Answer
Verified
399k+ views
Hint: The given equation $\dfrac{{xy}}{{x + y}} = a$ can be simplified as $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{a}$ . Let us assume the values $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}$ as $p,q,r$ respectively and substitute these values in the given relations to form linear equations in $p,q,r$. The linear equation can be solved to find the value of $p$ . The value of $x$ will be $\dfrac{1}{p}$ .
Complete step-by-step answer:
The given relations in the question are $\dfrac{{xy}}{{x + y}} = a$ , $\dfrac{{yz}}{{y + z}} = c$ and \[\dfrac{{xy}}{{x + z}} = b\].
The equation $\dfrac{{xy}}{{x + y}} = a$ can be simplified by dividing by $xy$ in the numerator and the denominator in the L.H.S. as
\[\dfrac{{\dfrac{{xy}}{{xy}}}}{{\dfrac{x}{{xy}} + \dfrac{y}{{xy}}}} = a\]
The above equation can be simplified as
$
\dfrac{1}{{\dfrac{1}{x} + \dfrac{1}{y}}} = a \\
\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{a} \\
$
Similarly, the equations $\dfrac{{yz}}{{y + z}} = c$ and \[\dfrac{{xy}}{{x + z}} = b\] can be written as
$\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{c}$
\[\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{b}\]
Let us assume the values $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}$ as $p,q,r$ respectively .
Substituting the values $p,q,r$ for $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}$ in the above equations, we get
$p + q = \dfrac{1}{a}$, equation (1)
$q + r = \dfrac{1}{c}$, equation (2)
$p + r = \dfrac{1}{b}$, equation (3)
Adding equations 1 and 3, we get
$p + q + p + r = \dfrac{1}{a} + \dfrac{1}{b}$
Substituting the value of $q + r$ from the equation 2 in the equation $p + q + p + r = \dfrac{1}{a} + \dfrac{1}{b}$, we get
$p + p + \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}$
The above equation can be simplified as
$
2p = \dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c} \\
\Rightarrow 2p = \dfrac{{bc + ac - ab}}{{abc}} \\
\Rightarrow p = \dfrac{{bc + ac - ab}}{{2abc}} \\
$
The value of $x$ can be found by the relation $\dfrac{1}{p} = x$
$
x = \dfrac{1}{{\dfrac{{bc + ac - ab}}{{2abc}}}} \\
\Rightarrow x = \dfrac{{2abc}}{{bc + ac - ab}} \\
$
Thus option D is the correct answer.
Note: Mistakes must be avoided while dealing with fractions. Careful analysis of the given equation is necessary to formulate the way of solving the questions. The values $x,y$ and $z$ are also non zero values as $a,b$ and $c$ are non-zero values.
Complete step-by-step answer:
The given relations in the question are $\dfrac{{xy}}{{x + y}} = a$ , $\dfrac{{yz}}{{y + z}} = c$ and \[\dfrac{{xy}}{{x + z}} = b\].
The equation $\dfrac{{xy}}{{x + y}} = a$ can be simplified by dividing by $xy$ in the numerator and the denominator in the L.H.S. as
\[\dfrac{{\dfrac{{xy}}{{xy}}}}{{\dfrac{x}{{xy}} + \dfrac{y}{{xy}}}} = a\]
The above equation can be simplified as
$
\dfrac{1}{{\dfrac{1}{x} + \dfrac{1}{y}}} = a \\
\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{a} \\
$
Similarly, the equations $\dfrac{{yz}}{{y + z}} = c$ and \[\dfrac{{xy}}{{x + z}} = b\] can be written as
$\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{c}$
\[\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{b}\]
Let us assume the values $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}$ as $p,q,r$ respectively .
Substituting the values $p,q,r$ for $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}$ in the above equations, we get
$p + q = \dfrac{1}{a}$, equation (1)
$q + r = \dfrac{1}{c}$, equation (2)
$p + r = \dfrac{1}{b}$, equation (3)
Adding equations 1 and 3, we get
$p + q + p + r = \dfrac{1}{a} + \dfrac{1}{b}$
Substituting the value of $q + r$ from the equation 2 in the equation $p + q + p + r = \dfrac{1}{a} + \dfrac{1}{b}$, we get
$p + p + \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}$
The above equation can be simplified as
$
2p = \dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c} \\
\Rightarrow 2p = \dfrac{{bc + ac - ab}}{{abc}} \\
\Rightarrow p = \dfrac{{bc + ac - ab}}{{2abc}} \\
$
The value of $x$ can be found by the relation $\dfrac{1}{p} = x$
$
x = \dfrac{1}{{\dfrac{{bc + ac - ab}}{{2abc}}}} \\
\Rightarrow x = \dfrac{{2abc}}{{bc + ac - ab}} \\
$
Thus option D is the correct answer.
Note: Mistakes must be avoided while dealing with fractions. Careful analysis of the given equation is necessary to formulate the way of solving the questions. The values $x,y$ and $z$ are also non zero values as $a,b$ and $c$ are non-zero values.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Two charges are placed at a certain distance apart class 12 physics CBSE
Difference Between Plant Cell and Animal Cell
What organs are located on the left side of your body class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The planet nearest to earth is A Mercury B Venus C class 6 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE