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Question
3x + y = 19
3x − y = 23
Solution
\[\text{ Given }: \hspace{0.167em} 3x + y = 19\]
\[ 3x - y = 23\]
Using Cramer's Rule, we get
\[D = \begin{vmatrix} 3 & 1 \\3 & - 1 \end{vmatrix} = - 3 - 3 = - 6\]
\[ D_1 = \begin{vmatrix} 19 & 1\\23 & - 1 \end{vmatrix} = - 19 - 23 = - 42\]
\[ D_2 = \begin{vmatrix} 3 & 19 \\3 & 23 \end{vmatrix} = \left( 3 \times 23 \right) - \left( 3 \times 19 \right) = 3 \times 4 = 12\]
Now,
\[x = \frac{D_1}{D} = \frac{- 42}{- 6} = 7\]
\[y = \frac{D_2}{D} = \frac{12}{- 6} = - 2\]
\[ \therefore x = 7 \text{ and }y = - 2\]
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