https://file.aiquickdraw.com/mathbot/file/1727110451144zenxsd51.jpg

The system of equations in the image is:

\[
\begin{aligned}
1) & \quad ax + by = (a - b) \\
2) & \quad bx - ay = (a + b)
\end{aligned}
\]

We need to solve this system of equations for \(x\) and \(y\). I'll solve it step by step:

### Step 1: Label the equations
\[
(1) \quad ax + by = a - b
\]
\[
(2) \quad bx - ay = a + b
\]

### Step 2: Eliminate one variable
We can use the method of elimination. First, multiply both equations by constants so that we can eliminate either \(x\) or \(y\).

- Multiply equation (1) by \(b\):
\[
b(ax + by) = b(a - b) \implies abx + b^2y = ab - b^2 \quad (3)
\]

- Multiply equation (2) by \(a\):
\[
a(bx - ay) = a(a + b) \implies abx - a^2y = a^2 + ab \quad (4)
\]

### Step 3: Subtract (4) from (3)
Now subtract equation (4) from equation (3):
\[
(abx + b^2y) - (abx - a^2y) = (ab - b^2) - (a^2 + ab)
\]

Simplify the left and right-hand sides:
\[
b^2y + a^2y = -b^2 - a^2
\]

Factor out \(y\) on the left:
\[
(a^2 + b^2)y = -(a^2 + b^2)
\]

### Step 4: Solve for \(y\)
If \(a^2 + b^2 \neq 0\), we can divide both sides by \(a^2 + b^2\):
\[
y = -1
\]

### Step 5: Substitute \(y = -1\) into one of the original equations
Substitute \(y = -1\) into equation (1):
\[
ax + b(-1) = a - b
\]
\[
ax - b = a - b
\]

Add \(b\) to both sides:
\[
ax = a
\]

Solve for \(x\):
\[
x = 1
\]

### Final Answer:
The solution to the system is:
\[
x = 1, \quad y = -1
\]

Let me know if you'd like any clarifications or more details! Here are some related questions:

1. Can we solve this system using substitution instead of elimination?
2. What happens if \(a^2 + b^2 = 0\)? Would the system still have a solution?
3. How does this system change if we swap the terms on the right-hand side of the equations?
4. Can you graph the system of equations and interpret the solution geometrically?
5. What happens if \(a = b\)? Does the solution change?

**Tip**: In systems of equations, elimination is often helpful when coefficients are already aligned for one variable, while substitution is easier when one equation is already solved for one variable.

Solve the system of equations: ax + by = (a - b), bx - ay = (a + b)

Learn how to solve the system of linear equations ax + by = (a - b) and bx - ay = (a + b) step-by-step using the elimination method. This example focuses on key algebraic concepts, suitable for students in Grades 9-12.

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