Q. 19 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $19$ and $20$, use Cramer’s Rule, if possible, to solve each system

$\left\{\begin{array}{l}4x+3y=-23\\ 3x-5y=19\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system is $(-2,-5)$

Step 1. Given information

System of equations is given as 

$\left\{\begin{array}{l}4x+3y=-23\\ 3x-5y=19\end{array}\right.$

Step 2. Checking for Cramer's rule

First take determinant of the coefficients,

$$\begin{gathered} D=\left|\begin{array}{cc}4& 3\\ 3& -5\end{array}\right| \\ =\left(4\right)(-5)-\left(3\right)\left(3\right) \\ =-20-9 \\ =-29 \end{gathered}$$

Since $D\ne 0$, we can use Cramer's rule which is $x=\frac{D_x}{D}$ and $y=\frac{D_y}{D}$

Step 3. Finding determinants

For D_x replace he entries in first column of D by the constants from other side and similarly for D_y, replace second column 

$$\begin{gathered} D_x=\left|\begin{array}{cc}-23& 3\\ 19& -5\end{array}\right| \\ =(-23)(-5)-\left(3\right)\left(19\right) \\ =115-57 \\ =58 \end{gathered}$$

and

$$\begin{gathered} D_y=\left|\begin{array}{cc}4& -23\\ 3& 19\end{array}\right| \\ =\left(4\right)\left(19\right)-(-23)\left(3\right) \\ =76-(-69) \\ =145 \end{gathered}$$

Step 4. Finding values

Now substitute the values of determinants as,

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{58}{-29} \\ x=-2 \end{gathered}$$

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{145}{-29} \\ y=-5 \end{gathered}$$