Find those values of c for which the equations: `2x+3y=3,(c+ 2)x + (c +4)y=(c+6),(c+2)^2x+(c+4)^2
y=(c+6)^2` are consistent.Also solve above equations for these values of c.

Answer

To find the values of \( c \) for which the equations are consistent, we will analyze the given equations:

1. \( 2x + 3y = 3 \)
2. \( (c + 2)x + (c + 4)y = c + 6 \)
3. \( (c + 2)^2x + (c + 4)^2y = (c + 6)^2 \)

### Step 1: Formulate the equations in matrix form

We can express the system of equations in matrix form as follows:

\[ \begin{bmatrix} 2 & 3 \\ c + 2 & c + 4 \\ (c + 2)^2 & (c + 4)^2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ c + 6 \\ (c + 6)^2 \end{bmatrix} \]

### Step 2: Set up the determinant

For the system of equations to be consistent, the determinant of the coefficient matrix must be zero. We denote the coefficient matrix as \( A \):

\[ A = \begin{bmatrix} 2 & 3 \\ c + 2 & c + 4 \\ (c + 2)^2 & (c + 4)^2 \end{bmatrix} \]

### Step 3: Calculate the determinant

We will calculate the determinant of \( A \) and set it to zero.

Using the determinant formula for a 3x2 matrix, we can expand it as follows:

\[ \text{det}(A) = 2 \cdot \begin{vmatrix} c + 4 & (c + 4)^2 \\ (c + 2) & (c + 2)^2 \end{vmatrix} - 3 \cdot \begin{vmatrix} c + 2 & (c + 2)^2 \\ (c + 2) & (c + 4)^2 \end{vmatrix} \]

### Step 4: Simplify the determinant

We will simplify each of the 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} c + 4 & (c + 4)^2 \\ c + 2 & (c + 2)^2 \end{vmatrix} = (c + 4)(c + 2)^2 - (c + 2)(c + 4)^2 \]

2. For the second determinant:
\[ \begin{vmatrix} c + 2 & (c + 2)^2 \\ c + 2 & (c + 4)^2 \end{vmatrix} = (c + 2)(c + 4)^2 - (c + 2)(c + 2)^2 = 0 \]

### Step 5: Set the determinant to zero

Now, we set the determinant to zero:

\[ 2 \cdot \left[(c + 4)(c + 2)^2 - (c + 2)(c + 4)^2\right] = 0 \]

This simplifies to:

\[ (c + 4)(c + 2)^2 - (c + 2)(c + 4)^2 = 0 \]

Factoring out \( (c + 2)(c + 4) \):

\[ (c + 2)(c + 4)\left[(c + 2) - (c + 4)\right] = 0 \]

This gives us:

\[ (c + 2)(c + 4)(-2) = 0 \]

### Step 6: Solve for \( c \)

Setting each factor to zero gives us:

1. \( c + 2 = 0 \) → \( c = -2 \)
2. \( c + 4 = 0 \) → \( c = -4 \)

### Step 7: Solve the equations for these values of \( c \)

**For \( c = -2 \):**

Substituting \( c = -2 \) into the equations:

1. \( 2x + 3y = 3 \)
2. \( 0x + 2y = 4 \) → \( y = 2 \)
3. \( 0x + 0y = 0 \) (consistent)

From \( 2x + 3(2) = 3 \):
\[ 2x + 6 = 3 \implies 2x = -3 \implies x = -\frac{3}{2} \]

**For \( c = -4 \):**

Substituting \( c = -4 \) into the equations:

1. \( 2x + 3y = 3 \)
2. \( -2x + 0y = 2 \) → \( x = -1 \)
3. \( 36x + 36y = 36 \) (consistent)

From \( 2(-1) + 3y = 3 \):
\[ -2 + 3y = 3 \implies 3y = 5 \implies y = \frac{5}{3} \]

### Final Result

The values of \( c \) for which the equations are consistent are \( c = -2 \) and \( c = -4 \). The corresponding solutions are:

- For \( c = -2 \): \( x = -\frac{3}{2}, y = 2 \)
- For \( c = -4 \): \( x = -1, y = \frac{5}{3} \)