The problem asks to find the real parameter $k$ such that the following system of linear equations has a unique solution: $(k+2)x + (k-2)y = 7$ $4x - 5y = 8+k$

1. Problem Description

The problem asks to find the real parameter $k$ such that the following system of linear equations has a unique solution:

$(k+2)x + (k-2)y = 7$

$4x - 5y = 8+k$

2. Solution Steps

For a system of two linear equations in two variables to have a unique solution, the determinant of the coefficient matrix must be non-zero. The system of equations is given by:

$(k+2)x + (k-2)y = 7$

$4x - 5y = 8+k$

The coefficient matrix is:

$\begin{bmatrix} k+2 & k-2 \\ 4 & -5 \end{bmatrix}$

The determinant of this matrix is:

$D = (k+2)(-5) - (k-2)(4)$

$D = -5k - 10 - 4k + 8$

$D = -9k - 2$

For a unique solution to exist, we require $D \neq 0$. Therefore,

$-9k - 2 \neq 0$

$-9k \neq 2$

$k \neq -\frac{2}{9}$

3. Final Answer

$k \neq -\frac{2}{9}$