Solve for x and y.
`ax+by=c`
`bx+ay=1+c`

Options

A. `x=(bc-ac+b)/(b^(2)-a^(2)),y=(bc-ac-a)/(b^(2)-a^(2))`
B. `x=(b)/(b^(2)-a^(2)),y=(a)/(b^(2)-a^(2))`
C. `x=a,y=b`
D. `x=b,y=(bc-ac+b)/(b^(2)-a^(2))`

Answer

To solve the given system of equations:

1. **Equations:**
\[ ax + by = c \quad \text{(1)} \]
\[ bx + ay = 1 + c \quad \text{(2)} \]

2. **Multiply Equation (1) by \(b\) and Equation (2) by \(a\):**
\[ b(ax + by) = bc \quad \Rightarrow \quad abx + b^2y = bc \quad \text{(3)} \]
\[ a(bx + ay) = a(1 + c) \quad \Rightarrow \quad abx + a^2y = a + ac \quad \text{(4)} \]

3. **Subtract Equation (4) from Equation (3):**
\[ (abx + b^2y) - (abx + a^2y) = bc - (a + ac) \]
This simplifies to:
\[ b^2y - a^2y = bc - a - ac \]
\[ (b^2 - a^2)y = (b - a)c - a \]

4. **Solve for \(y\):**
\[ y = \frac{(b - a)c - a}{b^2 - a^2} \]

5. **Now substitute \(y\) back into one of the original equations to find \(x\). We can use Equation (1):**
\[ ax + b\left(\frac{(b - a)c - a}{b^2 - a^2}\right) = c \]

6. **Multiply through by \(b^2 - a^2\) to eliminate the denominator:**
\[ ax(b^2 - a^2) + b((b - a)c - a) = c(b^2 - a^2) \]

7. **Rearranging gives:**
\[ ax(b^2 - a^2) = c(b^2 - a^2) - b((b - a)c - a) \]

8. **Now, isolate \(x\):**
\[ x = \frac{c(b^2 - a^2) - b((b - a)c - a)}{b^2 - a^2} \]

9. **Thus, the final solutions for \(x\) and \(y\) are:**
\[ x = \frac{c(b^2 - a^2) - b((b - a)c - a)}{b^2 - a^2} \]
\[ y = \frac{(b - a)c - a}{b^2 - a^2} \]