# algebra

How do I figure out this problem? You have 37 coins (all nickels , dimes, and quarters) adding up to \$5.50. You have 4 more quarters than nickels. How many dimes do you have?

Let q,d and n stand for the number of quarters, dimes and nickels you have respectively. Thus we know
(1) q + d + n = 37
(2) .25q + .10d + .05n = 5.50
(3) q = n + 4

We can use (3) to eliminate q in (1) and (2) to get
(n+4) + d + n= 37 or
(1') 2n +d =33

.25(n+4) + .10d + .05n = 5.50 or,
.25n + 1 +.1d +.05n = 5.50 so
(2').30n +.10d =4.50

From (1') d = 33-2n Use that for d in (2') to solve for n. Let us know if you need further help.

Thanks

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