Question

use cramers rule to solve each system.
(11) (4x + 2y=-16), (2x+3y = - 4)
(12) (-6x + 3y=-14), (6x + 2y=-14)
(13) (6x - y=17), (3x + y=-11)
(14) (-2x-2y = 6), (x + 6y=7)
(15) (-2x + 3y=-30), (3x + 4y=-6)
(16) (-5x-2y = 5), (-2x + y=6)
(17) (-4x-3y = 10), (-2x-4y = 10)
(18) (5x-5y=-30), (-2x + 5y=15)

Explanation:

Step1: Recall Cramer's Rule for a 2x2 system

For a system of linear equations

$$\begin{cases}a_1x + b_1y=c_1\\a_2x + b_2y=c_2\end{cases}$$

, the determinant of the coefficient - matrix $D=

$$\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}$$

=a_1b_2 - a_2b_1$, $D_x=

$$\begin{vmatrix}c_1&b_1\\c_2&b_2\end{vmatrix}$$

=c_1b_2 - c_2b_1$, and $D_y=

$$\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix}$$

=a_1c_2 - a_2c_1$. Then $x = \frac{D_x}{D}$ and $y=\frac{D_y}{D}$ (when $D
eq0$).

Step2: Solve the system

$$\begin{cases}6x - y=17\\3x + y=-11\end{cases}$$

First, calculate the determinant of the coefficient - matrix $D=

$$\begin{vmatrix}6&-1\\3&1\end{vmatrix}$$

=6\times1-3\times(-1)=6 + 3=9$.
Next, calculate $D_x=

$$\begin{vmatrix}17&-1\\-11&1\end{vmatrix}$$

=17\times1-(-11)\times(-1)=17 - 11 = 6$.
Then, calculate $D_y=

$$\begin{vmatrix}6&17\\3&-11\end{vmatrix}$$

=6\times(-11)-3\times17=-66 - 51=-117$.
So, $x=\frac{D_x}{D}=\frac{6}{9}=\frac{2}{3}$ and $y=\frac{D_y}{D}=\frac{-117}{9}=-13$.
As an example for one of the systems. For other systems, follow the same steps:

1. For

$$\begin{cases}-2x-2y = 6\\x + 6y=7\end{cases}$$

:

• $D=

$$\begin{vmatrix}-2&-2\\1&6\end{vmatrix}$$

=(-2)\times6-1\times(-2)=-12 + 2=-10$.

• $D_x=

$$\begin{vmatrix}6&-2\\7&6\end{vmatrix}$$

=6\times6-7\times(-2)=36 + 14 = 50$.

• $D_y=

$$\begin{vmatrix}-2&6\\1&7\end{vmatrix}$$

=(-2)\times7-1\times6=-14 - 6=-20$.

• $x=\frac{D_x}{D}=\frac{50}{-10}=-5$ and $y=\frac{D_y}{D}=\frac{-20}{-10}=2$.

1. For

$$\begin{cases}-2x + 3y=-30\\3x + 4y=-6\end{cases}$$

:

• $D=

$$\begin{vmatrix}-2&3\\3&4\end{vmatrix}$$

=(-2)\times4-3\times3=-8 - 9=-17$.

• $D_x=

$$\begin{vmatrix}-30&3\\-6&4\end{vmatrix}$$

=(-30)\times4-(-6)\times3=-120 + 18=-102$.

• $D_y=

$$\begin{vmatrix}-2&-30\\3&-6\end{vmatrix}$$

=(-2)\times(-6)-3\times(-30)=12 + 90 = 102$.

• $x=\frac{D_x}{D}=\frac{-102}{-17}=6$ and $y=\frac{D_y}{D}=\frac{102}{-17}=-6$.

1. For

$$\begin{cases}-5x-2y = 5\\-2x + y=6\end{cases}$$

:

• $D=

$$\begin{vmatrix}-5&-2\\-2&1\end{vmatrix}$$

=(-5)\times1-(-2)\times(-2)=-5 - 4=-9$.

• $D_x=

$$\begin{vmatrix}5&-2\\6&1\end{vmatrix}$$

=5\times1-6\times(-2)=5 + 12 = 17$.

• $D_y=

$$\begin{vmatrix}-5&5\\-2&6\end{vmatrix}$$

=(-5)\times6-(-2)\times5=-30 + 10=-20$.

• $x=\frac{D_x}{D}=-\frac{17}{9}$ and $y=\frac{D_y}{D}=\frac{20}{9}$.

1. For

$$\begin{cases}-4x-3y = 10\\-2x-4y = 10\end{cases}$$

:

• $D=

$$\begin{vmatrix}-4&-3\\-2&-4\end{vmatrix}$$

=(-4)\times(-4)-(-2)\times(-3)=16 - 6 = 10$.

• $D_x=

$$\begin{vmatrix}10&-3\\10&-4\end{vmatrix}$$

=10\times(-4)-10\times(-3)=-40 + 30=-10$.

• $D_y=

$$\begin{vmatrix}-4&10\\-2&10\end{vmatrix}$$

=(-4)\times10-(-2)\times10=-40 + 20=-20$.

• $x=-1$ and $y=-2$.

1. For

$$\begin{cases}5x-5y=-30\\-2x + 5y=15\end{cases}$$

:

• $D=

$$\begin{vmatrix}5&-5\\-2&5\end{vmatrix}$$

=5\times5-(-2)\times(-5)=25 - 10 = 15$.

• $D_x=

$$\begin{vmatrix}-30&-5\\15&5\end{vmatrix}$$

=(-30)\times5-15\times(-5)=-150 + 75=-75$.

• $D_y=

$$\begin{vmatrix}5&-30\\-2&15\end{vmatrix}$$

=5\times15-(-2)\times(-30)=75 - 60 = 15$.

• $x=-5$ and $y = 1$.

Answer:

For the system

$$\begin{cases}6x - y=17\\3x + y=-11\end{cases}$$

, $x=\frac{2}{3},y=-13$. For

$$\begin{cases}-2x-2y = 6\\x + 6y=7\end{cases}$$

, $x=-5,y = 2$. For

$$\begin{cases}-2x + 3y=-30\\3x + 4y=-6\end{cases}$$

, $x = 6,y=-6$. For

$$\begin{cases}-5x-2y = 5\\-2x + y=6\end{cases}$$

, $x=-\frac{17}{9},y=\frac{20}{9}$. For

$$\begin{cases}-4x-3y = 10\\-2x-4y = 10\end{cases}$$

, $x=-1,y=-2$. For

$$\begin{cases}5x-5y=-30\\-2x + 5y=15\end{cases}$$

, $x=-5,y = 1$.