document.write( "Question 1109822: The charge $C, for hiring a hall for an event is C = 150 + 2N, where N stands for the number of people at the event. Find:
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document.write( "a) the charge when 225 people are at the event
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document.write( "b) the number of people at the event when the charge is $394 \n" );
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Algebra.Com's Answer #724760 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! Part (a)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Plug in N = 225 to find C\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "C = 150 + 2N \n" ); document.write( "C = 150 + 2*225 \n" ); document.write( "C = 150 + 450 \n" ); document.write( "C = 600\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "It will cost $600\r \n" ); document.write( "\n" ); document.write( "====================================================================== \n" ); document.write( "Part (b)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Plug in C = 394 and solve for (or isolate) N. The idea is to undo PEMDAS.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "C = 150 + 2N \n" ); document.write( "394 = 150 + 2N \n" ); document.write( "150 + 2N = 394 \n" ); document.write( "2N + 150 = 394 \n" ); document.write( "2N + 150 - 150 = 394 - 150 subtract 150 from both sides \n" ); document.write( "2N = 244 \n" ); document.write( "2N/2 = 244/2 divide both sides by 2 \n" ); document.write( "N = 122\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "There are 122 people at the event. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |