SOLUTION: A newspaper carrier has $7.10 in change. He has ten more quarters than dimes but two times as many nickels as quarters. How many coins of each type does he have? This is what I

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Question 984043: A newspaper carrier has $7.10 in change. He has ten more quarters than dimes but two times as many nickels as quarters. How many coins of each type does he have?
This is what I got so far: x + 3 2(x+3)


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +n+ = number of nickels
Let +d+ = number of dimes
Let +q+ = number of quarters
------------------------------
(1) +5n+%2B+10d+%2B+25q+=+710+ ( in cents )
(2) +q+=+d+%2B+10+
(3) +n+=+2q+
--------------------
You have 3 equations and 3 unknowns,
so it's solvable
(2) +d+=+q+-+10+
Substitute (2) and (3) into (1)
(1) +5%2A2q+%2B+10%2A%28+q-10+%29+%2B+25q+=+710+
(1) +10q+%2B+10q-+100+%2B+25q+=+710+
(1) +45q+=+810+
(1) +q+=+18+
and
(3) +n+=+2q+
(3) +n+=+2%2A18+
(3) +n+=+36+
and
(2) +d+=+q+-+10+
(2) +d+=+18+-+10+
(2) +d+=+8+
--------------------
He has:
36 nickels
8 dimes
18 quarters
------------------
check:
(1) +5n+%2B+10d+%2B+25q+=+710+
(1) +5%2A36+%2B+10%2A8+%2B+25%2A18+=+710+
(1) +180+%2B+80+%2B+450+=+710+
(1) +710+=+710+
OK