Question
Solve the following simultaneous equations using Cramer’s rule.
4m + 6n = 54; 3m + 2n = 28
4m + 6n = 54; 3m + 2n = 28
✓
Answer
$4 m+6 n =54 $
$3 m+2 n =28$
$D =\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}=(4 \times 2)-(6 \times 3)=8-18=10$
$D _{ x }=\begin{bmatrix} 54 & 6 \\ 28 & 2\end{bmatrix}=(54 \times 2)-(6 \times 28)=108-168=60 $
$D _{ y }=\begin{bmatrix} 4 & 54 \\ 3 & 28\end{bmatrix}=(4 \times 28)-(54 \times 3)=112-162=50 $
$x =\frac{ D _{ x }}{ D }=\frac{60}{10}=6 y =\frac{ D _{ y }}{ D }=\frac{50}{10}=5$
$\therefore( x , y )=(6,5)$ is solution.
$3 m+2 n =28$
$D =\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}=(4 \times 2)-(6 \times 3)=8-18=10$
$D _{ x }=\begin{bmatrix} 54 & 6 \\ 28 & 2\end{bmatrix}=(54 \times 2)-(6 \times 28)=108-168=60 $
$D _{ y }=\begin{bmatrix} 4 & 54 \\ 3 & 28\end{bmatrix}=(4 \times 28)-(54 \times 3)=112-162=50 $
$x =\frac{ D _{ x }}{ D }=\frac{60}{10}=6 y =\frac{ D _{ y }}{ D }=\frac{50}{10}=5$
$\therefore( x , y )=(6,5)$ is solution.
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