Question
Solve the following simultaneous equations using Cramer’s rule.
$2 x+3 y=2 ; x-\frac{y}{2}=\frac{1}{2}$
$2 x+3 y=2 ; x-\frac{y}{2}=\frac{1}{2}$
✓
Answer
$2 x +3 y =2 $
$x -\frac{ y }{2}=\frac{1}{2} \Rightarrow 2 x - y =1$
$D =\left[\begin{array}{cc}
2 & 3 \\
2 & -1
\end{array}\right]=(2 \times-1)-(3 \times 2)=-2-6=-8 $
$D _{ x }=\left[\begin{array}{cc}
2 & 3 \\
1 & -1
\end{array}\right]=(2 \times-1)-(3 \times 1)=-2-3=-5 $
$D_y=\left[\begin{array}{ll}
2 & 2 \\
2 & 1
\end{array}\right]=(2 \times 1)-(2 \times 2)=2-4=(-2) $
$x =\frac{ D _{ x }}{ D }=\frac{-5}{-8}=\frac{5}{8} y =\frac{ D _{ y }}{ D }=\frac{-2}{-8}=\frac{1}{4}$
$\therefore( x , y )=\left(\frac{5}{8}, \frac{1}{4}\right)$ is solution.
$x -\frac{ y }{2}=\frac{1}{2} \Rightarrow 2 x - y =1$
$D =\left[\begin{array}{cc}
2 & 3 \\
2 & -1
\end{array}\right]=(2 \times-1)-(3 \times 2)=-2-6=-8 $
$D _{ x }=\left[\begin{array}{cc}
2 & 3 \\
1 & -1
\end{array}\right]=(2 \times-1)-(3 \times 1)=-2-3=-5 $
$D_y=\left[\begin{array}{ll}
2 & 2 \\
2 & 1
\end{array}\right]=(2 \times 1)-(2 \times 2)=2-4=(-2) $
$x =\frac{ D _{ x }}{ D }=\frac{-5}{-8}=\frac{5}{8} y =\frac{ D _{ y }}{ D }=\frac{-2}{-8}=\frac{1}{4}$
$\therefore( x , y )=\left(\frac{5}{8}, \frac{1}{4}\right)$ is solution.
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