Explanation
Given:
$$\dfrac{10}{x+y}+\dfrac{2}{x-y}=4$$
$$\dfrac{15}{x+y}+\dfrac{5}{x-y}=-2$$
let $$\dfrac{1}{x+y}=a \ and \ \dfrac{1}{x-y}=b$$
$$10a+2b=4.......eq1$$$$15a-5b=-2........eq2$$
Multiply eq1 with 3 and eq2 with 2 we get$$30a+6b=12......eq3$$$$30a-10b=-4......eq4$$
subtracting eq3 and eq4 we get$$16b=16$$$$b=1$$
Substituting $$b=1$$ in eq1 we get$$10a+2\times 1=4$$$$10a=2$$$$a=\dfrac{1}{5}$$
As $$\dfrac{1}{x+y}=a=\dfrac{1}{5}$$
$$x+y=5............eq5$$
and $$\dfrac{1}{x-y}=b=1$$
$$x-y=1...........eq6$$
subtracting eq5 and eq6 we get$$2x=6$$$$x=3$$
putting $$x=3$$ in eq 5 we get$$3+y=5$$$$y=2$$
hence $$x=3$$ and $$y=2$$
$$6x+3y=6xy$$ .........Given
Multiply by $$\dfrac{1}{xy}$$ we get
$$\dfrac{6x}{xy}+\dfrac{3y}{xy}=\dfrac{6xy}{xy}$$
$$\dfrac{6}{y}+\dfrac{3}{x}=6............eq1$$
$$2x+4y=5xy$$ ...........Given
Multiply with $$\dfrac{1}{xy}$$we get
$$\dfrac{2x}{xy}+\dfrac{4y}{xy}=\dfrac{5xy}{xy}$$
$$\dfrac{2}{y}+\dfrac{4}{x}=5........eq2$$
Let $$\dfrac{1}{x}=a \ and \ \dfrac{1}{y}=b$$
$$a+2b=2.......eq3$$$$4a+2b=5.....eq4$$
Subtracting eq3 and eq4 we get$$-3a=-3$$$$a=1$$
Substituting a=1 in eq3 we get $$1+2b=2$$$$2b=1$$$$b=\dfrac{1}{2}$$
Thus if $$a=\dfrac{1}{x}=1\Rightarrow x=1$$
If $$b=\dfrac{1}{y}=\dfrac{1}{2}\Rightarrow y=2$$
Hence $$x=1$$ and $$y=2$$
The given pair of equations are:$$3x+y = 1...(1)$$$$\left ( 2k-1 \right )x +\left ( k-1 \right )y = 2k+1..(2)$$Now rearranging eq1 and eq2 will get$$3x+y-1=0...(3)$$$$\left ( 2k-1 \right )x +\left ( k-1 \right )y -\left( 2k+1 \right )= 0..(4)$$Now compare with$$a_{1} = 3, b_{1} = 1, c_{1} = -1$$$$a_{2} = 2k-1, b_{2} = k-1, c_{2} = -\left( 2k+1 \right )$$Now we get$$ \dfrac{a_{1}}{a_{2}} =\dfrac{3}{2k-1} , \dfrac{b_{1}}{b_{2}} = \dfrac{1}{k-1},\dfrac{c_{1}}{c_{2}} = \dfrac{-1}{-\left( 2k+1 \right )} $$Now will take$$ \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} $$$$\Rightarrow \dfrac{3}{2k-1} = \dfrac{1}{k-1}$$$$\Rightarrow 3k-3= 2k-1$$$$\Rightarrow 3k-2k = -1+3$$$$\Rightarrow k = 2$$Hence $$ k =2 $$ is the value.
$${\textbf{Step -1: Framing equations}}$$
$$\text{Without loss of generality, suppose,}$$ $$ x$$ $${\text{and}}$$ $$ y$$ $${\text{be two numbers and }} x > y.$$
$${\text{Given that, sum of two numbers is 35}}{\text{.}}$$
$$ \Rightarrow x + y = 35$$ $$\ldots \left( 1 \right)$$
$${\text{Difference between these two numbers is 13}}{\text{.}}$$
$$ \Rightarrow x - y = 13$$ $$ \ldots \left( 2 \right) $$
$${\textbf{Step -2: Adding equation }}\left( \mathbf 1 \right)$$ $${\textbf{and equation }}\left(\mathbf 2 \right)\textbf .$$
$$ \Rightarrow x + y + \left( {x - y} \right) = 35 + 13$$
$$ \Rightarrow 2x = 48$$
$$ \Rightarrow x = 24$$
$${\text{Substituting the value of}}$$ $$x$$ $${\text{in equation }}\left( 1 \right)$$ $${\text{we get,}}$$
$$ \Rightarrow 24 + y = 35$$
$$ \Rightarrow y = 35 - 24$$
$$ \Rightarrow y = 11$$
$${\text{Hence, these two numbers are 24 and 11}}{\text{.}}$$
$${\textbf{ Hence, option (C) is correct answer.}}$$
$$\dfrac {1}{(3x+y)}+\dfrac {1}{(3x-y)}=\dfrac {3}{4}, \dfrac {1}{2(3x+y)}+\dfrac {1}{2(3x-y)}=\dfrac {-1}{8}$$
$$let \dfrac{1}{3x+y}=a and \dfrac{1}{3x-y}=b$$
Then$$a+b=\dfrac{3}{4}\Rightarrow 4a+4b=3........eq1$$
$$\dfrac{a}{2}-\dfrac{b}{2}=\dfrac{-1}{8}$$
$$\dfrac{a-b}{2}=\dfrac{-1}{8}$$
$$a-b=\dfrac{-2}{8}\Rightarrow a-b=\dfrac{-1}{4}$$
$$4a-4b=-1..............eq2$$
Adding eq1 and eq2 we get
$$8a=2$$
$$a=\dfrac{2}{8}\Rightarrow \dfrac{1}{4}$$
Substituting the value of a in eq1 we get
$$4\times \dfrac{1}{4}+4b=3$$
$$4b=2$$
$$b=\dfrac{2}{4}\Rightarrow \dfrac{1}{2}$$
Then $$\dfrac{1}{3x+y}=\dfrac{1}{4}$$
$$3x+y=4................eq3$$
$$\dfrac{1}{3x-y}=\dfrac{1}{2}$$
$$3x-y=2...............eq4$$
Adding eq3 and eq4 we get$$6x=6$$$$x=1$$
Substituting $$x=1$$ in eq3 we get$$3\times 1 +y=4$$$$y=1$$
Thus $$x=1$$ and $$y=1$$
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