MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 8

If the system of equations

$3x + y =1$

$\left(2k - 1\right)x + \left(k - 1\right)y = 2k + 1$ is inconsistent, then

$k =$

0%
$1$
0%
$-1$
0%
$-2$
0%
$2$

Explanation

The system of equations is inconsistent if

$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$

$\therefore$

$\displaystyle\frac{3}{2k - 1} = \displaystyle\frac{1}{k - 1} \neq \displaystyle\frac{1}{2k + 1}$

$\Rightarrow 3k - 3 = 2k - 1$

$\Rightarrow k = 2$

Solve:

$\dfrac {x+2y+1}{2x-y+1}=2\,;\,\, \dfrac {3x-y+1}{x-y+3}=5$

0%
$x = 13, y = 10$
0%
$x = 17, y = 5$
0%
$x = 9, y = -5$
0%
None of these

Explanation

$\Rightarrow \dfrac{x+2y+1}{2x-y+1}=2$

$\Rightarrow 4x-2y+2=x+2y+1$

$\Rightarrow 3x-4y=-1........eq1$

$\Rightarrow \dfrac{3x-y+1}{x-y+3}=5$

$\Rightarrow 5x-5y+15=3x-y+1$

$\Rightarrow 2x-4y=-14........eq2$
Substract eq1 and eq2

$\Rightarrow x=13$
put x=13 in eq1

$\Rightarrow 3\times13-4y=-1$

$\Rightarrow -4y=-40 \Rightarrow y=10$
Hence x=13 and y=10

The pair of linear equations

$2x + ky = k, 4x + 2y = k + 1$ has infinitely many solutions if

0%
$k=1$
0%
$k\neq 1$
0%
$k=2$
0%
$k=4$

Explanation

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

The value of k for which the system of linear equations:

$kx+4y=k-4$ ,

$16x+ky=k$ , has infinitely many solutions, is

0%
$0$
0%
$8$
0%
$-8$
0%
$4$

Explanation

$kx+4y=k-4$

$16x+ky=k$

${a}_{1}=k,{b}_{1}=4,{c}_{1}=-(k-4)$

${a}_{2}=16,{b}_{2}=k,{c}_{2}=-k$

Here condition is

$\displaystyle\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$

$\displaystyle\frac{k}{16}=\frac{4}{k}=\frac{k-4}{k}$

$\displaystyle\frac{k}{16}=\frac{4}{k}\Longrightarrow {k}^{2}=64\Longrightarrow k=\pm 8$

Also,

$\frac{4}{k}=\frac{k-4}{k}\Longrightarrow 4k={k}^{2}-4k$

$\Longrightarrow {k}^{2}-8k=0\Longrightarrow k(k-8)=0$

$k=0$ or

$k=8$ but

$k=0$ is not possible.
Therefore,

$k$ is

$8$ is correct value for infinitely many solutions.

The pair of linear equations

$x+y=3, 2x+5y=12$ has a unique solution

$x=x_1, y=y_1$ then value of

$x_1$ is?

0%
$1$
0%
$2$
0%
$-1$
0%
$-2$

Explanation

Given equations are

$x + y = 3$

$2x + 5y = 12$
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 equations, we get

$x_{1} + y_{1} = 3$ ....

$(1)$

$2 x_{1} +5y_{1} = 12$ ....

$(2)$
From equation

$(1)$ , we have

$y_1 = 3 - x_{1}$
Putting the value of

$y_{1}$ in equation

$(2)$ ,

$\Rightarrow 2x_{1} + 5\left ( 3 - x_{1} \right ) = 12$

$\Rightarrow 2x_{1} + 15 - 5x_{1} = 12$

$\Rightarrow 2x_{1} - 5x_{1} = 12 - 15$

$\Rightarrow - 3x_{1} = -3$

$\Rightarrow x_{1} = \displaystyle \frac{-3}{-3}$

$\Rightarrow x_{1} = 1$

From Delhi station, if we buy 2 tickets for station A and 3 tickets for station B, the total cost is Rs.But if we buy 3 tickets for station A and 5 tickets for station B, the total cost is Rs.What are the fares from Delhi to station A and to station B?

