To determine the number of nickels (denoted as [tex]\( N \)[/tex]) and the number of dimes (denoted as [tex]\( D \)[/tex]) that Daniel has, we can solve the following system of linear equations:
[tex]\[
\begin{cases}
0.05N + 0.10D = 2.20 \\
N + D = 23
\end{cases}
\][/tex]
Here's a detailed step-by-step solution:
1. Solve the second equation for one of the variables. Let's solve for [tex]\( N \)[/tex] in terms of [tex]\( D \)[/tex]:
[tex]\[
N + D = 23 \quad \Rightarrow \quad N = 23 - D
\][/tex]
2. Substitute this expression for [tex]\( N \)[/tex] into the first equation to eliminate [tex]\( N \)[/tex]:
[tex]\[
0.05(23 - D) + 0.10D = 2.20
\][/tex]
3. Distribute the 0.05 into the parentheses:
[tex]\[
0.05 \cdot 23 - 0.05D + 0.10D = 2.20 \quad \Rightarrow \quad 1.15 - 0.05D + 0.10D = 2.20
\][/tex]
4. Combine like terms (combine the terms with [tex]\( D \)[/tex]):
[tex]\[
1.15 + 0.05D = 2.20
\][/tex]
5. Isolate [tex]\( D \)[/tex] by subtracting 1.15 from both sides:
[tex]\[
0.05D = 2.20 - 1.15 \quad \Rightarrow \quad 0.05D = 1.05
\][/tex]
6. Divide both sides by 0.05 to solve for [tex]\( D \)[/tex]:
[tex]\[
D = \frac{1.05}{0.05} \quad \Rightarrow \quad D = 21
\][/tex]
7. Now that we have [tex]\( D \)[/tex], substitute it back into the equation [tex]\( N = 23 - D \)[/tex]:
[tex]\[
N = 23 - 21 \quad \Rightarrow \quad N = 2
\][/tex]
Thus, Daniel has:
Nickels [tex]\( N = 2 \)[/tex]
Dimes [tex]\( D = 21 \)[/tex]
So, Daniel has [tex]\( 2 \)[/tex] nickels and [tex]\( 21 \)[/tex] dimes.
