Solve: `a x+b y=c ,\ \ \ \ b x+a y=1+c`

Answer

To solve the system of equations given by:

1. \( ax + by = c \) (Equation 1)
2. \( bx + ay = 1 + c \) (Equation 2)

we will follow these steps:

### Step 1: Multiply the equations
We will multiply Equation 1 by \( a \) and Equation 2 by \( b \).

- Multiplying Equation 1 by \( a \):
\[ a(ax + by) = ac \implies a^2x + aby = ac \quad \text{(Equation 3)} \]

- Multiplying Equation 2 by \( b \):
\[ b(bx + ay) = b(1 + c) \implies b^2x + aby = b + bc \quad \text{(Equation 4)} \]

### Step 2: Subtract Equation 3 from Equation 4
Now, we will subtract Equation 3 from Equation 4:

\[ (b^2x + aby) - (a^2x + aby) = (b + bc) - ac \]

This simplifies to:
\[ (b^2 - a^2)x = b + bc - ac \]

### Step 3: Solve for \( x \)
Now, we can isolate \( x \):
\[ x = \frac{b + bc - ac}{b^2 - a^2} \]

### Step 4: Substitute \( x \) back into Equation 1
Now we will substitute the value of \( x \) back into Equation 1 to find \( y \):

\[ a\left(\frac{b + bc - ac}{b^2 - a^2}\right) + by = c \]

### Step 5: Simplify the equation for \( y \)
Rearranging gives:
\[ by = c - a\left(\frac{b + bc - ac}{b^2 - a^2}\right) \]

Now, we will simplify this expression:
\[ by = c - \frac{a(b + bc - ac)}{b^2 - a^2} \]

### Step 6: Solve for \( y \)
To isolate \( y \), we divide both sides by \( b \):
\[ y = \frac{c - \frac{a(b + bc - ac)}{b^2 - a^2}}{b} \]

### Final Values
Thus, we have:
\[ x = \frac{b + bc - ac}{b^2 - a^2} \]
\[ y = \frac{c - \frac{a(b + bc - ac)}{b^2 - a^2}}{b} \]