Q. 20 - 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. 

$\left\{\begin{array}{l}4x+5y=-3\\ -2y=-4\end{array}\right.$

Step-by-Step Solution

Answer

The solution is $x=-\frac{13}{4}, \mathrm{y}=2 \mathrm{and} \mathrm{in} \mathrm{ordered} \mathrm{pair} \mathrm{is} \left(-\frac{13}{4},2\right).$

Step 1. Given Information

The given system of equation is

$\left\{\begin{array}{l}4x+5y=-3\\ -2y=-4\end{array}\right.$

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

$$\begin{gathered} D=\left|\begin{array}{cc}4& 5\\ 0& -2\end{array}\right| \\ D=\left(4\right)\left(-2\right)-\left(5\right)\left(0\right) \\ D=-8-0 \\ D=-8 \end{gathered}$$

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

Step 3. Use Cramer's rule

Let's find $D_x \mathrm{and} D_y$

$$\begin{gathered} D_x=\left|\begin{array}{cc}-3& 5\\ -4& -2\end{array}\right| \\ {D}_x=\left(-3\right)\left(-2\right)-\left(5\right)\left(-4\right) \\ {D}_x=6+20 \\ {D}_x=26 \end{gathered}$$

And

$$\begin{gathered} D_y=\left|\begin{array}{cc}4& -3\\ 0& -4\end{array}\right| \\ {D}_y=\left(4\right)\left(-4\right)-\left(-3\right)\left(0\right) \\ {D}_y=-16+0 \\ {D}_y=-16 \end{gathered}$$

Step 4. Finding x and y

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

$$\begin{gathered} x=\frac{D_x}{D} and y=\frac{D_y}{D} \\ x=\frac{26}{-8} and y=\frac{-16}{-8} \\ x=-\frac{13}{4} and y=2 \end{gathered}$$