Pair of Linear Equations in Two Variables Test - 45

$$6x+3y=6xy$$      .........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.

$${\textbf{Step -1: Framing equations}}$$

               $$\text{Without loss of generality, suppose,}$$ $$ x$$ $${\text{and}}$$ $$ y$$ $${\text{be two numbers and }} x > y.$$

               $${\text{Given that, sum of two numbers is 35}}{\text{.}}$$

               $$ \Rightarrow x + y = 35$$                    $$\ldots \left( 1 \right)$$

               $${\text{Difference between these two numbers is 13}}{\text{.}}$$

               $$ \Rightarrow x - y = 13$$                    $$ \ldots \left( 2 \right)      $$

$${\textbf{Step -2: Adding equation }}\left( \mathbf 1 \right)$$ $${\textbf{and equation }}\left(\mathbf 2 \right)\textbf .$$

               $$ \Rightarrow x + y + \left( {x - y} \right) = 35 + 13$$

               $$ \Rightarrow 2x = 48$$

               $$ \Rightarrow x = 24$$

               $${\text{Substituting the value of}}$$ $$x$$ $${\text{in equation }}\left( 1 \right)$$ $${\text{we get,}}$$

               $$ \Rightarrow 24 + y = 35$$

               $$ \Rightarrow y = 35 - 24$$

               $$ \Rightarrow y = 11$$

               $${\text{Hence, these two numbers are 24 and 11}}{\text{.}}$$

$${\textbf{ Hence, option (C) is correct answer.}}$$