0%
$A = Rs. 12 ; B = Rs. 17$
0%
$A = Rs. 13 ; B = Rs. 17$
0%
$A = Rs. 18 ; B = Rs. 17$
0%
$A = Rs. 19 ; B = Rs. 17$

Explanation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The pair of linear equations

$3x+5y=3, 6x+ky=8$ do not have any solution if

0%
$k=5$
0%
$k=10$
0%
$k\neq 10$
0%
$k\neq 5$

Explanation

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

$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}$

$\Rightarrow \displaystyle \frac{3}{6}=\frac{5}{k}\neq \frac{-3}{-8}$

$\Rightarrow 3k=30\Rightarrow k=10$

If the sum of the ages of a father and his son in years is

$65$ and twice the difference of their ages in years is

$50$ , then the age of the father is:

0%
$45$ years
0%
$40$ years
0%
$50$ years
0%
$55$ years

Explanation

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

$3x - 5y + 1 = 0, 2x - y + 3 = 0$ has a unique solution

$x = x_1, y=y_1$ then

$y_1=$

0%
$1$
0%
$-1$
0%
$-2$
0%
$-4$

Explanation

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

$y_{1} =2\left ( -2 \right ) + 3$

$y_{1} = -1$

For what value of

$k$ , do the equations

$3x - y + 8 = 0$ and

$6x - ky = - 16$ represent coincident lines?

0%
$\displaystyle \frac{1}{2}$
0%
$-\displaystyle \frac{1}{2}$
0%
$2$
0%
$-2$

Explanation

The equations are

$3x-y+8=0$

$6x-ky+16=0$
Here,

$a_{1}=3,b_{1}=-1,c_{1}=8$
and

$a_{2}=6,b_{2}=-k,c_{2}=16$

The equation will represent coincident lines only when they have infinite number of solutions.

$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \displaystyle \frac{3}{6}=\frac{-1}{-k}=\frac{8}{16}$

Taking,

$\displaystyle \frac{3}{6}=\frac{-1}{-k}$

$\Rightarrow \displaystyle \frac{1}{2}=\displaystyle \frac{1}{k}$

$\Rightarrow k=2$

Taking,

$\displaystyle \frac{-1}{-k}=\frac{8}{16}$

$\Rightarrow \displaystyle \frac{1}{k}=\frac{1}{2}$

$\Rightarrow k=2$

So, the answer is

$k=2$

The pair of equations

$x + 2y + 5 = 0$ and

$- 3x - 6y + 1 = 0$ have

0%
A unique solution
0%
Exactly two solutions
0%
Infinitely many solutions
0%
No solution

Explanation

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.

The pair of linear equations

$x + 2y = 5, 3x + 12y = 10$ has

0%
Unique solution
0%
No solution
0%
More than two solutions
0%
Infinitely many solutions

Explanation

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

Six years hence, a man's age will be three times the age of his son and three years ago he was nine times as old as his son. The present age of the man is

