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.

$\left\{\begin{array}{l}x+3y=5\\ 2x-3y=-8\end{array}\right.$

Short answer

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

Step-by-step solution

Step 1. Given Information 

The given system of equation is

$\left\{\begin{array}{l}x+3y=5\\ 2x-3y=-8\end{array}\right.$

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

$$\begin{gathered} D=\left|\begin{array}{cc}1& 3\\ 2& -3\end{array}\right| \\ D=\left(1\right)\left(-3\right)-\left(3\right)\left(2\right) \\ D=-3-6 \\ D=-9 \end{gathered}$$

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

Step 3. Use Cramer's rule 

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

$$\begin{gathered} D_x=\left|\begin{array}{cc}5& 3\\ -8& -3\end{array}\right| \\ {D}_x=\left(5\right)\left(-3\right)-\left(3\right)\left(-8\right) \\ {D}_x=-15+24 \\ {D}_x=9 \end{gathered}$$

And

$$\begin{gathered} D_y=\left|\begin{array}{cc}1& 5\\ 2& -8\end{array}\right| \\ {D}_y=\left(1\right)\left(-8\right)-\left(5\right)\left(2\right) \\ {D}_y=-8-10 \\ {D}_y=-18 \end{gathered}$$

Step 4. Finding x and y 

By using Cramer's rule, the values of xand yare

$$\begin{gathered} x=\frac{D_x}{D}andy=\frac{D_y}{D} \\ x=\frac{9}{-9}andy=\frac{-18}{-9} \\ x=-1andy=2 \end{gathered}$$