`(a-b)x+ay=7,(a+b)x+2ay=a+11` has infinite many solutions then find value of `a` and `b`

Options

A. 3, 1
B. 3, 2
C. 4, 1
D. 4, 2

Answer

To find the values of \( a \) and \( b \) such that the system of equations

\[ (a-b)x + ay = 7 \]
\[ (a+b)x + 2ay = a + 11 \]

has infinitely many solutions, we can use the condition that the ratios of the coefficients of \( x \), \( y \), and the constant terms must be equal.

### Step 1: Write the equations in standard form

The given equations can be rewritten in the standard form:

1. \( (a-b)x + ay - 7 = 0 \)
2. \( (a+b)x + 2ay - (a + 11) = 0 \)

### Step 2: Identify coefficients

From the first equation, we have:
- Coefficient of \( x \) (let's call it \( A_1 \)): \( a - b \)
- Coefficient of \( y \) (let's call it \( B_1 \)): \( a \)
- Constant term (let's call it \( C_1 \)): \( -7 \)

From the second equation, we have:
- Coefficient of \( x \) (let's call it \( A_2 \)): \( a + b \)
- Coefficient of \( y \) (let's call it \( B_2 \)): \( 2a \)
- Constant term (let's call it \( C_2 \)): \( -(a + 11) \)

### Step 3: Set up the ratio condition

For the system to have infinitely many solutions, the following condition must hold:

\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

Substituting the coefficients we found:

\[ \frac{a-b}{a+b} = \frac{a}{2a} = \frac{-7}{-(a+11)} \]

### Step 4: Simplify the ratios

1. From \( \frac{a}{2a} = \frac{1}{2} \), we have:

\[ \frac{a-b}{a+b} = \frac{1}{2} \]

Cross-multiplying gives:

\[ 2(a-b) = a+b \]

Expanding and rearranging:

\[ 2a - 2b = a + b \implies 2a - a = 2b + b \implies a = 3b \tag{1} \]

2. Now, equate the constant terms:

\[ \frac{-7}{-(a+11)} = \frac{1}{2} \]

Cross-multiplying gives:

\[ -7 \cdot 2 = -(a + 11) \]

This simplifies to:

\[ -14 = -a - 11 \implies a = 3 \tag{2} \]

### Step 5: Substitute \( a \) into equation (1)

From equation (1), substituting \( a = 3 \):

\[ 3 = 3b \implies b = 1 \]

### Conclusion

Thus, the values of \( a \) and \( b \) are:

\[ \boxed{a = 3, b = 1} \]