document.write( "Question 134750: A city wants to buy some animals but they only have 100 dollars. Horses cost 10 dollars cows cost 3 dollars and ducks cost 50 cents. How many animals can they buy. \r
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document.write( "10x+3y+.5z+100 I simplified for x and came up with x=10+-.3y+.05z now do I simplify for Y and then put both statements into the equaltion? \r
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document.write( "Please solve \n" );
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Algebra.Com's Answer #98577 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A city wants to buy some animals but they only have 100 dollars. Horses cost 10 dollars cows cost 3 dollars and ducks cost 50 cents. How many animals can they buy. \r \n" ); document.write( "\n" ); document.write( "10x+3y+.5z <= 100 \r \n" ); document.write( "\n" ); document.write( "---------------------- \n" ); document.write( "Make a list of numbers of horses, cows, ducks \n" ); document.write( "There are no more than 10 horses so put the numbers 0 through 10 \n" ); document.write( "below horses. \n" ); document.write( "with zero horses list the possible numbers of cows from 0 to 33 \n" ); document.write( "with zero horses and zero cows you can buy 0 to 200 ducks \n" ); document.write( "------------- \n" ); document.write( "With one horse the possible # of cows and ducks change. \n" ); document.write( "Similarly with two horses, or three etc. \n" ); document.write( "----------------- \n" ); document.write( "Comment: there are a lot of possible combinations. \n" ); document.write( "======================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |