# Chapter 11 - Additional Topics - 11.2 - 3 x 3 Systems of Equations - Problem Set 11.2 - Page 484: 17

10 quarters, 8 dimes, and 2 nickels.

#### Work Step by Step

We are told: $d + q + n = 20$ And $.1d + .25q + .05n = 3.4$ And $d + n = q$ Substituting q for d + n in equation one, we obtain: $2q = 20 \\ q = 10$ Thus, we are left with: $.1d + .05n = .9$ And $d + n = 10$ Using d = 10 - n, we find: $.1(10 - n) + .05n = .9 \\ -.05n = -.1 \\ n = 2$ This means that the number of dimes is: $d = 10 -2 = 8$

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