We are given a system of two linear equations with two variables $x$ and $y$, and a parameter $a$: $(a+2)x + (a-2)y = 16$ $2x + 4y = a-2$ We are asked to solve this system using determinants. The image shows the calculation of the determinant $D$ of the coefficients matrix. We need to compute $D_x$ and $D_y$, and then find the solutions for $x$ and $y$.

1. Problem Description

We are given a system of two linear equations with two variables $x$ and $y$, and a parameter $a$:

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

$2x + 4y = a-2$

We are asked to solve this system using determinants. The image shows the calculation of the determinant $D$ of the coefficients matrix. We need to compute $D_x$ and $D_y$, and then find the solutions for $x$ and $y$.

2. Solution Steps

First, let's verify the calculation of the determinant $D$:

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

Next, we compute $D_x$:

$D_x = \begin{vmatrix} 16 & a-2 \\ a-2 & 4 \end{vmatrix} = (16)(4) - (a-2)(a-2) = 64 - (a^2 - 4a + 4) = 64 - a^2 + 4a - 4 = -a^2 + 4a + 60 = -(a^2 - 4a - 60) = -(a-10)(a+6)$.

Now, we compute $D_y$:

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

Using Cramer's rule, we have:

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

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

If $a \neq -6$, then we can simplify the expressions for $x$ and $y$:

$x = \frac{-(a-10)}{2} = \frac{-a+10}{2} = 5 - \frac{a}{2}$

$y = \frac{a-6}{2} = \frac{a}{2} - 3$

3. Final Answer

If $a \neq -6$, then

$x = 5 - \frac{a}{2}$

$y = \frac{a}{2} - 3$

$D = 2(a+6)$

$D_x = -(a-10)(a+6)$

$D_y = (a-6)(a+6)$

If $a = -6$, $D = 0$, so we cannot use Cramer's rule directly. In that case, the system becomes:

$-4x - 8y = 16$

$2x + 4y = -8$

Dividing the first equation by -4, we get:

$x + 2y = -4$

The second equation is the same: $2x + 4y = -8$, which gives $x + 2y = -4$. So the equations are linearly dependent.

The solution is $x = -4 - 2y$, where $y$ can be any real number.

Final Answer:

If $a \neq -6$, then

$x = 5 - \frac{a}{2}$

$y = \frac{a}{2} - 3$

If $a = -6$, then $x = -4 - 2y$, where $y$ can be any real number.

$D = 2(a+6)$

$D_x = -(a-10)(a+6)$

$D_y = (a-6)(a+6)$