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:

$x = \frac{2ac - b(c + d)}{2a^2}, \quad y = \frac{c + d}{2a}$

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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.