Question 2.1.16: A boy has a jar full of coins. Altogether there are 180 nick...

Let n be the number of nickels, d the number of dimes, and q the number of quarters. Then

n + d + q = 180

The second piece of information we are given is that

d = \frac{1}{2} (n + q)

We rewrite this into standard form for a linear equation:

n – 2d + q = 0

Finally, we have the value of the coins, in cents:

5n + 10d + 25q = 1600

Thus, n, d, and q satisfy the system of linear equations:

n \ \ \ \ \ + d \ \ \ \ \ + q = 180 \\

n \ \ \ -2d \ \ \ \ \ + q = 0 \\

5n + 10d + 25q = 1600

Write the augmented matrix and row reduce:

\left [ \begin{matrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 5 & 10 & 25 \end{matrix} \left | \begin{matrix} 180 \\ 0 \\ 1600 \end{matrix} \right.\right ] \begin{matrix} \\ R_{2} – R_{1} \\ R_{3} – 5R_{1} \end{matrix} \thicksim \left [ \begin{matrix} 1 & 1 & 1 \\ 0 & -3 & 0 \\ 0 & 5 & 20 \end{matrix} \left | \begin{matrix} 180 \\ -180 \\ 700 \end{matrix} \right.\right] \begin{matrix} \\ ({-1}/{3}) R_{2} \\ ({1}/{5}) R_{3} \end{matrix} \thicksim

\left [ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 4 \end{matrix} \left | \begin{matrix} 180 \\ 60 \\ 140 \end{matrix} \right.\right] \begin{matrix} \\ \\ R_{3} – R_{2} \end{matrix} \thicksim \left [ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{matrix} \left | \begin{matrix} 180 \\ 60 \\ 80 \end{matrix} \right.\right]

According to Theorem 2.1.3, the system is consistent with a unique solution. In particular, writing the final augmented matrix as a system of equations, we get

n + d + q = 180 \\

  \ \ \ \ \ \ \ \ \ \ \ \ \ \ d = 60 \\

\ \ \ \ \ \ \ \ \ \ \ \ 4q = 80

So, by back-substitution, we get q = 20, d = 60, n = 180 – d – q = 100. Hence, the boy has 100 nickels, 60 dimes, and 20 quarters.