MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 1

The condition for the pair of equations

$a_1x+b_1y+c_1=0$ and

$a_2x+b_2y+c_2 = 0$ to have a unique solution is -

0%
$a_1b_2-a_2b_1=0$
0%
$a_1b_2-a_2b_1 \neq 0$
0%
$a_1b_1 - a_2 b_2 =0$
0%
$a_1b_1 - a_2b_2 \neq 0$

Explanation

For unique solution

$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

$\Rightarrow a_1b_2 - a_2b_1 \neq 0$

$\cfrac { a }{ x+y } =\cfrac { b }{ y+z } =\cfrac { c }{ z-x }$ , then which of the following equations is true?

0%
$a=b+c$
0%
$c=a+b$
0%
$b=a\times c$
0%
$b=a+c$

Explanation

Let

$\cfrac { a }{ x+y } =\cfrac { b }{ y+z } =\cfrac { c }{ z-x } =k$

$\therefore$

$a=xk+yk........(1)$

$b=yk+zk...........(2)$

$c=zk-xk...........(3)$

equation

$(1)+(3)$

$a+c=zk-xk+xk+yk$

$=zk+yk$

$=b$

$\therefore$

$a+c=b$

Hence, option D is correct.

The point of intersection of the lines

$y=3x$ and

$x=3y$ is :

0%
(3, 0)
0%
(0, 3)
0%
(3, 3)
0%
(0, 0)

Explanation

The point of intersection of lines

$y = 3x$ (black line) and

$x = 3y$ (brown line) must satisfies both the equation.
From the graph, we can see that solution is (0,0).

Hence the point of intersection of given lines is at origin i.e.

$\left( 0, 0 \right)$ .

If,

$\displaystyle \frac{1}{x}+\frac{1}{y}=k$ and

$\displaystyle \frac{1}{x}-\frac{1}{y}=k$ , then the value of y .................

0%
$3$
0%
$-4$
0%
Does not exist.
0%
None of these

Explanation

Let

$\displaystyle \frac{1}{x}+\frac{1}{y}=k$ ......(1)

$\displaystyle \frac{1}{x}-\frac{1}{y}=k$ ......(2)

$(1)-(2) \Rightarrow \displaystyle \frac{2}{y}=0$

$\therefore$ the value of

$y$ does not exist.

If the pair of equations has no solution, then the pair of equations is :

0%
consistent
0%
inconsistent
0%
coincident
0%
none of these

Which of the following pairs of equations represent inconsistent system?

0%
$3x - 2y = 8$

$2x + 3y = 1$
0%
$3x - y = - 8$

$3x - y = 24$
0%
$x - y = m$

$x + my = 1$
0%
$5x - y = 10$

$10x - 6y = 20$

Explanation

For a pair of linear equations to be inconsistent, the equations must satisfy the given condition which is

$\cfrac{{a}_{1}}{{a}_{2}} = \cfrac{{b}_{1}}{{b}_{2}} \ne \cfrac{{c}_{1}}{{c}_{2}}$

Since, the equations

$3x - y = -8$ and

$3x - y = 24$ clearly satisfy the above condition

$\cfrac{3}{3} = \cfrac{-1}{-1} \ne \cfrac{-8}{24}$ , the pair of equations represents inconsistent system.
In options A, C and D,

$\cfrac{{a}_{1}}{{a}_{2}} \neq \cfrac{{b}_{1}}{{b}_{2}}$

If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is ...................

0%
inconsistent.
0%
consistent.
0%
Cannot be determined
0%
None of above

State whether the given statement is true or false:

A pair of linear equations is given by

$a_1x+b_1y+c_1=0$ and

$a_2x+b_2y +c_2=0$ also

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

In this case, the pair of linear equations is consistent.

0%
True
0%
False

Explanation

$\displaystyle \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ .
The lines are coincident and infinitely many solutions are obtained.
Therefore, in this case, the pair of linear equations is consistent.
The correct answer is True.

