Q 4.97 - Intermediate Algebra

Question

Solve using Cramer's rule $\left\{\begin{array}{l}3x+y=-3\\ 2x+3y=6\end{array}\right.$

Step-by-Step Solution

Answer

The value of      

 $$\begin{gathered} x=-\frac{15}{7} \mathrm{and} \\ y=\frac{24}{7} \end{gathered}$$

Step 1. Given information

The given equations are 

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

Step 2. Find the determinant D by using the coefficients of the variables.

$$\begin{gathered} D=\left|\begin{array}{cc}3& 1\\ 2& 3\end{array}\right| \\ = 3\times 3-1\times 2 \\ =9-2 \\ =7 \end{gathered}$$

Step 3.Evaluate the determinant D x and D y.

For the determinant  of D_x ,

we replace the coefficients of x , 3 and 2 with the constants, -3 and 6.

$$\begin{gathered} D_x=\left|\begin{array}{cc}-3& 1\\ 6& 3\end{array}\right| \\ = -3\times 3-1\times 6 \\ =-9-6 \\ =-15 \end{gathered}$$

For the determinant  of D_y ,

we replace the coefficients of y , 1 and 3 with the constants, -3 and 6.

$$\begin{gathered} D_y=\left|\begin{array}{cc}3& -3\\ 2& 6\end{array}\right| \\ = 3\times 6-(-3)\times 2 \\ =18+6 \\ =24 \end{gathered}$$

Step 4. Find x and y.

$$\begin{gathered} x=\frac{D_x}{D} \mathrm{and} \\ \mathrm{y}=\frac{{\mathrm{D}}_{\mathrm{y}}}{\mathrm{D}} \\ \mathrm{x}=\frac{-15 }{7}\mathrm{and} \\ \mathrm{y}=\frac{24}{7} \\ \mathrm{The} \mathrm{ordered} \mathrm{pair} \mathrm{is} (\frac{-15 }{7},\frac{24}{7}). \end{gathered}$$ 

Substitute the values in the linear equation.

As the points satisfies the equations, $Hence\left(\frac{-15}{7},\frac{24}{7}\right)$ it is the solution to the system

Step 5. Checking the equation

Substitute the value $x=-\frac{15}{7},y=\frac{24}{7}$

$$\begin{gathered} \mathrm{in} \mathrm{the} \mathrm{equation} 3\mathrm{x}+\mathrm{y}=-3 \\ 3(-\frac{15}{7})+\frac{24}{7}=-3 \\ -\frac{45}{7}+\frac{24}{7}=-3 \\ \frac{-21}{7}=-3 \\ -3=-3 \mathrm{True} \\ \mathrm{Also} \mathrm{substitute} \mathrm{the} \mathrm{value} \\ \mathrm{x}= -\frac{15}{7}, \\ \mathrm{y}=\frac{24}{7} \mathrm{in} \mathrm{the} \mathrm{equation} \\ 2\mathrm{x}+3\mathrm{y}=6 \\ 2(-\frac{15}{7})+3\left(\frac{24}{7}\right)=6 \\ \frac{-30}{7}+\frac{72}{7}=6 \\ \frac{42}{7}=6 \\ 6=6 \mathrm{True} \end{gathered}$$

Hence the solution to the equation is $x=-\frac{15}{7}, y=\frac{24}{7}$