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.

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

Short 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-by-step solution

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 determinantD 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 xand yare

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