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प्रश्न
Solve the following simultaneous equation.
\[\frac{2}{x} + \frac{2}{3y} = \frac{1}{6} ; \frac{3}{x} + \frac{2}{y} = 0\]
उत्तर
\[\frac{2}{x} + \frac{2}{3y} = \frac{1}{6} ; \frac{3}{x} + \frac{2}{y} = 0\]
Let \[\frac{1}{x} = u\text{ and }\frac{1}{y} = v\]
\[2u + \frac{2}{3}v = \frac{1}{6} \]
\[12u + 4v = 1 . . . . . \left( I \right)\]
\[3u + 2v = 0 . . . . . \left( II \right)\]
Multiply (II) with 2
\[6u + 4v = 0 . . . . . \left( III \right)\]
\[\left( I \right) - \left( III \right)\]
\[6u = 1\]
\[ \Rightarrow u = \frac{1}{6}\]
Putting the value of u in II.
\[3 \times \frac{1}{6} + 2v = 0\]
\[ \Rightarrow \frac{1}{2} + 2v = 0\]
\[ \Rightarrow v = \frac{- 1}{4}\]
\[\frac{1}{x} = u\]
\[ \Rightarrow x = 6\]
\[\frac{1}{y} = v\]
\[ \Rightarrow y = - 4\]
\[\left( x, y \right) = \left( 6, - 4 \right)\]
