Q 4.98 - Intermediate Algebra

Question

Solve using Cramer's rule 

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

Step-by-Step Solution

Answer

The value of      $$\begin{gathered} x=-2 \\ y=0 \end{gathered}$$

Step 1. Given information

The given equations are  $\left\{\begin{array}{l}-x+y=2\\ 2x+y=-4\end{array}\right.$

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

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

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

Fore the determinant  of D_x ,

we replace the coefficients of x , -1 and 2 with the constants, 2 and -4.

$$\begin{gathered} D_x=\left|\begin{array}{cc}2& 1\\ -4& 1\end{array}\right| \\ = 2\times 1-1\times -4 \\ =2+4 \\ =6 \end{gathered}$$

Fore the determinant  of D_y ,

we replace the coefficients of y , 1 and 1 with the constants, 2 and -4.

$$\begin{gathered} D_y=\left|\begin{array}{cc}-1& 2\\ 2& -4\end{array}\right| \\ = -1\times -4-2\times 2 \\ =4-4 \\ =0 \end{gathered}$$

Step 4. Find x and y.

$$\begin{gathered} x=\frac{D_x}{D} \mathrm{and} \\ \mathrm{y}=\frac{{\mathrm{D}}_{\mathrm{y}}}{\mathrm{D}} \\ \mathrm{x}=\frac{6 }{-3}\mathrm{and} \\ \mathrm{y}=\frac{0}{-3} \\ \mathrm{The} \mathrm{ordered} \mathrm{pair} \mathrm{is} (-2,0). \end{gathered}$$

Substitute the values in the linear equation.

As the points satisfy the equations, Hence$(-2,0)$. it is the solution to the system

Step 5. Checking the values

Substitute the value
$x=-2, y=0$

$$\begin{gathered} \mathrm{in} \mathrm{the} \mathrm{equation} -\mathrm{x}+\mathrm{y}=2 \\ -(-2)+0=2 \\ 2=2 \\ \mathrm{Also} \mathrm{substitute} \mathrm{the} \mathrm{value} \\ \mathrm{x}= -2, \\ \mathrm{y}=0 \\ \mathrm{in} \mathrm{the} \mathrm{equation} \\ 2\mathrm{x}+\mathrm{y}=-4 \\ 2(-2)+0=-4 \\ -4=-4 \mathrm{True} \end{gathered}$$

Hence the solution to the equation is

$x=-2, y=0$