Q. 25 - 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. 

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

Step-by-Step Solution

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 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 $D_x \mathrm{and} {\mathrm{D}}_{\mathrm{y}}$

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

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 x and y are 

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