document.write( "Question 83621: joe has a collection of nickels and dimes worth $7.05. if the number of dimes were doubled and the number of nickels were incresed by 5, the value of the coins would be $11.80 how many dimes does he have ? \n" ); document.write( "
Algebra.Com's Answer #60134 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
Write an equation for each statement: n=no. of nickels; d = no. of dimes
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\n" ); document.write( "\"joe has a collection of nickels and dimes worth $7.05\".
\n" ); document.write( " .05n + .10d = 7.05; call it eq1
\n" ); document.write( ":
\n" ); document.write( " \"if the number of dimes were doubled and the number of nickels were increased by 5, the value of the coins would be $11.80\"
\n" ); document.write( " 2(.10d) + .05(n+5) = 11.80
\n" ); document.write( ".2d + .05n + .25 = 11.80
\n" ); document.write( ".05n + .2d = 11.80 - .25; put it in the same order as eq1
\n" ); document.write( ".05n + .2d = 11.55; call it eq2
\n" ); document.write( ":
\n" ); document.write( "Subtract eq1 from eq2:
\n" ); document.write( ".05n + .20d = 11.55
\n" ); document.write( ".05n + .10d = 7.05
\n" ); document.write( "--------------------subtracting eliminates n
\n" ); document.write( "0 n + .10d = 4.50
\n" ); document.write( "d = 4.50/.10
\n" ); document.write( "d = 45 dimes
\n" ); document.write( ":
\n" ); document.write( "To check our solution, Find the number of nickels using eq1, substitute 45 for d:
\n" ); document.write( ".05n + .10(45) = 7.05
\n" ); document.write( ".05n + 4.50 = 7.05
\n" ); document.write( ".05n = 7.05 - 4.50
\n" ); document.write( ".05n = 2.55
\n" ); document.write( "n = 2.55/.05
\n" ); document.write( "n = 51 nickels
\n" ); document.write( ":
\n" ); document.write( "Check solution using eq2
\n" ); document.write( ".05(51) + .20(45) =
\n" ); document.write( "2.55 + 9.00 = 11.55
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