We are given a system of two linear equations: $(a+2)x + (a-2)y = 16$ $2x + 4y = a-2$ We are asked to solve the system for $x$ and $y$.

1. Problem Description

We are given a system of two linear equations:

$(a+2)x + (a-2)y = 16$

$2x + 4y = a-2$

We are asked to solve the system for $x$ and $y$.

2. Solution Steps

Let the given equations be

$(a+2)x + (a-2)y = 16$ (1)

$2x + 4y = a-2$ (2)

We can use the method of determinants to solve this system.

Let $D$ be the determinant of the coefficients of $x$ and $y$.

$D = \begin{vmatrix} a+2 & a-2 \\ 2 & 4 \end{vmatrix} = 4(a+2) - 2(a-2) = 4a+8-2a+4 = 2a+12 = 2(a+6)$

Let $D_x$ be the determinant obtained by replacing the $x$ coefficients with the constants.

$D_x = \begin{vmatrix} 16 & a-2 \\ a-2 & 4 \end{vmatrix} = 16(4) - (a-2)(a-2) = 64 - (a^2-4a+4) = 60+4a-a^2$

Let $D_y$ be the determinant obtained by replacing the $y$ coefficients with the constants.

$D_y = \begin{vmatrix} a+2 & 16 \\ 2 & a-2 \end{vmatrix} = (a+2)(a-2) - 16(2) = a^2-4-32 = a^2-36 = (a-6)(a+6)$

If $D \neq 0$, then the solution is given by

$x = \frac{D_x}{D} = \frac{60+4a-a^2}{2(a+6)} = \frac{-(a^2-4a-60)}{2(a+6)} = \frac{-(a-10)(a+6)}{2(a+6)}$

$y = \frac{D_y}{D} = \frac{a^2-36}{2(a+6)} = \frac{(a-6)(a+6)}{2(a+6)}$

If $a \neq -6$, then

$x = \frac{-(a-10)}{2} = \frac{10-a}{2}$

$y = \frac{a-6}{2}$

3. Final Answer

If $a \neq -6$, then $x = \frac{10-a}{2}$ and $y = \frac{a-6}{2}$.

If $a = -6$, the determinant $D$ is

0. The system has either infinitely many solutions or no solution. Substituting $a = -6$ into the equations, we get:

$(-6+2)x + (-6-2)y = 16 \implies -4x - 8y = 16 \implies x + 2y = -4$

$2x + 4y = -6-2 \implies 2x+4y = -8 \implies x+2y = -4$

Since both equations are equivalent to $x+2y = -4$, there are infinitely many solutions when $a = -6$.

Final Answer:

If $a \neq -6$, $x = \frac{10-a}{2}$, $y = \frac{a-6}{2}$.

If $a = -6$, $x+2y = -4$ which represents an infinite number of solutions.