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.

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

Short answer

The solution is$x=\frac{1}{2},\mathrm{y}=\frac{3}{4}\mathrm{and}\mathrm{in}\mathrm{ordered}\mathrm{pair}\mathrm{is}\left(\frac{1}{2},\frac{3}{4}\right).$

Step-by-step solution

Step 1. Given Information 

The given system of equation is

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

$$\begin{gathered} D=\left|\begin{array}{cc}2& -4\\ 3& 2\end{array}\right| \\ D=\left(2\right)\left(2\right)-\left(-4\right)\left(3\right) \\ D=\left(4\right)-\left(-12\right) \\ D=16 \end{gathered}$$

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

Step 3. Use Cramer's rule 

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

role="math" localid="1647430987257" $$\begin{gathered} D_x=\left|\begin{array}{cc}-2& -4\\ 3& 2\end{array}\right| \\ {D}_x=\left(-2\right)\left(2\right)-\left(-4\right)\left(3\right) \\ {D}_x=-4+12 \\ {D}_x=8 \end{gathered}$$

And

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

Step 4. Finding x and y  

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

$$\begin{gathered} x=\frac{D_x}{D}andy=\frac{D_y}{D} \\ x=\frac{8}{16}andy=\frac{12}{16} \\ x=\frac{1}{2}andy=\frac{3}{4} \end{gathered}$$