0%
$28$ years
0%
$30$ years
0%
$32$ years
0%
$34$ years

Explanation

Let the present age of man be

$x$ years and the present age of son be

$y$ years

After six years:
Man's age =

$(x+6)\ years$

Son's age =

$(y+6)\ years$

According to the condition

$\Rightarrow x+6=3\left (y+6 \right )$

$\Rightarrow x-3y=12$

$\Rightarrow x=12+3y$

$...(1)$

Three years ago:
Man's age =

$(x-3)\ years$
Son's age =

$(y-3)\ years$

According to the condition

$\Rightarrow x-3=9\left (y-3 \right )$

$\Rightarrow x-9y=-24$

$...(2)$

Put the value of

$x$ in equation

$\left ( 2 \right )$

$\Rightarrow 12+3y-9y=-24$

$\Rightarrow y=6$

Put the value of

$y$ in equation

$\left ( 1\right )$

$x-3\times6=12$

$\Rightarrow x=30$

So, Man's age = 30 years

The pair of equations

$x = 0$ and

$y = - 7$ has

0%
One solution
0%
Two solutions
0%
Infinitely many solutions
0%
No solution

The pair of linear equations

$x + 2y = 5, 7x + 3y = 13$ has a unique solution

0%
$x=1, y=2$
0%
$x=2, y=1$
0%
$x=3, y=1$
0%
$x=1, y=3$

Explanation

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)

$\Rightarrow x + 2\left ( 2 \right ) = 5$

$\Rightarrow x = 5 - 4$

$\Rightarrow x = 1$

$x = 1 , y = 2$

Three chairs and two tables cost Rs1,Five chairs and three tables cost Rs 2,Then the total cost of one chair and one table is

0%
Rs 800
0%
Rs 850
0%
Rs 900
0%
Rs 950

Explanation

Let cost of one chair be Rs

$x$ and cost of one table be Rs

$y$
Now as per the question,

$3x+2y = 1850$ .....(1)

$5x + 3y = 2850$ ......(2)
Now mutiplying equation (1) by 5 and equation (2) by 3

$15x+10y = 9250$ .....(3)

$15x+9y = 8550$ ......(4)
Now subtracting equation (4) from equation (3)

$\Rightarrow y = 700$
Put this value in eq (1),

$\Rightarrow 3x+2\left ( 700 \right ) = 1850$

$3x+1400 = 1850$

$3x = 450$

$x = 150$
Now total cost of 1 chair and 1 table =

$x + y$

$=150+700= 850$
So, total cost of 1 chair and 1 table is Rs 850

Find the value of

$k$ for which the given system of equation has no solution.

$kx + 2y - 1 = 0$

$5x - 3y + 2= 0$

0%
$k=-\dfrac15$
0%
$k=-\dfrac{10}{3}$
0%
$k=\dfrac95$
0%
$k=\dfrac65$

Explanation

The equations are

$\Rightarrow kx+2y-1=0\\$

$\Rightarrow a_{1}=k,b_{1}=2,c_{1}=-1\\$

$\Rightarrow 5x-3y+2=0\\$

$\Rightarrow a_{2}=5,b_{2}=-3,c_{2}=2\\$

The given system of equations has no solution

$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\Rightarrow \displaystyle \frac{k}{5}=\frac{2}{-3}\Rightarrow k=-\frac{10}{3}$

On comparing the ratios

$\frac {a_1}{a_2}, \frac {b_1}{b_2}$ and

$\frac {c_1}{c_2}$ find out whether the following pair of linear equations are consistent or inconsistent.

$x - 3y = 4 ; 3x + 2y = 1$

0%
consistent
0%
inconsistent
0%
Ambiguous
0%
None of these

Explanation

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

$2x - 3y = 7$ and

$(a + b) x - (a + b - 3) y = 4a +b$ have infinite solutions then

$(a,b) =$

0%
$(-5, -1)$
0%
$(-5, 1)$
0%
$(5, 1)$
0%
$(5, - 1)$

Explanation

The equations are

$2x-3y-7=0$

$\left (a +b \right )x-\left (a+b-3 \right )y-(4a+b)=0$
Here,

$a_{1}=2,b_{1}=-3,c_{1}=-7$

$a_{2}=a+b ,b_{2}=-\left (a+b-3 \right ),c_{2}=-(4a+b)$
The system of linear equations has infinite solutions

$\therefore \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \displaystyle \frac{2}{a+b}=\frac{-3}{-\left (a+b-3 \right )}=\frac{-7}{-\left (4a+b \right )}$

