document.write( "Question 67475: Jill has $3.50 in nickels and dimes. If she has 50 coins, how many of each type of coin does she have?\r
\n" ); document.write( "\n" ); document.write( "Thanks for any help you can provide for me. \n" ); document.write( "
Algebra.Com's Answer #48021 by bam878s(77)\"\" \"About 
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Let x = the number of nickels she has and y = the number of dimes she has. From the information provided, we can see that:
\n" ); document.write( "$.05(x) + $.10(y) = $3.50 and
\n" ); document.write( "x + y = 50. \r
\n" ); document.write( "\n" ); document.write( "From the second of these equations we can solve for x
\n" ); document.write( "x + y = 50 ==> x = 50 - y (by subtracting y from both sides)
\n" ); document.write( "Now we can plug this value of x back into the first equation above.
\n" ); document.write( ".05(50-y) + .10y = 3.50 (multiply the .05 over 50 - y)
\n" ); document.write( "[.05(50) - .05(y)] + .10y = 3.50
\n" ); document.write( "2.5 - .05y + .10y = 3.50 (add the y terms)
\n" ); document.write( "2.5 + .05y = 3.50 (now subtract both sides by 2.5)
\n" ); document.write( ".05y = 3.50 - 2.50
\n" ); document.write( ".05y = 1 (divide both sides by .05)
\n" ); document.write( "y = 20 dimes!\r
\n" ); document.write( "\n" ); document.write( "Now we can use the second of the original equations to solve for x!\r
\n" ); document.write( "\n" ); document.write( "x + y = 50
\n" ); document.write( "x + 20 = 50
\n" ); document.write( "x = 50 - 20
\n" ); document.write( "x = 30.
\n" ); document.write( "So we have 30 nickels!\r
\n" ); document.write( "\n" ); document.write( "Lets check!
\n" ); document.write( "$0.05(30) + $0.10(20) = $1.50 + $2.00 = $3.50\r
\n" ); document.write( "\n" ); document.write( "There are 30 nickels and 20 dimes.\r
\n" ); document.write( "\n" ); document.write( "Hope this helps you! Hang in there and don't get discouraged! \n" ); document.write( "
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