SOLUTION: Can someone please walk me through the steps to better understand this problem? Thank you Joe has a collection of nickels and dimes that is worth $8.55. If the number of dimes w

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Question 677008: Can someone please walk me through the steps to better understand this problem? Thank you
Joe has a collection of nickels and dimes that is worth $8.55. If the number of dimes was tripled and the number of nickels was decreased by 2, the value of the coins would be $23.45. How many nickels and dimes does he have?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +n+ = the number of nickels
Let +d+ = the number of dimes
given:
(1) +5n+%2B+10d+=+855+ ( in cents )
(2) +5%2A%28+n+-+2+%29+%2B+10%2A%28+3d+%29+=+2345+ ( in cents )
--------------------------------
(2) +5n+-+10+%2B+30d++=+2345+
(2) +5n+%2B+30d+=+2355+
(2) +n+%2B+6d+=+471+
and
(1) +n+%2B+2d+=+171+
-------------------
Subtract (1) from (2)
(2) +n+%2B+6d+=+471+
(1) +-n+-+2d+=+-171+
+4d+=+300+
+d+=+75+
and, since
(1) +n+%2B+2d+=+171+
(1) +n+%2B+2%2A75+=+171+
(1) +n+=+171+-+150+
(1) +n+=+21+
Joe has 21 nickels and 75 dimes
check:
(2) +5%2A%28+n+-+2+%29+%2B+10%2A%28+3d+%29+=+2345+
(2) +5%2A%28+21+-+2+%29+%2B+10%2A%28+3%2A75+%29+=+2345+
(2) +5%2A19+%2B+10%2A225+=+2345+
(2) +95+%2B+2250+=+2345+
(2) +2345+=+2345+
and
(1) +5n+%2B+10d+=+855+
(1) +5%2A21+%2B+10%2A75+=+855+
(1) +105+%2B+750+=+855+
(1) +855+=+855+
OK