The problem is to solve a system of two linear equations with two variables, $x$ and $y$, using Cramer's rule (the determinant method). The system of equations is: $(1+2k)x + 5y = 7$ $(2+k)x + 4y = 8$

1. Problem Description

The problem is to solve a system of two linear equations with two variables, $x$ and $y$, using Cramer's rule (the determinant method). The system of equations is:

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

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

2. Solution Steps

Cramer's rule involves calculating determinants.

The general form of a system of two linear equations is:

$a_1x + b_1y = c_1$

$a_2x + b_2y = c_2$

The determinant of the coefficient matrix, $D$, is:

$D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1$

The determinant $D_x$ is obtained by replacing the first column of $D$ with the constants $c_1$ and $c_2$:

$D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1b_2 - c_2b_1$

The determinant $D_y$ is obtained by replacing the second column of $D$ with the constants $c_1$ and $c_2$:

$D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1c_2 - a_2c_1$

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

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

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

In our case, we have:

$a_1 = 1 + 2k, b_1 = 5, c_1 = 7$

$a_2 = 2 + k, b_2 = 4, c_2 = 8$

First, we calculate $D$:

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

Next, we calculate $D_x$:

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

Then, we calculate $D_y$:

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

Finally, we find $x$ and $y$:

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

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

3. Final Answer

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

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