$\Rightarrow \displaystyle \frac{2}{a+b}=\frac{3}{\left (a+b-3 \right )}$

$\Rightarrow 2a+2b-6=3a+3b$

$\Rightarrow b=-a-6$
Again, we have

$\Rightarrow \displaystyle \frac{3}{\left (a+b-3 \right )}=\frac{7}{\left (4a+b \right )}$

$\Rightarrow 12a+3b=7a+7b-21$

$\Rightarrow 5a-4b=-21$
Putting

$b=-a-6$

$\Rightarrow 5a-4\left ( -a-6 \right )=-21\Rightarrow 9a=-45$

$\Rightarrow a=-5$
Putting

$a=-5 \ in \ b=-a-6$

$\Rightarrow b=-\left (-5 \right )-6\Rightarrow b=-1$

$\Rightarrow a=-5,b=-1$

The graphic representation of the pair of equations

$2x + 4y - 15 = 0$ and

$x + 2y - 4 = 0$ gives a pair of

0%
Parallel lines
0%
Intersecting lines
0%
Coincident lines
0%
None of these

Explanation

The equations are

$2x+4y-15=0$

$x+2y-4=0$
Here,

$a_{1}=2,b_{1}=4,c_{1}=-15$
and

$a_{2}=1,b_{2}=2,c_{2}=-4$

$\displaystyle \frac{a_{1}}{a_{2}}=\frac{2}{1}=2$

$\displaystyle \frac{b_{1}}{b_{2}}=\frac{4}{2}=2$

$\displaystyle \frac{c_{1}}{c_{2}}=\frac{-15}{-4}$

$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}$

$\Rightarrow$ The system of equations has no solution.

$\therefore$ Graphical representation of the given pair of equations is parallel lines

$5$ pencils and

$7$ pens together costs Rs.

$50$ whereas

$7$ pencils and

$5$ pens together costs Rs.

$46$ . Thus the cost of one pencil and one pen respectively is:

0%
Rs. $5$ , Rs. $3$
0%
Rs. $3$ , Rs. $5$
0%
Rs. $4$ , Rs. $4$
0%
Rs. $2$ , Rs. $6$

Explanation

Let the cost of one pencil be Rs.

$x$ and the cost of one pen be Rs.

$y$
According to the first condition

$5x+7y=50....eq\left ( 1\right )$

$\Rightarrow x=\displaystyle \frac{50-7y}{5}$
According to the second condition

$7x+5y=46....eq\left ( 2 \right )$
Put the value of

$x$ in

$eq\left ( 2 \right )$

$\Rightarrow \displaystyle \frac{7\left ( 50-7y \right )}{5}+5y=46$

$\Rightarrow 350-49y+25y=230\Rightarrow y=5$

Resubstitute the value of

$y$ to find the value of

$x$

$\Rightarrow x=\displaystyle \frac{50-7(5)}{5}= \frac{50-35}{5}\Rightarrow x=3$

$\therefore$ cost of one pencil

$=$ Rs.

$3$ and cost of one pen

$=$ Rs.

$5$

Find the value of k for which the given system of equations has no solution.

$kx + 3y = k - 3; 12x + ky = k$

0%
$k=-5$
0%
$k=-2$
0%
$k=-3$
0%
None of these

Explanation

A system of equations,

$a_1x+b_1y+c_1=0$ and

$a_2x+b_2y+c_2=0$ has no solution for the following condition:

$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

The equations in the given question are:

$kx+3y-k+3=0$ and

$a_{1}=k,b_{1}=3,c_{1}=3$

$12x+ky-k=0$ and

$a_{2}=12,b_{2}=k,c_{2}=-k$

Substituting the above values in the condition mentioned above for no solution:

$\dfrac{a_{1}}{b_{1}}=\dfrac{a_{2}}{b_{2}}$

$\dfrac{k}{12}=\dfrac{3}{k}$

$k^{2}=36$

$\Rightarrow k=6$

On comparing the ratios

$\dfrac {a_1}{a_2}, \dfrac {b_1}{b_2}$ and

$\dfrac {c_1}{c_2}$ find out whether the following pair of linear equations are consistent or inconsistent.

