CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 8 - MCQExams.com

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $$=45 \text{ years}$$

Explanation:

Graphically x = 0 is a straight line which is the y - axis and y = - 7 is a straight line parallel to x - axis, 7 units below the origin therefore lines x = 0 and y = - 7 intersect at one point only.

Since there is only one point of intersection, there is one solution.

Thus, option A is correct.

The equations are

$$\ x-3y-4=0$$
$$\Rightarrow a_{1}=1,b_{1}=-3,c_{1}=-4$$
$$\ 3x+2y-1=0$$
$$\Rightarrow a_{2}=3,b_{2}=2,c_{2}=-1$$

 comparing  $$\frac{a_{1}}{a_{2}} and \frac{b_{1}}{b_{2}}$$

We have $$\ \frac{1}{3}\neq\frac{-3}{2}$$

$$\Rightarrow  \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$$

$$\Rightarrow   System  \  has \   unique \   solution$$

If the system has unique solution then it is said to be consistent system 

The equation are
$$\  x-2y-3=0$$
$$\Rightarrow a_{1}=1,b_{1}=-2,c_{1}=-3$$
$$\ 3x-6y-1=0$$
$$\Rightarrow a_{2}=3,b_{2}=-6,c_{2}=-1$$

comparing  $$\rightarrow      \dfrac{a_{1}}{a_{2}},\dfrac{b_{1}}{b_{2}},\dfrac{c_{1}}{c_{2}}$$

We have $$\ \dfrac{1}{3}=\dfrac{2}{6}\neq\dfrac{-3}{-1}$$
$$\Rightarrow \dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}$$

from the  above relation we can say that this $$\ System  \  has \   no \   solution$$

If the system has no solution, then it is said to be the inconsistent system.
 

The equations are
$$\ x-ky-2=0 $$
$$\Rightarrow a_{1}=1,b_{1}=-k,c_{1}=-2$$
$$\ 3x+2y+5=0$$
$$\Rightarrow a_{2}=3,b_{2}=2,c_{2}=5$$
The   given   system   of   equations   has   a   unique   solution
$$\Rightarrow \dfrac{a_{1}}{a_{2}}\neq \dfrac{b_{1}}{b_{2}}$$
$$\Rightarrow  \dfrac{1}{3}\neq -\dfrac{k}{2}$$
$$\Rightarrow$$$$k\neq-\dfrac{2}{3}$$