document.write( "Question 1138522: Ben was describing his coin collection to his friends. He said I have 1.45 pesos more in 5 centavos (nickels) than in 10 centavos (dimes). Ii I had two 5 centavos (nickels) less, I would have 5 times as many 5 centavo (nickels) as 10 centavos (dimes). How many coins of each kind does he have? \n" ); document.write( "

Algebra.Com's Answer #756419 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "Work the problem using n for 5-centavo coins (nickels) and d for 10-centavo coins (dimes). The peso is the dollar; so 1.45 pesos is 145 centavos (cents). Then

\n" ); document.write( "(1) I have 1.45 pesos more in 5 centavos (nickels) than in 10 centavos (dimes):
\n" ); document.write( " $\"5n-10d+=+145\"$

\n" ); document.write( "(2) If I had two 5 centavos (nickels) less, I would have 5 times as many 5 centavos (nickels) as 10 centavos (dimes):
\n" ); document.write( " $\"n-2+=+5d\"$

\n" ); document.write( "There are two equations in n and d; there are many different ways to solve the problem.

\n" ); document.write( "With \"10d\" in one equation and \"5d\" in the other, here is the way I chose to go:

\n" ); document.write( "multiply the second equation by 2:
\n" ); document.write( " $\"2n-4+=+10d\"$

\n" ); document.write( "substitute \"2n-4\" for \"10d\" in the first equation:
\n" ); document.write( " $\"5n-%282n-4%29+=+145\"$
\n" ); document.write( " $\"5n-2n%2B4+=+145\"$
\n" ); document.write( " $\"3n+=+141\"$
\n" ); document.write( " $\"n+=+47\"$

\n" ); document.write( "substitute n=47 in either of the original equations to find d:
\n" ); document.write( " $\"2%2847%29-4+=+10d\"$
\n" ); document.write( " $\"90+=+10d\"$
\n" ); document.write( " $\"d+=+9\"$

\n" ); document.write( "Ben has 47 5-centavo coins and 9 10-centavo coins. \n" ); document.write( "

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