Solve: (i) $4 x + 5 y - 62 = 0$
(ii) $3 x + 2 y - 36 = 0$

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Solution

$4 x + 5 y - 62 = 0 \Rightarrow 4 x + 5 y = 62 ⟶ (1) 3 x + 2 y - 36 = 0 \Rightarrow 3 x + 2 y = 36 ⟶ (2) ∵ a_{1} = 4; b_{1} = 5; c_{1} = 62 a_{2} = 3; b_{2} = 2; c_{2} = 36 ∴ x = \frac{∣ ∣ ∣ \begin{matrix} c_{1} & b_{1} c_{2} & b_{2} \end{matrix} ∣ ∣ ∣}{∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{2} & b_{2} \end{matrix} ∣ ∣ ∣} = \frac{∣ ∣ ∣ \begin{matrix} 62 & 5 36 & 2 \end{matrix} ∣ ∣ ∣}{∣ ∣ ∣ \begin{matrix} 4 & 5 3 & 2 \end{matrix} ∣ ∣ ∣} = 8 y = \frac{∣ ∣ ∣ \begin{matrix} a_{1} & c_{1} a_{2} & c_{2} \end{matrix} ∣ ∣ ∣}{∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{2} & b_{2} \end{matrix} ∣ ∣ ∣} = \frac{∣ ∣ ∣ \begin{matrix} 4 & 62 3 & 36 \end{matrix} ∣ ∣ ∣}{∣ ∣ ∣ \begin{matrix} 4 & 5 3 & 2 \end{matrix} ∣ ∣ ∣} = 6$

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