Answer
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Hint: To solve a pair of equations that are reducible to linear form, find the expressions that repeat in both the equations.
Give them a simpler form, say $ x $ and $ y $ . Solve the new pair of linear equations for the new variables.
Complete step-by-step answer:
Given equations are,
$ \Rightarrow $ $ \dfrac{1}{x} + \dfrac{1}{y} = 12,\dfrac{3}{x} - \dfrac{2}{y} = 1 $
Substituting $ \dfrac{1}{x} = a $ and $ \dfrac{1}{y} = b $ in both above equations.
$ \Rightarrow a + b = 12 $ _ _ _ _ _ _ _ _ _ _ $ \left( 1 \right) $
$ \Rightarrow 3a - 2b = 1 $ _ _ _ _ _ _ _ _ _ _ $ \left( 2 \right) $
Multiply equation $ \left( 1 \right) $ by $ 2 $ and solving with $ \left( 2 \right) $ ,
$ \Rightarrow 2a + 2b = 24 $ _ _ _ _ _ _ _ _ _ _ _ $ \left( 3 \right) $
$ \Rightarrow 3a - 2b = 1 $
________________
$ \Rightarrow 5a = 25 $
$ \Rightarrow a = \dfrac{{25}}{5} = 5 $
Substituting $ a = 5 $ in equation $ \left( 1 \right) $ ,
$
\Rightarrow a + b = 12 \\
\Rightarrow 5 + b = 12 \\
\Rightarrow b = 12 - 5 = 7 \;
$
Therefore,
$
\Rightarrow \dfrac{1}{x} = a \\
\Rightarrow x = \dfrac{1}{5}\;
$
$
\Rightarrow \dfrac{1}{y} = b \\
\Rightarrow y = \dfrac{1}{7} \;
$
So, the correct answer is “$x = \dfrac{1}{5}$ and $y = \dfrac{1}{7}$”.
Note: $ \Rightarrow $ Equations in which the powers of all the variables involved are one are called linear equations. The degree of a linear equation is always one.
$ \Rightarrow $ Identify unknown quantities and denote them by variables.
$ \Rightarrow $ Represent the relationships between quantities in a mathematical form, replacing the unknowns with variables.
Give them a simpler form, say $ x $ and $ y $ . Solve the new pair of linear equations for the new variables.
Complete step-by-step answer:
Given equations are,
$ \Rightarrow $ $ \dfrac{1}{x} + \dfrac{1}{y} = 12,\dfrac{3}{x} - \dfrac{2}{y} = 1 $
Substituting $ \dfrac{1}{x} = a $ and $ \dfrac{1}{y} = b $ in both above equations.
$ \Rightarrow a + b = 12 $ _ _ _ _ _ _ _ _ _ _ $ \left( 1 \right) $
$ \Rightarrow 3a - 2b = 1 $ _ _ _ _ _ _ _ _ _ _ $ \left( 2 \right) $
Multiply equation $ \left( 1 \right) $ by $ 2 $ and solving with $ \left( 2 \right) $ ,
$ \Rightarrow 2a + 2b = 24 $ _ _ _ _ _ _ _ _ _ _ _ $ \left( 3 \right) $
$ \Rightarrow 3a - 2b = 1 $
________________
$ \Rightarrow 5a = 25 $
$ \Rightarrow a = \dfrac{{25}}{5} = 5 $
Substituting $ a = 5 $ in equation $ \left( 1 \right) $ ,
$
\Rightarrow a + b = 12 \\
\Rightarrow 5 + b = 12 \\
\Rightarrow b = 12 - 5 = 7 \;
$
Therefore,
$
\Rightarrow \dfrac{1}{x} = a \\
\Rightarrow x = \dfrac{1}{5}\;
$
$
\Rightarrow \dfrac{1}{y} = b \\
\Rightarrow y = \dfrac{1}{7} \;
$
So, the correct answer is “$x = \dfrac{1}{5}$ and $y = \dfrac{1}{7}$”.
Note: $ \Rightarrow $ Equations in which the powers of all the variables involved are one are called linear equations. The degree of a linear equation is always one.
$ \Rightarrow $ Identify unknown quantities and denote them by variables.
$ \Rightarrow $ Represent the relationships between quantities in a mathematical form, replacing the unknowns with variables.
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