Q. 29 - Precalculus Enhanced with Graphing Utilities

Question

Use Cramer’s Rule, if applicable, to solve each system.

$\left\{\begin{array}{l}x-2y=4\\ 3x+2y=4\end{array}\right.$

Step-by-Step Solution

Answer

Ordered pair is $(2,-1)$

Step 1. Given information is

$\left\{\begin{array}{l}x-2y=4\\ 3x+2y=4\end{array}\right.$

Step 2. Finding determinants we get,

$$\begin{gathered} D=\left|\begin{array}{cc}1& -2\\ 3& 2\end{array}\right| \\ D=\left(1\right)\left(2\right)-(-2)\left(3\right) \\ D=2+6 \\ D=8 \\ D\ne 0 \end{gathered}$$

Using crammer's rule we get $x=\frac{D_x}{D}, y=\frac{D_y}{D}$

Step 3. Solving for D x ,   D y we get

$$\begin{gathered} D_x=\left|\begin{array}{cc}4& -2\\ 4& 2\end{array}\right| \\ {D}_x=4\left(2\right)-(-2)\left(4\right) \\ {D}_x=8+8 \\ {D}_x=16 \end{gathered}$$

$$\begin{gathered} D_y=\left|\begin{array}{cc}1& 4\\ 3& 4\end{array}\right| \\ {D}_y=\left(1\right)\left(4\right)-\left(4\right)\left(3\right) \\ {D}_y=4-12 \\ {D}_y=-8 \end{gathered}$$

Step 4. Now value of x and y

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{16}{8} \\ x=2 \\ y=\frac{D_y}{D} \\ y=\frac{-8}{8} \\ y=-1 \end{gathered}$$

The ordered pair we have is $(2,-1)$