Acollection of 30 coins worth $5.50 consists of nickels, dimes, and quarters. there are twice as many dimes as nickels. how many quarters are there?

Three unknowns requires three equations to solve . . .

1) n + d + q = 30
2) d = 2n
3) (0.05)n + (0.10)d + (0.25)q = 5.50

* plug 2) into 1) and solve for q

n + 2n + q = 30

3n + q = 30

q = 30 - 3n

* now we have variables d and q expressed in terms of n, so we plug those equations into 3) and solve for n . . .

0.05n + 0.10(2n) + 0.25(30 - 3n) = 5.50

0.05n + 0.20n + 7.50 - 0.75n = 5.50

(0.05 + 0.20 - 0.75)n + 7.50 = 5.50

-0.50n + 7.50 = 5.50

-0.50n = 5.50 - 7.50

-0.50n = -2.00

n = -2.00/-0.50 = 4

d = 2n = 2(4) = 8

q = 30 - 3n = 30 - 3(4) = 30 - 12 = 18

The answer is . . .

n = 4 nickels
d = 8 dimes
q = 18 quarters

. . . for a total of (4 + 8 + 18) = 30 coins
. . . 8 dimes is twice 4 nickels
. . . $0.05(4) + $0.10(8) + $0.25(18) = $5.50

The graph of the given function is increasing in the interval, 2 to 5

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