The problem asks us to solve the following system of linear equations for $x$ and $y$ using Cramer's rule: $(1+2k)x + 5y = 7$ $(2+k)x + 4y = 8$

1. Problem Description

The problem asks us to solve the following system of linear equations for $x$ and $y$ using Cramer's rule:

$(1+2k)x + 5y = 7$

$(2+k)x + 4y = 8$

2. Solution Steps

First, we find the determinant of the coefficient matrix:

$D = \begin{vmatrix} 1+2k & 5 \\ 2+k & 4 \end{vmatrix} = (1+2k)(4) - (5)(2+k) = 4 + 8k - 10 - 5k = 3k - 6$

Next, we find the determinant $D_x$ by replacing the first column of the coefficient matrix with the constants:

$D_x = \begin{vmatrix} 7 & 5 \\ 8 & 4 \end{vmatrix} = (7)(4) - (5)(8) = 28 - 40 = -12$

Then, we find the determinant $D_y$ by replacing the second column of the coefficient matrix with the constants:

$D_y = \begin{vmatrix} 1+2k & 7 \\ 2+k & 8 \end{vmatrix} = (1+2k)(8) - (7)(2+k) = 8 + 16k - 14 - 7k = 9k - 6$

Using Cramer's rule, we find $x$ and $y$ as follows:

$x = \frac{D_x}{D} = \frac{-12}{3k-6} = \frac{-12}{3(k-2)} = \frac{-4}{k-2}$

$y = \frac{D_y}{D} = \frac{9k-6}{3k-6} = \frac{3(3k-2)}{3(k-2)} = \frac{3k-2}{k-2}$

3. Final Answer

$x = \frac{-4}{k-2}$

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