Q. 16 - 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.

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

Step-by-Step Solution

Answer

The solution is $x=\frac{11}{3}, \mathrm{y}=\frac{2}{3} \mathrm{and} \mathrm{in} \mathrm{ordered} \mathrm{pair} \mathrm{is} \left(\frac{11}{3},\frac{2}{3}\right).$

Step 1. Given Information

The given system of equation is

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

We have to solve the system of equations using Cramer's rule if it is applicable.

Step 2. Finding the determinant

The determinant D of the coefficients of the variables is

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

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

Step 3. Use Cramer's rule

Let's find $D_x \mathrm{and} D_y$

$$\begin{gathered} D_x=\left|\begin{array}{cc}5& 2\\ 3& -1\end{array}\right| \\ {D}_x=\left(5\right)\left(-1\right)-\left(2\right)\left(3\right) \\ {D}_x=-5-6 \\ {D}_x=-11 \end{gathered}$$

And

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

Step 4. Finding x and y

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

$$\begin{gathered} x=\frac{D_x}{D} and y=\frac{D_y}{D} \\ x=\frac{-11}{-3} and y=\frac{-2}{-3} \\ x=\frac{11}{3} and y=\frac{2}{3} \end{gathered}$$