );
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document.write( "The critical points divide the entire number line in four intervals\r\n" );
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document.write( "    1)  (\"-infinity\",\"-1\"),   2)  (\"-1\",\"-2%2F3\"),   3)   (\"-2%2F3\",\"0\")  and   4)  [\"0\",\"infinity\").\r\n" );
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document.write( "In the interval #1, all three linear factors are negative, so the entire rational function is negative.\r\n" );
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document.write( "Thus the entire interval  (\"-infinity\",\"-1\")  IS the solution to inequality (2) and, hence, for the original inequality, too.\r\n" );
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document.write( "In the interval #2, the linear factor  (x+1)  is positive, while two other factors are negative, \r\n" );
document.write( "so the entire rational function is positive.\r\n" );
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document.write( "Thus the interval  (\"-1\",\"-2%2F3\")  is NOT the solution to inequality (2) and, hence, is NOT the solution for the original inequality.\r\n" );
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document.write( "In the interval #3, the linear factors  (x+1)  and  \"x%2B3%2F2\"  are both positive, while third factor  x   is negative, \r\n" );
document.write( "so the entire rational function is negative.\r\n" );
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document.write( "Thus the interval  (\"-2%2F3\",\"0\")  IS the solution to inequality (2) and, hence, IS the solution for the original inequality.\r\n" );
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document.write( "In the interval #4, all three linear factors  are positive, so the entire rational function in (2) is positive.\r\n" );
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document.write( "Thus the interval  [\"0\",\"infinity\")  is NOT the solution to inequality (2) and, hence, is not the solution for the original inequality.\r\n" );
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document.write( "ANSWER.  The solution set for the given inequality is the union of two intervals  (\"-infinity\",\"-1\") U (\"-2%2F3\",\"0\").\r\n" );
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document.write( "Please notice and pay special attention that all endpoints are treated correctly in my solution and in my answer.\r\n" );
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\n" ); document.write( "\n" ); document.write( "The problem is solved,  and solved correctly.\r
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\n" ); document.write( "\n" ); document.write( "The way on how I solved it,  is a  STANDARD  (but not a unique)  way solving such inequalities for rational functions.\r
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\n" ); document.write( "\n" ); document.write( "Now,  I promised to show where is the error in the solution by @MathLover1.\r
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\n" ); document.write( "\n" ); document.write( "It is in the  4-th line of her post,  where she multiplies both sides of the inequality by the factor  (x+1)\r
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\n" ); document.write( "\n" ); document.write( "Doing in this way,  she gets  NON-EQUIVALENT  inequality and comes to the wrong resulting answer.\r
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\n" ); document.write( "\n" ); document.write( "It is very common error in solving such inequalities for rational functions - so, be aware (!)\r
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\n" ); document.write( "\n" ); document.write( "At this point,  I completed my post.\r
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\n" ); document.write( "\n" ); document.write( "From it,  learn on how to solve similar problems correctly.\r
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\n" ); document.write( "\n" ); document.write( "If you want to see many other similar solved problems,  look into the lessons\r
\n" ); document.write( "\n" ); document.write( "    - Solving inequalities for rational functions with numerator and denominator factored into a product of linear binomials \r
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\n" ); document.write( "\n" ); document.write( "Consider these lessons as your handbook, textbook, tutorials and (free of charge) home teacher.\r
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