document.write( "Question 68585This question is from textbook An Incremental Development
\n" ); document.write( ": Please help to solve...\r
\n" ); document.write( "\n" ); document.write( "There are 26 nickels, dimes and quarters in all, and their value was $2.25. how many coins of each type were there if there were 10 times as many nickels as quarters?
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Algebra.Com's Answer #48826 by ptaylor(2198)\"\" \"About 
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\n" ); document.write( "Let x= number of quarters
\n" ); document.write( "Then 10x=number of nickels
\n" ); document.write( "And 26-x-10x or 26-11x=number of dimes\r
\n" ); document.write( "\n" ); document.write( "Now lets deal in pennies to reduce confusion\r
\n" ); document.write( "\n" ); document.write( "We are told that:\r
\n" ); document.write( "\n" ); document.write( "25x+5(10x)+10(26-11x)=225 clear parens\r
\n" ); document.write( "\n" ); document.write( "25x+50x+260-110x=225 subtract 260 from both sides
\n" ); document.write( "75x-110x=225-260
\n" ); document.write( "-35x=-35
\n" ); document.write( "x=1-----------------------------number of quarters
\n" ); document.write( "10x=10(1)=10---------------------number of nickels
\n" ); document.write( "26-11x=26-11=15---------------------number of dimes\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "1(25)+10(5)+15(10)=225
\n" ); document.write( "25+50+150=225
\n" ); document.write( "225=225\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps-----ptaylor
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