State whether the given statement is true or false:

A pair of linear equations is given by

$a_1x + b_1y+c_1=0$ and

$a_2x+b_2y+c_2=0$ and

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

In this case, the pair of linear equations is inconsistent.

0%
True
0%
False

Explanation

$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$ .
The lines are parallel and no solution is obtained.
Therefore, in this case, the pair of linear equations is inconsistent.
So the correct answer is

$A$ .

State true\false:

A pair of linear equations is given by

$a_1x+b_1y+{ c }_{ 1 }=0$ and

$a_2x+b_2y+c_2=0$ and

$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ .

In this case, the pair of linear equations is consistent.

0%
True
0%
False

Explanation

$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ . The lines intersect and a unique solution is obtained.
Therefore, in this case, the pair of linear equations is consistent.
The correct answer is A.

If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is .................

0%
Inconsistent
0%
No solution
0%
Consistent
0%
None of the Above

State true or false:

The graph of the linear equation

$x + 2y = 7$ passes through the point

$(0, 7).$

0%
True
0%
False

Explanation

For this, we put

$(0,7)$ in

$x+2y=7$ .

L.H.S.

$=$

$x+2y = 0+2\times7 = 14$

R.H.S.

$= 7$

Since, L.H.S is not equal to R.H.S. , above equation does not pass through

$(0,7)$ .

Solve, graphically, the following pairs of equations:

$x - 5 = 0$

$y+ 4 = 0$

0%
$x = 1, y = -8$
0%
$x = 5, y = -4$
0%
$x = 4, y = -1$
0%
$x = 0, y = -1$

Explanation

Given pair of equations is

$x-5=0$ and

$y+4=0$

Plotting these equations in the graph, we get the above image.

Clearly from the graph, we can find the point of intersection and that point is
the point of intersection is

$(5,-4)$ .

Hence,

$x=5, y=-4$ .

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

Let

$kx-y-2=0 ......(1)$

$6x-2y-3=0 ......(2)$
For the equation to be inconsistent

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

$\Rightarrow \dfrac{k}{6}=\dfrac{-1}{-2}\ne\dfrac{-2}{-3}$

$\therefore k=3$
Therefore the Assertion is correct.
Clearly, the Reason is correct.
Since both the Assertion and Reason are correct and Reason is a correct explanation of Assertion.
The correct answer is A.

If a pair of linear equations are consistent, then the lines will be

0%
Parallel
0%
Always Coincident
0%
Intersecting
0%
Coincident

Explanation

${\textbf{Step-1: Take two equations and check conditions.}}$

$\text{Lets consider two linear equations}$

${a_1}x + {b_1}y = {c_1}$

${a2}x + {b_2}y = {c_2}$

${\text{Generally, a system of pair of straight lines are represented by,}}$

${a_1}x + {b_1}y + {c_1} = 0$

${\text{We check the pair of lines coincide or not using the condition that,}}$

$\dfrac{a_1}{b_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

${\text{If the above condition satisfies then the lines are considered as consistent as}}$

${\text{it means the pair of lines are coincident.}}$

$\dfrac{a_1}{b_2} \ne \dfrac{b_1}{b_2}$

${\text{If the above condition satisfies then the lines are considered as consistent as it means the pair of}}$

${\text{lines are intersecting.}}$

${\text{Hence, if the pair of linear equations is consistent, then the lines either intersect or coincident.}}$

${\textbf{Therefore, they are Intersecting or coincident}}$

Value of

$\alpha$ for which equations

$\alpha x+ 3y = \alpha -3; 12x + \alpha y = \alpha$ has unique solution.

