document.write( "Question 612274: I have two algebra I word problems i need help setting up. I would like if you can explain the steps of setting up.
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\n" ); document.write( "Let w= width in units.
\n" ); document.write( "L=(2 1/2)w= 5w/2 = length in units
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Algebra.Com's Answer #852638 by MathTherapy(10579)\"\" \"About 
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document.write( "I have two algebra I word problems i need help setting up. I would like if you can explain the steps of setting up.\r\n" );
document.write( "Here it is:\r\n" );
document.write( "Problem #1\r\n" );
document.write( "The length of a rectangle is 2 1/2 times its width. Its area is 90 square units. What are its dimensions? (Hint: length times width = area)\r\n" );
document.write( "Let w= width in units.\r\n" );
document.write( "L=(2 1/2)w= 5w/2 = length in units\r\n" );
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document.write( "The person who responded is WRONG!! Width is NOT 3 units!\r\n" );
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document.write( "As indicated by you, let width be w\r\n" );
document.write( "Then length (L), as you stated is \"2%261%2F2w\", or 2.5w\r\n" );
document.write( "As it's given that area = 90 square units, we get: 2.5w(w) = 90\r\n" );
document.write( "                                                     2.5w2 = 90\r\n" );
document.write( "                                                        \"w%5E2+=+90%2F2.5\"\r\n" );
document.write( "                                                        \"w%5E2+=+36\"\r\n" );
document.write( "                           Width of rectangle, or \r\n" );
document.write( "                                          Length: 2.5(6) = 15 units
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