If the equations are `x+y-3=0`
`(1+beta)x+(2+beta)y-8=0 and x-(1+beta)y+(2+beta)=0`, then the value of `beta` for the consistent solution, is

Options

A. 2
B. `(16)/(3)`
C. `(-5)/(3)`
D. `(5)/(3)`

Answer

To find the value of \( \beta \) for which the given equations have a consistent solution, we need to set up the determinant of the coefficients of the equations and set it equal to zero.

The equations given are:
1. \( x + y - 3 = 0 \)
2. \( (1 + \beta)x + (2 + \beta)y - 8 = 0 \)
3. \( x - (1 + \beta)y + (2 + \beta) = 0 \)

### Step 1: Write the equations in standard form
The equations can be rewritten as:
1. \( 1x + 1y - 3 = 0 \)
2. \( (1 + \beta)x + (2 + \beta)y - 8 = 0 \)
3. \( 1x - (1 + \beta)y + (2 + \beta) = 0 \)

### Step 2: Form the coefficient matrix
The coefficients of \( x \), \( y \), and the constants can be arranged in the following matrix:

\[ \begin{vmatrix} 1 & 1 & -3 \\ 1 + \beta & 2 + \beta & -8 \\ 1 & -(1 + \beta) & 2 + \beta \end{vmatrix} \]

### Step 3: Calculate the determinant
To find the determinant, we can use the formula for a 3x3 determinant:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where:
- \( a = 1 \), \( b = 1 \), \( c = -3 \)
- \( d = 1 + \beta \), \( e = 2 + \beta \), \( f = -8 \)
- \( g = 1 \), \( h = -(1 + \beta) \), \( i = 2 + \beta \)

Calculating the determinant step by step:

1. **Calculate \( ei - fh \)**:
\[ ei = (2 + \beta)(2 + \beta) = (2 + \beta)^2 = 4 + 4\beta + \beta^2 \]
\[ fh = (-8)(-(1 + \beta)) = 8(1 + \beta) = 8 + 8\beta \]
\[ ei - fh = (4 + 4\beta + \beta^2) - (8 + 8\beta) = \beta^2 - 4\beta - 4 \]

2. **Calculate \( di - fg \)**:
\[ di = (1 + \beta)(2 + \beta) = 2 + 3\beta + \beta^2 \]
\[ fg = (-8)(1) = -8 \]
\[ di - fg = (2 + 3\beta + \beta^2) + 8 = \beta^2 + 3\beta + 10 \]

3. **Calculate \( dh - eg \)**:
\[ dh = (1 + \beta)(-(1 + \beta)) = -(1 + \beta)^2 = -1 - 2\beta - \beta^2 \]
\[ eg = (2 + \beta)(1) = 2 + \beta \]
\[ dh - eg = (-1 - 2\beta - \beta^2) - (2 + \beta) = -3 - 3\beta - \beta^2 \]

4. **Putting it all together**:
\[ D = 1(\beta^2 - 4\beta - 4) - 1(\beta^2 + 3\beta + 10) - 3(-3 - 3\beta - \beta^2) \]
\[ = (\beta^2 - 4\beta - 4) - (\beta^2 + 3\beta + 10) + 9 + 9\beta + 3\beta^2 \]
\[ = \beta^2 - 4\beta - 4 - \beta^2 - 3\beta - 10 + 9 + 9\beta + 3\beta^2 \]
\[ = (3\beta^2) + (9\beta - 4\beta - 3\beta) + (-4 - 10 + 9) = 3\beta^2 + 2\beta - 5 \]

### Step 4: Set the determinant to zero
For the system to have a consistent solution, we set the determinant to zero:
\[ 3\beta^2 + 2\beta - 5 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = 2 \), and \( c = -5 \).

\[ \beta = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \]
\[ = \frac{-2 \pm \sqrt{4 + 60}}{6} \]
\[ = \frac{-2 \pm \sqrt{64}}{6} \]
\[ = \frac{-2 \pm 8}{6} \]

Calculating the two possible values:
1. \( \beta = \frac{6}{6} = 1 \)
2. \( \beta = \frac{-10}{6} = -\frac{5}{3} \)

### Final Answer
The values of \( \beta \) for which the system has a consistent solution are \( \beta = 1 \) and \( \beta = -\frac{5}{3} \).