document.write( "Question 155544: On the worksheet it shows the equation:
\n" ); document.write( "\"1%2Fz-1%2F2z-1%2F5z=10%2F%28z%2B1%29\" The first thing i tried to do was get rid of the fraction by mutiplying by the LCD, which i figured would be (z)(2z)(5z)(z+1). however it ended up being a jumbled mess and left me very confused and now I have no idea what to do. \n" ); document.write( "
Algebra.Com's Answer #114582 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
Edwin's solution:
\n" ); document.write( "On the worksheet it shows the equation:
\n" ); document.write( "\"1%2Fz-1%2F2z-1%2F5z=10%2F%28z%2B1%29\" The first thing i tried to do was get rid of the fraction by mutiplying by the LCD, which i figured would be (z)(2z)(5z)(z+1). however it ended up being a jumbled mess and left me very confused and now I have no idea what to do.
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document.write( "You are a little confused on how to find the LCD.\r\n" );
document.write( "You don't need to repeat the factor z in the LCD. \r\n" );
document.write( "You only need it ONCE in the LCD, not three times!!!\r\n" );
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document.write( "Let's put parentheses around every term:\r\n" );
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document.write( "\"%281%2Fz%29-%281%2F%282z%29%29-%281%2F%285z%29%29=%2810%2F%28z%2B1%29%29\"\r\n" );
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document.write( "The factors of the denominators are z, 2, 5, and z+1\r\n" );
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document.write( "The factor z appears 1 time in the first fraction, 1 time in the second\r\n" );
document.write( "fraction, 1 time in the third fraction and 0 times in the fourth fraction.\r\n" );
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document.write( "The most number of times that z appears in any ONE denominator is\r\n" );
document.write( "1 time.  Therefore z needs to appear only 1 time in the LCD.\r\n" );
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document.write( "---\r\n" );
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document.write( "The factor 2 appears 0 times in the first fraction, 1 time in the second\r\n" );
document.write( "fraction, 0 times in the third fraction and 0 times in the fourth fraction.\r\n" );
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document.write( "The most number of times that 2 appears in any ONE denominator is\r\n" );
document.write( "1 time.  Therefore 2 needs to appear only 1 time in the LCD.\r\n" );
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document.write( "---\r\n" );
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document.write( "The factor 5 appears 0 times in the first fraction, 0 times in the second\r\n" );
document.write( "fraction, 1 time in the third fraction and 0 times in the fourth fraction.\r\n" );
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document.write( "The most number of times that 5 appears in any ONE denominator is\r\n" );
document.write( "1 time.  Therefore 5 needs to appear only 1 time in the LCD.\r\n" );
document.write( "\r\n" );
document.write( "---\r\n" );
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document.write( "The factor z+1 appears 0 times in the first fraction, 0 times in the second\r\n" );
document.write( "fraction, 0 times in the third fraction and 1 time in the fourth fraction.\r\n" );
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document.write( "The most number of times that z+1 appears in any ONE denominator is\r\n" );
document.write( "1 time.  Therefore z+1 needs to appear only 1 time in the LCD.\r\n" );
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document.write( "---\r\n" );
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document.write( "Therefore, the LCD is (z)(2)(5)(z+1) or 10z(z+1).  We put the LCD\r\n" );
document.write( "over 1 like this \"%2810z%28z%2B1%29%29%2F1\" and multiply it by every term:\r\n" );
document.write( "  \r\n" );
document.write( "\"%281%2Fz%29-%281%2F%282z%29%29-%281%2F%285z%29%29=%2810%2F%28z%2B1%29%29\"\r\n" );
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document.write( "Now we cancel away the denominators:\r\n" );
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document.write( "              \"5\"               \"2\"\r\n" );
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document.write( "\"10%28z%2B1%29-5%28z%2B1%29-2%28z%2B1%29=10z%2A10\"\r\n" );
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document.write( "\"10z%2B10-5z-5-2z-2=100z\"\r\n" );
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document.write( "\"3z%2B3=100z\"\r\n" );
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document.write( "\"3z-100z=-3\"\r\n" );
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document.write( "\"-97z=-3\"\r\n" );
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document.write( "\"z=%28-3%29%2F%28-97%29\"\r\n" );
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document.write( "\"z=3%2F97\"\r\n" );
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document.write( "That's an ugly solution, but it's correct!\r\n" );
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document.write( "Edwin

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