Pair of Linear Equations in Two Variables Test - 37

$${\textbf{Step -1: Writing mathematical equations according to conditions given in question}}$$

                   $${\text{Let the price of ticket for station A be Rs}}{\text{. x and for station B be Rs}}{\text{. y,}}$$

                  $${\text{According to question, }}$$

                  $${\text{2x  +  3y  =  77 }} \to {\text{ Equation 1}}$$

                  $${\text{3x  +  5y  =  124 }} \to {\text{ Equation 2}}$$

$${\textbf{Step -2: Calculating fares for station A and station B .}}$$

                  $${\text{Solving Equation 1 and Equation 2,}}$$

                  $${\text{Multiplying Equation 1 by 3,}}$$

                  $${\text{6x  +  9y  =  231 }} \to {\text{ Equation 3}}$$

                  $${\text{Multiplying Equation 2 by 2,}}$$

                  $${\text{6x  +  10y  =  248 }} \to {\text{ Equation 4}}$$

                  $${\text{Subtracting Equation 3 from Equation 4,}}$$

                  $$ \Rightarrow {\text{ y  =  248  -  231}}$$

                  $$ \Rightarrow {\text{ y  =  17}}$$

                  $${\text{Substituting value of y in Equation 1,}}$$

                  $$ \Rightarrow {\text{ 2x  +  3(17)  =  77}}$$

                  $$ \Rightarrow {\text{ 2x  +  51  =  77}}$$

                  $$ \Rightarrow {\text{ 2x  =  26}}$$

                  $$ \Rightarrow {\text{ x  =  13}}$$

$${\textbf{Thus, the fare to station A is Rs}}{\textbf{. 13 and to station B is Rs}}{\textbf{.17 .}}$$

The equation are
$$ 3x+5y-3=0$$     ....(1)
$$ 6x+ky-8=0$$     .....(2)
Here, $$ a_{1}=3,b_{1}=5,c_{1}=-3$$
and $$ a_{2}=6,b_{2}=k,c_{2}=-8$$
The   given   system   has   no   solution   if
$$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}$$
$$\Rightarrow \displaystyle \frac{3}{6}=\frac{5}{k}\neq \frac{-3}{-8}$$
$$\Rightarrow 3k=30\Rightarrow k=10$$  

Given equations are
$$3x - 5y + 1= 0$$
$$2x - y + 3= 0 $$
Since, the system has unique solution at $$x = x_1$$ and $$y=y_1$$
That means the lines intersect at only one point $$(x_{1},y_{1})$$
By putting $$x =x_{1}$$ and $$y = y_{1}$$ in given equations
$$ 3x_{1} -5 y_{1} + 1 = 0 $$     .....(1)
$$2 x_{1} -y_{1} + 3 = 0 $$        .....(2)
From equation(2), we have 
$$ y_{1} = 2 x_{1} + 3$$      .....(3)
Putting the value of $$y_{1}$$ in  equation(1)
$$ 3x_{1} - 5\left ( 2x_{1} + 3 \right ) + 1 = 0 $$
$$ \Rightarrow  3x_{1} - 10x_{1} - 15 + 1 = 0$$
$$ \Rightarrow  -7x_{1} - 14 = 0$$
$$ \Rightarrow  -7x_{1} = 14$$
$$\Rightarrow  x_{1}  = -2$$
Put this value in equation (3), we get
$$   y_{1} =2\left ( -2 \right ) + 3$$
$$   y_{1} = -1$$

Let father's age $$=x$$, son's age $$=y$$    

 As per question,

$$x+y=65\Rightarrow y=65-x $$         .....(1)                                                

$$2\left( x-y \right )=50\Rightarrow2x-2y=50 $$       .......(2)
$$\Rightarrow 2x-2\left (65-x  \right )=50$$    $$[$$ substituting value of $$y$$ from equation (1) in equation (2) $$]$$

$$\Rightarrow2x-130+2x=50$$
$$\Rightarrow x=45 $$

So father's age $$=45 \text{ years}$$

The pair of linear equations
$$ 13x+ky-k=0$$
$$ 39x+6y-(k+4)=0$$

Here, $$a_{1}=13,b_{1}=k,c_{1}=-k$$
$$ a_{2}=39,b_{2}=6,c_{2}=-(k+4)$$

The given equations are:
$$x+2y+5= 0 $$
$$-3x-6y+1 = 0$$
From the given equations we have:
$$ \displaystyle\frac{a_{1}}{a_{2}} = \frac{1}{-3}$$
$$ \displaystyle \frac{b_{1}}{b_{2}} = \frac{2}{-6} = \frac{1}{-3} $$
$$\displaystyle\frac{c_{1}}{c_{2}} =\displaystyle \frac{5}{1}$$
$$\Rightarrow \displaystyle\frac{a_{1}}{a_{2}} = \displaystyle \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$
Hence the given pair of equations have no solution.

Given equations are:
$$x+2y = 5$$
$$3x+12y=10$$
From the equations we have:
$$\Rightarrow a_{1}= 1 , b_{1}= 2 $$
$$\Rightarrow a_{2}= 3 , b_{2}= 12$$
$$ \displaystyle \frac{a_{1}}{a_{2}} =  \displaystyle \frac{1}{3}$$
$$ \displaystyle \frac{b_{1}}{b_{2}} =  \displaystyle \frac{2}{12}=\frac{1}{6}$$

$$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}} \neq    \frac{b_{1}}{b_{2}}$$
Hence system of equations has a unique solution

Let the present age of man be $$x$$ years and the present age of son be $$y$$ years

Given equations are:
$$ x + 2y = 5$$      ....(1)
$$7x+3y = 13$$    .....(2)
Multiplying equation (1) by 7
$$7x+14y = 35 $$    .....(3)
Subtracting equation (2) from equation (3), we have
$$ 11y = 22$$
$$ y = 2 $$
Now putting $$y$$ value in equation (1) 
$$\Rightarrow x + 2\left ( 2 \right ) = 5$$
$$\Rightarrow x = 5 - 4$$
$$\Rightarrow x = 1$$
$$ x = 1 , y = 2 $$

The equation are
$$2x+ky-k=0$$
$$4x+2y-(k+1)=0$$
Here, $$ a_{1}=2,b_{1}=k,c_{1}=-k$$
and $$ a _{2}=4,b_{2}=2,c_{2}=-(k+1)$$
For the system to have infinite solutions,
$$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$
$$\Rightarrow \displaystyle \frac{2}{4}=\frac{k}{2}=\frac{-k}{-(k+1)}$$

Taking, 
$$\displaystyle \frac{2}{4}=\frac{k}{2}$$

$$\Rightarrow 4k=4$$

$$\Rightarrow k=1$$   

Taking,
$$\displaystyle\frac{k}{2}=\frac{-k}{-(k+1)}$$

$$\Rightarrow k+1=2 \Rightarrow k=1$$

So, $$k=1$$ is the answer