Q. 26 - 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. 

$\left\{\begin{array}{l}3x+3y=3\\ 4x+2y=\frac{8}{3}\end{array}\right.$

Step-by-Step Solution

Answer

The solution is $x=\frac{1}{3}, \mathrm{y}=\frac{2}{3} \mathrm{and} \mathrm{in} \mathrm{ordered} \mathrm{pair} \mathrm{is} \left(\frac{1}{3},\frac{2}{3}\right).$

Step 1. Given Information

The given system of equation is

$\left\{\begin{array}{l}3x+3y=3\\ 4x+2y=\frac{8}{3}\end{array}\right.$

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

$$\begin{gathered} D=\left|\begin{array}{cc}3& 3\\ 4& 2\end{array}\right| \\ D=\left(3\right)\left(2\right)-\left(3\right)\left(4\right) \\ D=\left(6\right)-\left(12\right) \\ D=-6 \end{gathered}$$

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

Step 3. Use Cramer's rule

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

$$\begin{gathered} D_x=\left|\begin{array}{cc}3& 3\\ \frac{8}{3}& 2\end{array}\right| \\ {D}_x=\left(3\right)\left(2\right)-\left(3\right)\left(\frac{8}{3}\right) \\ {D}_x=6-8 \\ {D}_x=-2 \end{gathered}$$

And

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

Step 4. Finding x and y

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

$$\begin{gathered} x=\frac{D_x}{D} and y=\frac{D_y}{D} \\ x=\frac{-2}{-6} and y=\frac{-4}{-6} \\ x=\frac{1}{3} and y=\frac{2}{3} \end{gathered}$$