document.write( "Question 175668This question is from textbook Algebra and Trigonometry
\n" ); document.write( ": u+1/1+1/u
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\n" ); document.write( "\n" ); document.write( "Okay, it sounds likee, u plus 1 over 1 plus 1 over u. alll of that over (divided by) , 1 over u+1\r
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\n" ); document.write( "\n" ); document.write( "I can't figure out how to do it.\r
\n" ); document.write( "\n" ); document.write( "Thank youuu.
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Algebra.Com's Answer #130777 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
\"%28%28u+%2B+1%29%2F%281+%2B+%281%2Fu%29%29%29%2F%281%2F%28u+%2B+1%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "Start by inverting the denominator fraction ( \"1%2F%28u+%2B+1%29\" ) and multiplying times the numerator fraction.\r
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\n" ); document.write( "\n" ); document.write( "Now your denominator has addition of an integer to a fraction. The LCD is \"u\", so \"1+%2B+%281%2Fu%29+=+%28u+%2B+1%29%2Fu\". Replace the denominator with this new expression:\r
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\n" ); document.write( "\n" ); document.write( "Again, invert and multiply:\r
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\n" ); document.write( "\n" ); document.write( "\"%28%28%28u+%2B+1%29%28u+%2B+1%29%29%2F1%29%28u%2F%28u%2B1%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "Remove like terms from numerator and denominator:\r
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\n" ); document.write( "\n" ); document.write( "\"%28%28cross%28%28u+%2B+1%29%29%28u+%2B+1%29%29%2F1%29%28u%2Fcross%28u%2B1%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"u%28u+%2B+1%29\" or \"u%5E2+%2B+u\" if you prefer.\r
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\n" ); document.write( "\n" ); document.write( "Check the answer: You can't absolutely prove the answer this way, but you can get a pretty good idea of whether you did the manipulations correctly. Pick a number. Best is a small whole number other than zero or one. Let's try 2. If u is 2, then plugging 2 into the original expression should yield 2 X (2 + 1) = 6.\r
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\n" ); document.write( "\n" ); document.write( "\"%28%282+%2B+1%29%2F%281+%2B+%281%2F2%29%29%29%2F%281%2F%282+%2B+1%29%29\". You get to do the arithmetic to see if my solution is correct.
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