Q. 30 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 15 – 42, solve each system of equations using Cramer’s Rule if it is applicable. If Cramer’s Rule is not applicable, say so. 

$\left\{\begin{array}{l}\frac{1}{2}x+y=-2\\ x-2y=8\end{array}\right.$

Step-by-Step Solution

Answer

The solution is $x=2, \mathrm{y}=-3 \mathrm{and} \mathrm{in} \mathrm{ordered} \mathrm{pair} \mathrm{is} \left(2,-3\right).$

Step 1. Given Information

The given system of equation is

$\left\{\begin{array}{l}\frac{1}{2}x+y=-2\\ x-2y=8\end{array}\right.$

We have to solve the system of equations using Cramer's rule if it is applicable.

Step 2. Finding the determinant

The determinant D of the coefficients of the variables is

$$\begin{gathered} D=\left|\begin{array}{cc}\frac{1}{2}& 1\\ 1& -2\end{array}\right| \\ D=\left(\frac{1}{2}\right)\left(-2\right)-\left(1\right)\left(1\right) \\ D=\left(-1\right)-\left(1\right) \\ D=-2 \end{gathered}$$

Since $D\ne 0$ Cramer's rule is applicable.

Step 3. Use Cramer's rule

Let's find $D_x \mathrm{and} {\mathrm{D}}_{\mathrm{y}}$

$$\begin{gathered} D_x=\left|\begin{array}{cc}-2& 1\\ 8& -2\end{array}\right| \\ {D}_x=\left(-2\right)\left(-2\right)-\left(1\right)\left(8\right) \\ {D}_x=4-8 \\ {D}_x=-4 \end{gathered}$$

And

$$\begin{gathered} D_y=\left|\begin{array}{cc}\frac{1}{2}& -2\\ 1& 8\end{array}\right| \\ {D}_y=\left(\frac{1}{2}\right)\left(8\right)-\left(-2\right)\left(1\right) \\ {D}_y=4+2 \\ {D}_y=6 \end{gathered}$$

Step 4. Finding x and y

By using Cramer's rule, the values of x and y are 

$$\begin{gathered} x=\frac{D_x}{D} and y=\frac{D_y}{D} \\ x=\frac{-4}{-2} and y=\frac{6}{-2} \\ x=2 and y=-3 \end{gathered}$$