Question
Solve the following simultaneous equations using Cramer’s rule.
3x – 4y = 10; 4x + 3y = 5
3x – 4y = 10; 4x + 3y = 5
✓
Answer
$\therefore(x, y)=(2,-1)$ is the solution
$3 x-4 y=10 $
$4 x+3 y=5 $
$D=\left|\begin{array}{cc}3 & -4 \\4 & 3 \end{array}\right|=(3 \times 3)-(-4 \times 4)=9+16=25$
$D_x=\left[\begin{array}{cc} 10 & -4 \\ 5 & 3 \end{array}\right]=(10 \times 3)-(-4 \times 5)=30+20=50$
$D_y=\left[\begin{array}{cc}3 & 10 \\ 4 & 5\end{array}\right]=(3 \times 5)-(10 \times 4)=15-40=-25$
$x =\frac{ D _{ x }}{ D }=\frac{50}{25}=2 y =\frac{ D _{ y }}{ D }=-\frac{25}{25}=-1$
$3 x-4 y=10 $
$4 x+3 y=5 $
$D=\left|\begin{array}{cc}3 & -4 \\4 & 3 \end{array}\right|=(3 \times 3)-(-4 \times 4)=9+16=25$
$D_x=\left[\begin{array}{cc} 10 & -4 \\ 5 & 3 \end{array}\right]=(10 \times 3)-(-4 \times 5)=30+20=50$
$D_y=\left[\begin{array}{cc}3 & 10 \\ 4 & 5\end{array}\right]=(3 \times 5)-(10 \times 4)=15-40=-25$
$x =\frac{ D _{ x }}{ D }=\frac{50}{25}=2 y =\frac{ D _{ y }}{ D }=-\frac{25}{25}=-1$
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