document.write( "Question 185243: Can you please help me with this word problem for some reason Im not understanding the concept of figuring these problems.\r
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document.write( "The lenght of a rectangular picture is 5 inches greater thatn twice the width. If the perimeter is 112 inches, find the dimensions of the frame.\r
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document.write( "thanks, \n" );
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Algebra.Com's Answer #138958 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Let x = the width \n" ); document.write( "\"L\" = the length \n" ); document.write( ": \n" ); document.write( "Just write an equation for what it says: \n" ); document.write( ": \n" ); document.write( "\"The length of a rectangular picture is 5 inches greater that twice the width. \n" ); document.write( "L = 2x + 5 \n" ); document.write( ": \n" ); document.write( "\" If the perimeter is 112 inches,\" \n" ); document.write( "twice the length + twice the width = 112 \n" ); document.write( "2L + 2x = 112 \n" ); document.write( "Simplify, divide equation by 2; the results: \n" ); document.write( "L + x = 56 \n" ); document.write( ": \n" ); document.write( "From the 1st statement, replace L with (2x+5) \n" ); document.write( "(2x+5) + x = 56 \n" ); document.write( "subtract 5 from both sides \n" ); document.write( "2x + x = 56 - 5 \n" ); document.write( "3x = 51 \n" ); document.write( "x = \n" ); document.write( "x = 17 in is the width \n" ); document.write( "then \n" ); document.write( "L = 2(17) + 5 \n" ); document.write( "L = 34 + 5 \n" ); document.write( "L = 39 in is the length \n" ); document.write( "; \n" ); document.write( ": \n" ); document.write( "Check solution by finding the perimeter \n" ); document.write( "2(39) + 2(17) = \n" ); document.write( "78 + 34 = 112, confirms our solution \n" ); document.write( "; \n" ); document.write( "Did this make sense? \r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |