 Solve the linear system
 ax +by =c,  −bx+ay =d

To solve the system of linear equations:

\[
ax + by = c \tag{1}
\]
\[
-bx + ay = d \tag{2}
\]

We will use the method of substitution or elimination. Let’s use elimination by adding the two equations together to eliminate \( x \).

### Step 1: Add the two equations
When we add equation (1) and equation (2), we get:
\[
(ax + by) + (-bx + ay) = c + d
\]
This simplifies to:
\[
(a + a)y + (b - b)x = c + d
\]
\[
2ay = c + d
\]

### Step 2: Solve for \( y \)
Now, solving for \( y \):
\[
y = \frac{c + d}{2a}
\]

### Step 3: Substitute \( y \) back into one of the original equations
Now that we know \( y \), substitute it into one of the original equations, say equation (1):
\[
ax + b\left(\frac{c + d}{2a}\right) = c
\]
Multiply out the terms:
\[
ax + \frac{b(c + d)}{2a} = c
\]
Multiply through by \( 2a \) to clear the denominator:
\[
2a^2x + b(c + d) = 2ac
\]
Now, solve for \( x \):
\[
2a^2x = 2ac - b(c + d)
\]
\[
x = \frac{2ac - b(c + d)}{2a^2}
\]

Thus, the solution to the system is:
\[
x = \frac{2ac - b(c + d)}{2a^2}, \quad y = \frac{c + d}{2a}
\]

### Summary of Solution:
\[
x = \frac{2ac - b(c + d)}{2a^2}, \quad y = \frac{c + d}{2a}
\]

---

Would you like further details or clarifications on any part? Here are 5 related questions:

1. How would the solution change if \( a = 0 \)?
2. Can we use matrices to solve this system of equations?
3. What if the coefficients \( b \) and \( d \) were also equal?
4. How can we graph this system of equations?
5. Can this method be extended to solve nonlinear systems?

**Tip:** Always check for special cases (e.g., if a coefficient equals zero) when solving systems of equations.

Solve the linear system ax + by = c, −bx + ay = d

Learn how to solve the linear system of equations ax + by = c and −bx + ay = d using the elimination method. This step-by-step guide covers key linear algebra concepts and provides formulas for finding the solution. Suitable for high school students in Grades 10-12.

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