Q. 29 - 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. 

$\left\{\begin{array}{l}2x+3y=6\\ x-y=\frac{1}{2}\end{array}\right.$

Step-by-Step Solution

Answer

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

Step 1. Given Information

The given system of equation is

$\left\{\begin{array}{l}2x+3y=6\\ x-y=\frac{1}{2}\end{array}\right.$

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

$$\begin{gathered} D=\left|\begin{array}{cc}2& 3\\ 1& -1\end{array}\right| \\ D=\left(2\right)\left(-1\right)-\left(3\right)\left(1\right) \\ D=\left(-2\right)-\left(3\right) \\ D=-5 \end{gathered}$$

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

Step 3. Use Cramer's rule

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

$$\begin{gathered} D_x=\left|\begin{array}{cc}6& 3\\ \frac{1}{2}& -1\end{array}\right| \\ {D}_x=\left(-1\right)\left(6\right)-\left(3\right)\left(\frac{1}{2}\right) \\ {D}_x=-6-\frac{3}{2} \\ {D}_x=-\frac{15}{2} \end{gathered}$$

And

$$\begin{gathered} D_y=\left|\begin{array}{cc}2& 6\\ 1& \frac{1}{2}\end{array}\right| \\ {D}_y=\left(2\right)\left(\frac{1}{2}\right)-\left(1\right)\left(6\right) \\ {D}_y=1-6 \\ {D}_y=-5 \end{gathered}$$

Step 4. Finding x and y

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

$$\begin{gathered} x=\frac{D_x}{D} and y=\frac{D_y}{D} \\ x=\frac{-{\displaystyle \raisebox{1ex}{$15$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{-5} and y=\frac{-5}{-5} \\ x=\frac{3}{2} and y=1 \end{gathered}$$