document.write( "Question 74264This question is from textbook intermediate algebra
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document.write( ": A manufacturer builds racing bikes and mountain bikes, with the per unit manufacturing costs shown in table . The company has bugeted $15,900 for labor and $13,075 for materials .How many bicycles of each type can be built?\r
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document.write( "model cost of materials cost of labor\r
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document.write( "racing $55 $60
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document.write( "mountain $70 $90
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Algebra.Com's Answer #53382 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A manufacturer builds racing bikes and mountain bikes, with the per unit manufacturing costs shown in table . The company has bugeted $15,900 for labor and $13,075 for materials .How many bicycles of each type can be built?\r \n" ); document.write( "\n" ); document.write( "model cost of materials cost of labor\r \n" ); document.write( "\n" ); document.write( "racing $55 $60 \n" ); document.write( "mountain $70 $90 \n" ); document.write( "-------------------- \n" ); document.write( "Let r be # of racing bikes and m be # of mountain bikes: \n" ); document.write( "60r+90m = 15900 \n" ); document.write( "55r+70m = 13075 \n" ); document.write( "-------------- \n" ); document.write( "Rewrite to get smaller numbers: \n" ); document.write( "2r + 3m = 530 \n" ); document.write( "11r+ 14m =2615 \n" ); document.write( "---------- \n" ); document.write( "Multiply 1st by 11 and 2nd by 2 to get: \n" ); document.write( "22r + 33m =11*530 \n" ); document.write( "22r + 28m =2*2615 \n" ); document.write( "------------- \n" ); document.write( "Subtract 2nd from 1st to get: \n" ); document.write( "5m = 600 \n" ); document.write( "m=120 (# of mountain bikes) \n" ); document.write( "Substitute in 2r+3m = 530 to solve for \"r\": \n" ); document.write( "2r+3*120 = 530 \n" ); document.write( "2r = 170 \n" ); document.write( "r = 85 (# of racing bikes) \n" ); document.write( "-------------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |