document.write( "Question 1094535: Hi! I'm really having difficulty solving these problems. :( How do you solve for x in these problems? Hope someone would help me. Thank you in advance! \r
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Algebra.Com's Answer #709119 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "\"%28x%2B1%29%2F%283x-6%29+=+%285x%29%2F6+%2B+1%2F%28x-2%29\" \r\n" );
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document.write( "Factor the first denominator by factoring out 3\r\n" );
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document.write( "\"%28x%2B1%29%2F%283%28x-2%29%29+=+%285x%29%2F6+%2B+1%2F%28x-2%29\"\r\n" );
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document.write( "The LCD is 6(x-2) so we multiply every term\r\n" );
document.write( "through by \"red%286%28x-2%29%2F1%29\"\r\n" );
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document.write( "We cancel whatever will cancel:\r\n" );
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document.write( "What we have left is:\r\n" );
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document.write( "\"2%28x%2B1%29=%28x-2%29%285x%29%2B6\"\r\n" );
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document.write( "\"2x%2B2=5x%28x-2%29%2B6\"\r\n" );
document.write( "\"2x%2B2=5x%5E2-10x%2B6\"\r\n" );
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document.write( "Get 0 on the right:\r\n" );
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document.write( "\"-5x%5E2%2B8x-4=0\"\r\n" );
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document.write( "Multiply through by -1 so x2 term will be positive:\r\n" );
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document.write( "\"5x%5E2-8x%2B4=0\"\r\n" );
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document.write( "Factor:\r\n" );
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document.write( "\"%285x-2%29%28x-2%29=0\"\r\n" );
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document.write( "5x-2 = 0;   x-2 = 0\r\n" );
document.write( "  5x = 2      x = 2\r\n" );
document.write( "   x = \"2%2F5\"\r\n" );
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document.write( "Two answers are \"2%2F5\" and \"2\".\r\n" );
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document.write( "However, the original problem has denominators that contain \r\n" );
document.write( "variables.  Therefore we must check for extraneous answers \r\n" );
document.write( "that are not solutions because they cause the denominator to \r\n" );
document.write( "equal 0.  No denominator can ever equal to 0.  \r\n" );
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document.write( "The denominators in the original equation are 3x-6 and x-2\r\n" );
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document.write( "Substituting \"2%2F5\" for x in denominator 3x-6 is\r\n" );
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document.write( "\"3%282%2F5%29-6\"\"%22%22=%22%22\"\"6%2F5-6\"\"%22%22=%22%22\"\"6%2F5-30%2F5\"\"%22%22=%22%22\"\"-24%2F5\" which is not 0\r\n" );
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document.write( "Substituting \"2%2F5\" for x in denominator x-2 is\r\n" );
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document.write( "\"2%2F5-2%29\"\"%22%22=%22%22\"\"2%2F5-10%2F5\"\"%22%22=%22%22\"\"-8%2F5\" which is not 0 \r\n" );
document.write( "either, so \"2%2F5\" is a solution..\r\n" );
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document.write( "Substituting \"2\" for x in denominator 3x-6 is\r\n" );
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document.write( "\"3%282%29-6\"\"%22%22=%22%22\"\"6-6\"\"%22%22=%22%22\"\"0\" so we must\r\n" );
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document.write( "discard the answer x = 2, for it is extraneous and is not a\r\n" );
document.write( "solution.   Therefore there is only one solution, x = \"2%2F5\"\r\n" );
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document.write( "The second one:\r\n" );
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document.write( "\"%283x%29%2F%28x%2B1%29+-+5%2F%282x%29+=+3%2F%282x%29\"\r\n" );
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document.write( "Add \"%22%22%2B5%2F%282x%29\" to both sides:\r\n" );
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document.write( "\"%283x%29%2F%28x%2B1%29+=+3%2F%282x%29+%2B+5%2F%282x%29\"\r\n" );
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document.write( "Since both terms on the right have the same denominator\r\n" );
document.write( "we can just add the numerators and place it over their\r\n" );
document.write( "common denominator:\r\n" );
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document.write( "\"%283x%29%2F%28x%2B1%29+=+8%2F%282x%29\"\r\n" );
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document.write( "Simplify the right side by dividing top and bottom by 4:\r\n" );
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document.write( "\"%283x%29%2F%28x%2B1%29+=+4%2Fx\"\r\n" );
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document.write( "We could get an LCD as we did in the other one but since \r\n" );
document.write( "there is only a fraction on each side, we can just \r\n" );
document.write( "cross-multiply:\r\n" );
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document.write( "\"3x%28x%29+=+4%28x%2B1%29\"\r\n" );
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document.write( "\"3x%5E2+=+4x%2B4\"\r\n" );
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document.write( "Get 0 on the right:\r\n" );
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document.write( "\"3x%5E2-4x-4=0\"\r\n" );
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document.write( "\"%283x%2B2%29%28x-2%29=0\"\r\n" );
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document.write( "3x+2 = 0;    x-2 = 0\r\n" );
document.write( "  3x = -2;     x = 2\r\n" );
document.write( "   x = \"-2%2F3\"\r\n" );
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document.write( "Two answers are \"-2%2F3\" and \"2\".\r\n" );
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document.write( "However, the original problem also has denominators that contain \r\n" );
document.write( "variables.  Therefore we must check for extraneous answers \r\n" );
document.write( "that are not solutions because they cause the denominator to \r\n" );
document.write( "equal 0.  No denominator can ever equal to 0.  \r\n" );
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document.write( "The denominators in the original equation are x-1 and 2x.\r\n" );
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document.write( "Substituting \"-2%2F3\" for x in denominator x-1 is\r\n" );
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document.write( "\"-2%2F3-1\"\"%22%22=%22%22\"\"-2%2F3-3%2F3\"\"%22%22=%22%22\"\"-5%2F3\"\r\n" );
document.write( "which is not 0\r\n" );
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document.write( "Substituting \"-2%2F3\" for x in denominator 2x is\r\n" );
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document.write( "\"2%28-2%2F3%29\"\"%22%22=%22%22\"\"-4%2F3\" which is not 0 either, so \r\n" );
document.write( "\"-2%2F3\" is a solution.\r\n" );
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document.write( "Substituting \"2\" for x in denominator x-1 is\r\n" );
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document.write( "\"2-1\"\"%22%22=%22%22\"\"1\"which is not 0\r\n" );
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document.write( "Substituting \"2\" for x in denominator 2x is\r\n" );
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document.write( "\"2%282%29\"\"%22%22=%22%22\"\"4\" which is not 0 either, so \r\n" );
document.write( "\"2\" is also a solution.\r\n" );
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document.write( "So there are two solutions  \"-2%2F3\" and 2. \r\n" );
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document.write( "But as you see we must make sure that any answer we get\r\n" );
document.write( "really is a solution by substituting into each denominator\r\n" );
document.write( "to show that no answer causes any denominator in the original\r\n" );
document.write( "equation to be 0.\r\n" );
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document.write( "Edwin
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