$x - 2y = 3 ; 3x - 6y = 1$

0%
consistent
0%
Inconsistent
0%
Ambiguous
0%
Data insufficient

Explanation

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

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

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

$x-y=0.9$ and

$\displaystyle \frac {11}{x+y}=2$ have the solution

0%
$x=5, y=1$
0%
$x=3.2$ and $y=2.3$
0%
$x=3$ and $y=2$
0%
none of these

Explanation

The equation are

$x-y=0.9$

$\Rightarrow x=0.9+y$ .....(1)

$\displaystyle \frac{11}{x+y}=2$

$\Rightarrow 2x+2y=11$ .....(2)
Put the value of

$x$ in eq(2)

$\Rightarrow 2\left (.9+y \right )+2y=11$

$\Rightarrow 1.8+2y+2y=11$

$\Rightarrow 4y=9.2$

$\Rightarrow y=2.3$

$\Rightarrow x=.9+y\Rightarrow x=.9+2.3=3.2$

$\therefore x=3.2 ,y=2.3$

Find the value of k for which the given system of equations has a unique solution.

$(k- 3)x + 3y = k; kx + ky = 12$

0%
$k\neq3$
0%
$k\neq 9$
0%
$k\neq 6$
0%
$k\neq 2$

Explanation

Comparing

$\left ( k-3 \right )x +3y = k$ with

$a_1x+b_1y+c_1=0$ ,

and

$kx+ky =12$ with

$a_2x+b_2y+c_2=0$ ,

we get,

$a_{1} = \left ( k-3 \right ) , a_{2} = k , b_{1} = 3, b_{2}= k$

Now,

Given system of equations has unique solution

$\Rightarrow \dfrac{a_1{}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}$

$\dfrac{\left ( k-3 \right )}{k}\neq \dfrac{3}{k}$

$\Rightarrow k^{2}- 3k \neq 3k$

$\Rightarrow k^{2} \neq 6k$

$\Rightarrow k^{2} -6k\neq 0$

$\Rightarrow k(k-6) \neq 0$

$\Rightarrow k \neq 0$ and

$\ k \neq 6$

$k \neq 6$ is the correct option among given options.

Find the value of

$k$ for which the given system of equations has a unique solution:

$x - ky = 2$ ;

$3x + 2y = - 5$

0%
$k\neq -\dfrac23$
0%
$k\neq \dfrac12$
0%
$k\neq \dfrac32$
0%
$k\neq -\dfrac52$

Explanation

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
$\Rightarrow \dfrac{a_{1}}{a_{2}}\neq \dfrac{b_{1}}{b_{2}}$
$\Rightarrow \dfrac{1}{3}\neq -\dfrac{k}{2}$
$\Rightarrow$ $k\neq-\dfrac{2}{3}$

Solve graphically the following pair of linear equations:

$2x + 3y - 12 = 0, 2x - y - 4 = 0$ .

0%
$x=0, y=1$
0%
$x=4, y=3$
0%
$x=3, y=2$
0%
$x=0, y=5$

Explanation

$2x+3y-12=0$

$\Rightarrow 2x=12-3y$

$\Rightarrow x=\cfrac{12-3y}{2}$

Let,

$y=2$ , then

$x=\cfrac{12-3(2)}{2}=3$

Similarly, when

$y=0, x=6$

$2x-y-4=0$

$\Rightarrow 2x=4+y$

$\Rightarrow x=\cfrac{4+y}{2}$

Let,

$y=2$ , then

$x=\cfrac{4+2}{2}=3$

Similarly, when

$y=0, x=2$

Plotting the points above, the intersecting point is

$(3,2)$

$\therefore x=3,y=2$

Find the value of a and b for which the given system of linear equation has an infinite number of solutions.

