document.write( "Question 1011499: Brainstorming !! I send a famer to buy 100 animals with $100.00 dollars,how many off each he can buy? Ducks $10 each, pigs $3.00 each, chicks .50¢ each??
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Algebra.Com's Answer #627157 by macston(5194)\"\" \"About 
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\n" ); document.write( "C+D+P=100
\n" ); document.write( "C=100-D-P\r
\n" ); document.write( "\n" ); document.write( ".
\n" ); document.write( "$10D+$3P+$0.50C=$100
\n" ); document.write( "$10D+$3P+$0.50(100-D-P)=$100
\n" ); document.write( "$10D+$3P+$50-$0.50D-$0.50P=$100
\n" ); document.write( "$9.5D+$2.5P=$50
\n" ); document.write( "$19D+$5P=$100
\n" ); document.write( "$5P=$100-$19D
\n" ); document.write( "P=20-(19D/5)
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\n" ); document.write( "Since we know P must be a positive integer (or zero,
\n" ); document.write( "no partial pigs), we know 19D/5 must be a
\n" ); document.write( "positive integer (or zero), so 19D must be evenly divisible
\n" ); document.write( "by 5. (i.e. , D must be a multiple of 5 - or zero)
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\n" ); document.write( "For D=0:
\n" ); document.write( "P=20-19(0)/5=20-0=20
\n" ); document.write( "We have 20 pigs and no ducks.
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\n" ); document.write( "For D=5:
\n" ); document.write( "P=20-(19(5)/5)=20-19=1
\n" ); document.write( "We have 1 pig and 5 ducks.
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\n" ); document.write( "For D=10:
\n" ); document.write( "P=20-(19(10)/5)=20-38=-18
\n" ); document.write( "Since P is negative, this cannot be a solution
\n" ); document.write( "(no negative pigs, no anti-pigs)
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\n" ); document.write( "So we have either 20 pigs and no ducks or 1 pig and 5 ducks.
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\n" ); document.write( "For P=20, D=0:
\n" ); document.write( "C=100-D-P
\n" ); document.write( "C=100-0-20
\n" ); document.write( "C=80
\n" ); document.write( "ANSWER 1: We have 80 chicks, 20 pigs, and no ducks.
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\n" ); document.write( "For P=1, D=5:
\n" ); document.write( "C=100-D-P
\n" ); document.write( "C=100-1-5
\n" ); document.write( "C=94
\n" ); document.write( "ANSWER 2: We have 94 chicks, 1 pig, and 5 ducks.
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\n" ); document.write( "CHECK 1:
\n" ); document.write( "C=80; P=20; D=0
\n" ); document.write( "$0.50C+$3P+$10D=$100
\n" ); document.write( "$0.50(80)+$3(20)+$10(0)=$100
\n" ); document.write( "$40+$60=$100
\n" ); document.write( "$100=$100
\n" ); document.write( "Answer 1 checks.
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\n" ); document.write( "CHECK 2:
\n" ); document.write( "C=94; P=1; D=5
\n" ); document.write( "$0.50C+$3P+$10D=$100
\n" ); document.write( "$0.50(94)+$3(1)+$10(5)=$100
\n" ); document.write( "$47+$3+$50=$100
\n" ); document.write( "$100=$100
\n" ); document.write( "Answer 2 checks, so we have our two solutions.
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