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Question
3x + ay = 4
2x + ay = 2, a ≠ 0
Solution
\[\text{ Given} : 3x + ay = 4\]
\[ 2x + ay = 2 \]
Using Cramer's rule, we get
\[ D = \begin{vmatrix} 3 & a\\2 & a \end{vmatrix} = 3a - 2a = a\]
\[ D_1 = \begin{vmatrix} 4 & a\\2 & a \end{vmatrix} = 4a - 2a = 2a\]
\[ D_2 = \begin{vmatrix} 3 & 4 \\2 & 2 \end{vmatrix} = 6 - 8 = - 2\]
Now,
\[x = \frac{D_1}{D} = \frac{2a}{a} = 2\]
\[y = \frac{D_2}{D} = \frac{- 2}{a} = - \frac{2}{a}\]
\[ \therefore x = 2\text{ and }y = - \frac{2}{a}\]
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