Answer
See proof
Work Step by Step
Consider the system:
$\begin{cases}
(a_1+a_2)x+(b_1+b_2)y=c_1+c_2\\
a_2x+b_2y=c_2
\end{cases}$
Compute the determinant $D_{new}$ of the coefficients of this system:
$D_{new}=\begin{vmatrix}a_1+a_2&b_1+b_2\\a_2&b_2\end{vmatrix}=(a_1+a_2)b_2-a_2(b_1+b_2)$
$=a_1b_2+a_2b_2-a_2b_1-a_2b_2$
$=a_1b_2-a_2b_1$
The determinant $D_{old}$ of the original system is:
$D_{old}=\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}=a_1b_2-a_2b_1$
We notice that $D_{new}=D_{old}$.
Compute $D_{x_{new}}$ and $D_{x_{old}}$:
$D_{x_{new}}=\begin{vmatrix}c_1+c_2&b_1+b_2\\c_2&b_2\end{vmatrix}$
$=b_2(c_1+c_2)-(b_1+b_2)c_2$
$=b_2c_1+b_2c_2-b_1c_2-b_2c_2$
$=b_2c_1-b_1c_2$
$D_{x_{old}}=\begin{vmatrix}c_1&b_1\\c_2&b_2\end{vmatrix}$
$=b_2c_1-b_1c_2$
We notice that $D_{x_{old}}=D_{x_{new}}$. In the same way we can prove that $D_{y_{old}}=D_{y_{new}}$.
This means we have:
$\dfrac{D_{x_{new}}}{D_{new}}=\dfrac{D_{x_{old}}}{D_{old}}$
$\dfrac{D_{y_{new}}}{D_{new}}=\dfrac{D_{y_{old}}}{D_{old}}$
Therefore the two systems have the same solutions.
