document.write( "Question 158174This question is from textbook Algebra and Trigonometry: Structure and Method, book 2
\n" ); document.write( ": The three sides of an equilateral triangle are increased by 20 cm, 30 cm, and 40 cm, respectively. The perimeter of the resulting triangle is between twice and three times the perimeter of the original triangle. What can you conclude about the length of a side of the original triangle? \n" ); document.write( "
Algebra.Com's Answer #116544 by jim_thompson5910(35256)\"\" \"About 
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Let s=length of side of equilateral triangle\r
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\n" ); document.write( "\n" ); document.write( "So the perimeter of the equilateral triangle is \"P=s%2Bs%2Bs=3s\"\r
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\n" ); document.write( "\n" ); document.write( "Because the sides \"are increased by 20 cm, 30 cm, and 40 cm\", this tells us that the new sides are \"s%2B20\", \"s%2B30\", and \"s%2B40\". \r
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\n" ); document.write( "\n" ); document.write( "The perimeter of the new triangle is \"P=%28s%2B20%29%2B%28s%2B30%29%2B%28s%2B40%29=3s%2B90\"\r
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\n" ); document.write( "\n" ); document.write( "Since the \"perimeter of the resulting triangle is between twice and three times the perimeter of the original triangle\", this means that \r
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\n" ); document.write( "\n" ); document.write( "\"2%283s%29%3C=3s%2B90%3C=3%283s%29\" Note: remember, the term \"3s\" is the original perimeter\r
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\n" ); document.write( "\n" ); document.write( "\"6s%3C=3s%2B90%3C=9s\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "Break up the compound inequality to get:\r
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\n" ); document.write( "\n" ); document.write( "\"6s%3C=3s%2B90\" AND \"3s%2B90%3C=9s\"\r
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\n" ); document.write( "\n" ); document.write( "So let's solve the first inequality \"6s%3C=3s%2B90\"\r
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\n" ); document.write( "\n" ); document.write( "\"6s%3C=3s%2B90\" Start with the given inequality.\r
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\n" ); document.write( "\n" ); document.write( "\"6s-3s%3C=90\" Subtract \"3s\" from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"3s%3C=90\" Combine like terms on the left side.\r
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\n" ); document.write( "\n" ); document.write( "\"s%3C=%2890%29%2F%283%29\" Divide both sides by \"3\" to isolate \"s\". \r
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\n" ); document.write( "\n" ); document.write( "\"s%3C=30\" Reduce.\r
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\n" ); document.write( "\n" ); document.write( "Now let's solve the second inequality \"3s%2B90%3C=9s\"\r
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\n" ); document.write( "\n" ); document.write( "\"3s%2B90%3C=9s\" Start with the given inequality.\r
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\n" ); document.write( "\n" ); document.write( "\"3s%3C=9s-90\" Subtract \"90\" from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"3s-9s%3C=-90\" Subtract \"9s\" from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"-6s%3C=-90\" Combine like terms on the left side.\r
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\n" ); document.write( "\n" ); document.write( "\"s%3E=%28-90%29%2F%28-6%29\" Divide both sides by \"-6\" to isolate \"s\". note: Remember, the inequality sign flips when we divide both sides by a negative number. \r
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\n" ); document.write( "\n" ); document.write( "\"s%3E=15\" Reduce.\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 solution is \"s%3C=30\" and \"s%3E=15\".\r
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\n" ); document.write( "\n" ); document.write( "Recombine the inequalities to get \"15%3C=s%3C=30\"\r
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\n" ); document.write( "\n" ); document.write( "So the side lengths of the original triangle are between 15 and 30 cm
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