Answer
$C = \frac{a_k}{r^k} \;-\; k\,D
\quad\text{or}\quad
C
= \frac{a_m}{r^m} \;-\; m\,D$
$D
= \frac{\tfrac{a_m}{r^m} - \tfrac{a_k}{r^k}}{m - k}.
$
Work Step by Step
We want to show that for any nonzero real \(r\) and distinct integers \(k\neq m\), and for any real numbers \(a_k\) and \(a_m\), there is a **unique** pair \((C,D)\in\mathbb{R}^2\) satisfying
\[
\begin{cases}
C\,r^k \;+\; k\,D\,r^k = a_k,\\[6pt]
C\,r^m \;+\; m\,D\,r^m = a_m.
\end{cases}
\]
---
## 1. Reformulate the Equations
Notice each equation can be factored by pulling out \(r^k\) or \(r^m\):
1. \(r^k(C + kD) = a_k.\)
2. \(r^m(C + mD) = a_m.\)
Since \(r\neq 0\), we can rewrite this system as
\[
\begin{cases}
C + kD = \dfrac{a_k}{r^k},\\[6pt]
C + mD = \dfrac{a_m}{r^m}.
\end{cases}
\]
So the unknowns \(C\) and \(D\) satisfy a 2×2 linear system:
\[
\begin{pmatrix}
1 & k\\[4pt]
1 & m
\end{pmatrix}
\begin{pmatrix}
C \\[4pt] D
\end{pmatrix}
=
\begin{pmatrix}
\dfrac{a_k}{r^k}\\[4pt]\dfrac{a_m}{r^m}
\end{pmatrix}.
\]
---
## 2. Existence and Uniqueness from a Nonzero Determinant
The coefficient matrix is
\[
M = \begin{pmatrix}1 & k\\[2pt]1 & m\end{pmatrix}.
\]
Its determinant is
\[
\det(M) = 1\cdot m \;-\; (1\cdot k) = m - k.
\]
Because \(k\) and \(m\) are distinct integers, \(m-k \neq 0\). Thus \(\det(M)\neq 0\), so \(M\) is invertible.
An invertible 2×2 matrix guarantees:
1. **Existence:** There is at least one solution \((C,D)\).
2. **Uniqueness:** There is at most one solution.
Hence there is exactly one solution \((C,D)\).
---
## 3. Explicit Formulas (Optional)
You can solve directly by elimination or by using the inverse of \(M\). For instance, from
\[
\begin{cases}
C + kD = \dfrac{a_k}{r^k},\\
C + mD = \dfrac{a_m}{r^m},
\end{cases}
\]
subtracting the first equation from the second gives
\[
(m - k)\,D
= \frac{a_m}{r^m} \;-\; \frac{a_k}{r^k},
\]
so
\[
D
= \frac{\tfrac{a_m}{r^m} - \tfrac{a_k}{r^k}}{m - k}.
\]
Then
\[
C
= \frac{a_k}{r^k} \;-\; k\,D
\quad\text{or}\quad
C
= \frac{a_m}{r^m} \;-\; m\,D
\]
(either form is valid). Thus \(C\) and \(D\) are uniquely determined.
---
## Conclusion
Because \(r\neq 0\) and \(k\neq m\), the linear system has a nonzero determinant and therefore exactly one solution \((C,D)\). This proves that there **exist unique** real numbers \(C\) and \(D\) satisfying
\[
\begin{cases}
C\,r^k + k\,D\,r^k = a_k,\\[2pt]
C\,r^m + m\,D\,r^m = a_m.
\end{cases}
\]
