Question
Solve the following simultaneous equations using Cramer’s rule.
$x + 2y = –1; 2x – 3y = 12$
$x + 2y = –1; 2x – 3y = 12$
✓
Answer
$x+2 y=-1 $
$2 x-3 y=12 $
$D=\left[\begin{array}{cc}1 & 2 \\ 2 & -3\end{array}\right]=(1 \times-3)-(2 \times 2)=-3-4=-7 $
$D_x=\left[\begin{array}{cc} -1 & 2 \\ 12 & -3\end{array}\right]=(-1 \times-3)-(2 \times 12)=3-24=-21$
$D_y=\left[\begin{array}{cc} 1 & -1 \\ 2 & 12 \end{array}\right]=(1 \times 12)-(-1 \times 2)=12+2=14$
$x=\frac{D_x}{D}=-\frac{21}{-7}=3 y=\frac{D_y}{D}=\frac{14}{-7}=-2$
$\therefore( x , y )=(3,-2)$ is solution.
$2 x-3 y=12 $
$D=\left[\begin{array}{cc}1 & 2 \\ 2 & -3\end{array}\right]=(1 \times-3)-(2 \times 2)=-3-4=-7 $
$D_x=\left[\begin{array}{cc} -1 & 2 \\ 12 & -3\end{array}\right]=(-1 \times-3)-(2 \times 12)=3-24=-21$
$D_y=\left[\begin{array}{cc} 1 & -1 \\ 2 & 12 \end{array}\right]=(1 \times 12)-(-1 \times 2)=12+2=14$
$x=\frac{D_x}{D}=-\frac{21}{-7}=3 y=\frac{D_y}{D}=\frac{14}{-7}=-2$
$\therefore( x , y )=(3,-2)$ is solution.
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