{  █(2x+y=b………..(1)@  bx-(a-b)y=0………….(2))┤

To solve the given system of equations, we have:

1. \( 2x + y = b \)  \(\quad\quad\quad(1)\)
2. \( bx - (a - b)y = 0 \) \(\quad(2)\)

Let's solve this system step by step.

### Step 1: Solve Equation (2) for \( x \) in terms of \( y \)

Equation (2) is:
\[ bx - (a - b)y = 0 \]

Rearrange to solve for \( x \):
\[ bx = (a - b)y \]
\[ x = \frac{(a - b)y}{b} \quad\quad\quad(3) \]

### Step 2: Substitute \( x \) from Equation (3) into Equation (1)

Substitute \( x = \frac{(a - b)y}{b} \) into Equation (1):
\[ 2\left(\frac{(a - b)y}{b}\right) + y = b \]

Simplify:
\[ \frac{2(a - b)y}{b} + y = b \]
\[ \frac{2(a - b)y + by}{b} = b \]
\[ 2(a - b)y + by = b^2 \]
\[ (2a - 2b + b)y = b^2 \]
\[ (2a - b)y = b^2 \]

### Step 3: Solve for \( y \)

\[ y = \frac{b^2}{2a - b} \]

### Step 4: Solve for \( x \)

Substitute \( y = \frac{b^2}{2a - b} \) back into Equation (3):
\[ x = \frac{(a - b)y}{b} \]
\[ x = \frac{(a - b)\left(\frac{b^2}{2a - b}\right)}{b} \]
\[ x = \frac{(a - b)b^2}{b(2a - b)} \]
\[ x = \frac{(a - b)b}{2a - b} \]

### Final Solution
The solution to the system is:
\[ x = \frac{(a - b)b}{2a - b} \]
\[ y = \frac{b^2}{2a - b} \]

Would you like more details on any of the steps or have any questions?

### Related Questions
1. How would the solution change if \( a = 2b \)?
2. What if \( b = 0 \), how does it affect the solution?
3. Can you solve the system using matrix methods?
4. How can we verify the solution by substituting back into the original equations?
5. What is the geometric interpretation of these equations?
6. How do the solutions change if \( a \) and \( b \) are negative?
7. What happens if \( a \) is zero?
8. Can we find the determinant of the coefficient matrix of this system?

### Tip
When dealing with systems of linear equations, substituting one equation into another is a powerful method to reduce the number of variables and simplify the problem.

Learn how to solve the system of equations 2x + y = b and bx - (a - b)y = 0 with a detailed step-by-step solution. This problem involves linear equations and systems of equations, suitable for advanced high school students or college-level algebra courses.

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