Q. 19 - 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.  

$\left\{\begin{array}{l}3x=24\\ x+2y=0\end{array}\right.$

Step-by-Step Solution

Answer

The solution is $x=8, \mathrm{y}=-4 \mathrm{and} \mathrm{in} \mathrm{ordered} \mathrm{pair} \mathrm{is} \left(8,-4\right).$

Step 1. Given Information

The given system of equation is

$\left\{\begin{array}{l}3x=24\\ x+2y=0\end{array}\right.$

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

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

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

Step 3. Use Cramer's rule

Let's find $D_x \mathrm{and} D_y$

$$\begin{gathered} D_x=\left|\begin{array}{cc}24& 0\\ 0& 2\end{array}\right| \\ {D}_x=\left(24\right)\left(2\right)-\left(0\right)\left(0\right) \\ {D}_x=48-0 \\ {D}_x=48 \end{gathered}$$

And

$$\begin{gathered} D_y=\left|\begin{array}{cc}3& 24\\ 1& 0\end{array}\right| \\ {D}_y=\left(3\right)\left(0\right)-\left(24\right)\left(1\right) \\ {D}_y=0-24 \\ {D}_y=-24 \end{gathered}$$

Step 4. Finding x and y

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

$$\begin{gathered} x=\frac{D_x}{D} and y=\frac{D_y}{D} \\ x=\frac{48}{6} and y=\frac{-24}{6} \\ x=8 and y=-4 \end{gathered}$$