Answer
See explanation
Work Step by Step
We want to prove that for real numbers \(r\), \(s\), \(a_0\), \(a_1\) with \(r \neq s\), there exist **unique** real numbers \(C\) and \(D\) satisfying:
\[
\begin{cases}
C + D = a_0,\\
Cr + Ds = a_1.
\end{cases}
\]
---
## 1. Reformulate as a Linear System
The conditions
\[
C + D = a_0,
\quad
Cr + Ds = a_1
\]
can be seen as the 2×2 linear system
\[
\begin{pmatrix}
1 & 1\\[6pt]
r & s
\end{pmatrix}
\begin{pmatrix}
C \\[4pt]
D
\end{pmatrix}
=
\begin{pmatrix}
a_0 \\[4pt]
a_1
\end{pmatrix}.
\]
---
## 2. Existence and Uniqueness via the Nonzero Determinant
Consider the matrix
\[
M = \begin{pmatrix}
1 & 1\\
r & s
\end{pmatrix}.
\]
Its determinant is
\[
\det(M) = 1\cdot s \;-\; 1\cdot r = s - r.
\]
Since \(r \neq s\), we have \(\det(M) \neq 0\).
A 2×2 matrix with a nonzero determinant is invertible, which guarantees:
1. **Existence:** There is at least one solution \(\begin{pmatrix}C \\ D\end{pmatrix}\) to the system \(M \begin{pmatrix}C\\D\end{pmatrix} = \begin{pmatrix}a_0\\a_1\end{pmatrix}\).
2. **Uniqueness:** There is no more than one solution. In fact, the inverse of \(M\) is
\[
M^{-1}
= \frac{1}{s-r}
\begin{pmatrix}
s & -1\\
-r & 1
\end{pmatrix}.
\]
Thus we can explicitly solve for \(\begin{pmatrix}C\\D\end{pmatrix}\) by multiplying both sides by \(M^{-1}\).
---
## 3. Explicit Formulas for \(C\) and \(D\)
Using Cramer's Rule or matrix inversion:
\[
\begin{pmatrix}
C\\[4pt]
D
\end{pmatrix}
=
\frac{1}{s-r}
\begin{pmatrix}
s & -1\\
-r & 1
\end{pmatrix}
\begin{pmatrix}
a_0\\[2pt]
a_1
\end{pmatrix}.
\]
Hence,
\(C = \tfrac{1}{s-r}\,\bigl(s\,a_0 - a_1\bigr)\),
\(D = \tfrac{1}{s-r}\,\bigl(-r\,a_0 + a_1\bigr)\).
Alternatively, one can do a direct elimination approach:
1. From \(C + D = a_0\), get \(D = a_0 - C\).
2. Substitute into \(Cr + Ds = a_1\):
\[
Cr + (a_0 - C)s = a_1
\quad\Longrightarrow\quad
Cr + a_0 s - C s = a_1
\quad\Longrightarrow\quad
C(r - s) = a_1 - a_0 s.
\]
Since \(r \neq s\), we can solve:
\[
C = \frac{a_1 - a_0 s}{\,r - s\,}.
\]
Then \(D = a_0 - C\) yields
\[
D = a_0 - \frac{a_1 - a_0 s}{r - s}
= \frac{(r - s)a_0 - (a_1 - a_0 s)}{\,r - s\,}
= \frac{r\,a_0 - s\,a_0 - a_1 + a_0 s}{\,r - s\,}
= \frac{r\,a_0 - a_1}{\,r - s\,}.
\]
Either way, \(C\) and \(D\) exist and are uniquely determined.
---
## Conclusion
Because the determinant \(s - r\) is nonzero (due to \(r \neq s\)), the linear system
\[
\begin{cases}
C + D = a_0,\\
Cr + Ds = a_1
\end{cases}
\]
has a **unique** solution \((C,D)\). This proves there exist unique real numbers \(C\) and \(D\) satisfying the two given equations whenever \(r \neq s\).