$2x + 3y = 7$ and

$(a - b) x + (a + b)y = 3a + b - 2$

0%
$a=2, b=3$
0%
$a=5, b=1$
0%
$a=-5, b=3$
0%
$a=-1, b=2$

Explanation

Comparing

$2x+3y =7$ with

$a_1x+b_1y+c_1=0$ ,

we get,

$a_1=2, b_1=3,c_1=-7$

Comparing

$\left ( a-b \right )x + \left ( a +b \right )y = 3a + b - 2$ with

$a_2x+b_2y+c_2=0$ ,

we get,

$a_2=\left ( a-b \right ), b_2=\left ( a +b \right ), c_2=-(3a + b - 2)$

Given system of equation have infinite solutions:

$\Rightarrow \dfrac{a_1{}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$

$\Rightarrow \dfrac{2}{a-b}= \dfrac{3}{a+b}= \dfrac{7}{3a + b -2}$

From

$\dfrac{a_1{}}{a_{2}} = \dfrac{b_{1}}{b_{2}}$

$\Rightarrow \dfrac{2}{a-b}= \dfrac{3}{a+b}$

$\Rightarrow 2a + 2b = 3a-3b$

$\Rightarrow a-5b = 0 ......... (I)$

Now from

$\dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$

$\Rightarrow \dfrac{3}{a+b}= \dfrac{7}{3a + b -2}$

$\Rightarrow9a + 3b -6 = 7a + 7b$

$\Rightarrow 9a-7a+3b-7b =6$

$\Rightarrow 2a-4b = 6 ...... (II)$

Now, for solving eq

$I$ and eq

$II$

$a-5b = 0$

$2a-4b = 6$

Multiplying eq

$I$ by

$2$

$2a - 10b =0$

$2a - 4b =6$

on subtraction we get

$\Rightarrow -6b = -6$

$\Rightarrow b = 1$

Now, substituting

$b$ value in eq

$I$

$\Rightarrow 2a-10\left ( 1 \right ) = 0$

$2a - 10 = 0$

$2a = 10$

$a = \dfrac{10}{2}$

$a = 5$

$\Rightarrow a = 5 , b = 1$

Solve the following pair of linear equations by the substitution method.

$0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3$

0%
$x=1, y=-2$
0%
$x=6, y=-7$
0%
$x=5, y=1$
0%
$x=2, y=3$

Explanation

The given equations are

$\Rightarrow .2x+.3y=1.3........eq1$

$\Rightarrow x=\frac{1.3-.3y}{.2}.......eq2$

and

$\Rightarrow .4x+.5y=2.3.......eq3$
Substitute the value of x in eq3

$\Rightarrow .4\times\left (\frac{1.3-.3y}{.2} \right)+.5y=2.3$

$\Rightarrow 2.6-.6y+.5y=2.3$

$\Rightarrow -.1y=-.3 \Rightarrow y=3$
Substitute

$y=3$ in eq 2

$\Rightarrow x=\frac{1.3-.9}{.2}\Rightarrow x=2$
Hence x=2 and y=3

Determine whether the following system of linear equations have no solution, infinitely many solution or unique solutions.

$x + 2y = 3, 2x + 4y = 15$

0%
No solution
0%
Infinitely many solution
0%
Unique solution
0%
Cannot be determined

Explanation

The equations are

$\Rightarrow x+2y-3=0$

$\Rightarrow a_{1}=1,b_{1}=2,c_{1}=-3$

$\Rightarrow 2x+4y-15=0$

$\Rightarrow a_{2}=2,b_{2}=4,c_{2}=-15$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{1}{2}$

$\Rightarrow \frac{b_{1}}{b_{2}}=\frac{2}{4}=\frac{1}{2}$

$\Rightarrow \frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}} \neq\dfrac{c_{1}}{c_{2}}$
Hence, the given system of equations has no solution

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

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Practice Class 10 Maths Quiz Questions and Answers

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

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Practice Class 10 Maths Quiz Questions and Answers

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