We are asked to solve the following system of linear equations using Cramer's rule: $(1+2k)x + 5y = 7$ $(2+k)x + 4y = 8$

1. Problem Description

We are asked to solve the following system of linear equations using Cramer's rule:

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

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

2. Solution Steps

Cramer's rule states that for a system of linear equations:

$ax + by = e$

$cx + dy = f$

The solutions for $x$ and $y$ are given by:

$x = \frac{D_x}{D}$

$y = \frac{D_y}{D}$

Where:

$D = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$

$D_x = \begin{vmatrix} e & b \\ f & d \end{vmatrix} = ed - bf$

$D_y = \begin{vmatrix} a & e \\ c & f \end{vmatrix} = af - ec$

In our case:

$a = 1+2k$

$b = 5$

$c = 2+k$

$d = 4$

$e = 7$

$f = 8$

First, we calculate $D$:

$D = (1+2k)(4) - (5)(2+k) = 4 + 8k - 10 - 5k = 3k - 6$

Next, we calculate $D_x$:

$D_x = (7)(4) - (5)(8) = 28 - 40 = -12$

Then, we calculate $D_y$:

$D_y = (1+2k)(8) - (7)(2+k) = 8 + 16k - 14 - 7k = 9k - 6$

Now, we can find $x$ and $y$:

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

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

3. Final Answer

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

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