document.write( "Question 1015511: Brendan has $6.50, and it is made up of 10 cent and 59 cent coins. If he has 25 coins altogether, how many coins of each kind does he have? \n" ); document.write( "
Algebra.Com's Answer #632082 by macston(5194)\"\" \"About 
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\n" ); document.write( "There is no answer to this question as given.
\n" ); document.write( "If he has 10 cent coins and 50 cent coins (not 59 cent coins):
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\n" ); document.write( "T=number of 10 cent coins; F=number of fifty cent coins
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\n" ); document.write( "T+F=25
\n" ); document.write( "T=25-F . Use this to substitute for T later.
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\n" ); document.write( "$0.10T+$0.50F=$6.50 . Substitute for T.
\n" ); document.write( "$0.10(25-F)+$0.50F=$6.50
\n" ); document.write( "$2.50-$0.10F+$0.50F=$6.50
\n" ); document.write( "$0.40F=$4.00
\n" ); document.write( "F=10
\n" ); document.write( "ANSWER 1: He has ten fifty cent coins.
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\n" ); document.write( "T=25-F=25-10=15
\n" ); document.write( "ANSWER 2: He has fifteen 10 cent coins
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\n" ); document.write( "CHECK:
\n" ); document.write( "$0.10T+$0.50F=$6.50
\n" ); document.write( "$0.10(15)+$0.50(10)=$6.50
\n" ); document.write( "$1.50+$5.00-$6.50
\n" ); document.write( "$6.50=$6.50
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