document.write( "Question 84951This question is from textbook elementary and intermediate algebra
\n" ); document.write( ": Sean has 103 coins consisting of nickels,dimes and quarters. the number of dimes is one less than twice the number of nickels and the number of quartes is two more than three times the number of nickels. how many ocins of each kind does he have? \n" ); document.write( "
Algebra.Com's Answer #61280 by tutorcecilia(2152)\"\" \"About 
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Nickels + Dimes + Quarters = 103
\n" ); document.write( "Nickels = = N
\n" ); document.write( "Dimes = (2) x Nickels - 1 = 2N-1
\n" ); document.write( "Quarters = (3) x Nickles + 2 = 3N + 2
\n" ); document.write( "Total coins = 103
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\n" ); document.write( "(N) + (2N-1) + (3N+2) = 103 [combine like-terms]
\n" ); document.write( "N+2N-1+3N-1=103
\n" ); document.write( "6N+1=103 [solve for the N-term]
\n" ); document.write( "6N=103-1
\n" ); document.write( "6N=102
\n" ); document.write( "N=17
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\n" ); document.write( "Nickels = 17
\n" ); document.write( "Dimes = 2(N)-1=(2)(17)-1=33
\n" ); document.write( "Quarters = (3)(17)+ 2=53\r
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\n" ); document.write( "So, 17+33+53=103
\n" ); document.write( "103=103 [checks out]\r
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