College Algebra, Chapter 1, 1.2, Section 1.2, Problem 34

Jan has $\$ 3.00$ in nickels, dimes and quarters. If he has twice as many dimes as quarters and five more nickels than dimes, how many coins of each type does he have?

If we translate words into equations, we get

$
\begin{array}{|c|c|}
\hline\\
\text{In words} & \text{In algebra} \\
\hline\\
\text{Twice as many dimes as quarters} & d = 2q \\
\hline\\
\text{Five more nickels than dimes} & n = 5 + d\\
\hline
\end{array}
$

Recall that nickels have value of $\$ 0.05$,


$
\begin{equation}
\begin{aligned}

\text{dimes} =& \$ 0.10
\\
\\
\text{quarters} =& \$ 0.25

\end{aligned}
\end{equation}
$


So, we got


$
\begin{equation}
\begin{aligned}

\$ 3.00 =& \$ 0.05 (n) + \$ 0.10 (d) + \$ 0.25 (q) \qquad \text{Multiply both sides by } 100
\\
\\
300 =& 5n + 10d + 25q \qquad \text{Equation 1}

\end{aligned}
\end{equation}
$


From the table, we got

$d = 2q$ and $n = 5 + d$

Notice that we can rewrite the equations as


$
\begin{equation}
\begin{aligned}

0 =& 0 + d - 2q
&& \text{Equation 2}
\\
\\
5 =& n - d + 0
&& \text{Equation 3}

\end{aligned}
\end{equation}
$


By using Equations 1, 2 and 3 simultaneously, we get..


$
\begin{equation}
\begin{aligned}

n =& 15
\\
\\
d =& 10
\\
\\
q =& 5

\end{aligned}
\end{equation}
$


There are 15 nickels, 10 dimes and 5 quarters