Q 279 - Intermediate Algebra

Question

Explain the steps for solving a system of equations
using Cramer’s rule.

Step-by-Step Solution

Answer

For the system of equations $\left\{\begin{array}{l}{a}_{1}x+b_{1}y=k_1\\ a_{2}x+b_{2}y=k_2\end{array}\right.$, the solution $(x,y)$ can be determined by, $x=\frac{D_x}{D}$,

$y=\frac{D_y}{D}$

Where, $D=\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|$

           $D_x=\left|\begin{array}{cc}{k}_1& b_1\\ k_2& b_2\end{array}\right|$

           $D_y=\left|\begin{array}{cc}{a}_1& k_1\\ a_2& k_2\end{array}\right|$

Step 1. To explain

To explain the steps for solving the system of equations by using Cramer's rule.

Step 2. Consider the linear equation

Consider the system of linear equations,

$\left\{\begin{array}{l}-2x+3y=3\\ x+3y=12\end{array}\right.$

Step 3. Find the value of D

Evaluate the determinant $D$ by using the coefficients of variables.

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

   $=-6-3$

   $=-9$

The value of $D$ is $-9$

Step 4. Find D x

Evaluate $D_x$ by replacing the coefficient of $x,-2,1$ with the constants $3,12$

$D_x=\left|\begin{array}{cc}3& 1\\ 12& 3\end{array}\right|$

     $=9-12$

     $=-3$

The value of $D_x$ is $-3$

Step 5. Find D y

Evaluate $D_y$ by replacing the coefficient of $y,3,3$ by the constants $3,12$

$D_y=\left|\begin{array}{cc}-2& 3\\ 1& 12\end{array}\right|$

     $=-24-3$

     $=-27$

The value of $D_y$ is $-27$

Step 6. Find ( x , y )

Find the solution $x,y$

$x=\frac{D_x}{D}$

  $=\frac{-3}{-9}$

  $=\frac{1}{3}$

$y=\frac{D_y}{D}$

  $=\frac{-27}{-9}$

  $=3$

The solution $(x,y)$ is $(\frac{1}{3},3)$