document.write( "Question 146882This question is from textbook Glencoe Algebra 1
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\n" ); document.write( "\n" ); document.write( "3/r+4-1/r=1/r
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Algebra.Com's Answer #107266 by jim_thompson5910(35256)\"\" \"About 
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\"3%2Fr%2B4-1%2Fr=1%2Fr\" Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( " Multiply every term by the LCD r to clear the fractions.\r
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\n" ); document.write( "\n" ); document.write( "\"3%2B4r-1=1\" Multiply and simplify.\r
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\n" ); document.write( "\n" ); document.write( "\"4r%2B2=1\" Combine like terms on the left side.\r
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\n" ); document.write( "\n" ); document.write( "\"4r=1-2\" Subtract \"2\" from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"4r=-1\" Combine like terms on the right side.\r
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\n" ); document.write( "\n" ); document.write( "\"r=%28-1%29%2F%284%29\" Divide both sides by \"4\" to isolate \"r\".\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So the answer is \"r=-1%2F4\" \r
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\n" ); document.write( "\n" ); document.write( "Which approximates to \"r=-0.25\" \r
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