document.write( "Question 291137: \"%28a%2Bb%29%2F%28a-b%29+-+%28a-b%29%2F%28a%2Bb%29\" \n" ); document.write( "
Algebra.Com's Answer #210607 by Edwin McCravy(20055)\"\" \"About 
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\"%28a%2Bb%29%2F%28a-b%29+-+%28a-b%29%2F%28a%2Bb%29\"
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document.write( "Put parentheses around all binomials:\r\n" );
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document.write( "\"%28%28a%2Bb%29%29%2F%28%28a-b%29%29+-+%28%28a-b%29%29%2F%28%28a%2Bb%29%29\"\r\n" );
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document.write( "The LCD is \"red%28%28a-b%29%28a%2Bb%29%29\"\r\n" );
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document.write( "The first denominator \"%28a-b%29\" needs \"red%28%28a%2Bb%29%29\" to become the LCD\r\n" );
document.write( "\"red%28%28a-b%29%28a%2Bb%29%29\", so we multiply top and bottom of the first fraction\r\n" );
document.write( "by \"red%28%28a%2Bb%29%29\"\r\n" );
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document.write( "The second denominator \"%28a%2Bb%29\" needs \"red%28%28a-b%29%29\" to become the LCD\r\n" );
document.write( "\"red%28%28a-b%29%28a%2Bb%29%29\", so we multiply top and bottom of the second fraction\r\n" );
document.write( "by \"red%28%28a-b%29%29\"\r\n" );
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document.write( "The denominators are now the same, so we make just one fraction by\r\n" );
document.write( "indicating the subtraction of the two numerators over the common\r\n" );
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document.write( "FOIL out the binomials in the top and keep parentheses around them but DO NOT\r\n" );
document.write( "FOIL out the binomial in the denominator!\r\n" );
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document.write( "\"%28%28a%5E2%2B2ab%2Bb%5E2%29+-+%28a%5E2-2ab%2Bb%5E2%29%29%2F%28%28a-b%29%28a%2Bb%29%29\"\r\n" );
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document.write( "Now remove the parentheses:\r\n" );
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document.write( "\"%28a%5E2%2B2ab%2Bb%5E2+-+a%5E2%2B2ab-b%5E2%29%2F%28%28a-b%29%28a%2Bb%29%29\"\r\n" );
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document.write( "Cancel the \"a%5E2\" and the \"-a%5E2\"\r\n" );
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document.write( "Cancel the \"b%5E2\" and the \"-b%5E2\"\r\n" );
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document.write( "\"%284ab%29%2F%28%28a-b%29%28a%2Bb%29%29\"\r\n" );
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document.write( "Edwin
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