0%
All value of $\alpha$ (except 4 and -4)
0%
All value of $\alpha$ (except 6 and -6)
0%
All value of $\alpha$ (except 2 and -2)
0%
All value of $\alpha$ (except 3 and -3)

Explanation

For a unique solution, we must have

$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
i.e.,

$\displaystyle \dfrac{\alpha}{12} \neq \dfrac{3}{\alpha}$

$\Rightarrow \alpha^2 \neq 36$

$\Rightarrow \alpha \neq 6\ or\ -6$
Hence, for all valus of

$\alpha$ (except

$6$ and

$-6$ ).

The given pair of linear equations has a unique solution.

Solve the following pair of linear equations using Graphical method:

$x\, +\, y\, =\, 8; \quad x\, -\, y\, =\, 2$ .

Then

$(x, y)$ is equal to

0%
$(5, 3)$
0%
$(7, 6)$
0%
$(4, 1)$
0%
$(2, 2)$

Explanation

The given simultaneous equations are

$x+y=8 ..... (i) \text{ and } x-y=2 ..... (ii)$

From equation

$(i)$

$x$	$0$	$3$	$5$	$8$
$y$	$8$	$5$	$3$	$0$

From equation

$(ii)$

$x$	$0$	$2$	$4$	$5$
$y$	$-2$	$0$	$2$	$3$

Plotting

$x+y=8$ (blue line) and

$x-y=2$ (green line) in the graph, we get the point of intersection as

$(5,3)$ .

Hence,

$(x,y)=(5,3)$ is the correct answer.

A system of linear equations is given as follows :

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

Condition for two lines to have a unique solution is,

0%
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}$
0%
$\displaystyle \frac{a_1}{a_2}\neq \frac{b_1}{b_2}$
0%
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$
0%
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Explanation

$Two\quad equations\quad will\quad have\quad a\quad unique\quad solution\quad if,\quad after\quad \\ solving\quad them,\quad the\quad solution\quad is\quad valid\quad i.e\quad the\quad lines\quad actually\quad \\ intersect\quad at\quad a\quad point.$

$\\ Let\quad there\quad be\quad equations\quad \quad { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0..........(i)\quad and$

$\\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0.........(ii).\quad \quad$

$Multiplying\quad (i)\quad by\quad { a }_{ 2 }\quad \& \quad (ii)\quad by\quad { a }_{ 1 }\\ and\quad eliminating\quad x\quad by\quad subtraction\quad we\quad get\quad,$

$\\ y=\dfrac { { a }_{ 1 }{ c }_{ 2 }-{ a }_{ 2 }{ c }_{ 1 } }{ { a }_{ 2 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 2 } } .$

$\\ Again\quad Multiplying\quad (i)\quad by\quad { b }_{ 2 }\quad \& \quad (ii)\quad by\quad { b }_{ 1 }\\ and\quad eliminating\quad y\quad by\quad subtraction\quad we\quad get\quad$

$\\ x=\dfrac { { b }_{ 2 }{ c }_{ 1 }-{ b }_{ 1 }{ c }_{ 2 } }{ { a }_{ 2 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 2 } } .$

$\\ It\quad is\quad clear\quad that\quad the\quad solution\quad \left( x,y \right) =\left( \dfrac { { b }_{ 1 }{ c }_{ 2 }-{ b }_{ 2 }{ c }_{ 1 } }{ { a }_{ 2 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 2 } } ,\dfrac { { b }_{ 2 }{ c }_{ 1 }-{ b }_{ 1 }{ c }_{ 2 } }{ { a }_{ 2 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 2 } } \right) \\ will\quad be\quad valid\quad if\quad { a }_{ 2 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 2 }\neq 0\Longrightarrow \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \dfrac { { b }_{ 1 } }{ { b }_{ 2 } } .$

$\\ Ans-\quad Option\quad B.$

A system of linear equations is given as follows :

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

Both the lines are parallel only if,

0%
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}$
0%
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$
0%
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
0%
None of these

Explanation

$Let\quad us\quad take\quad two\quad equations\quad as,\quad$

$\\ { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0........(i)\quad and\\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0.......(ii).$

$\\ If\quad the\quad the\quad lines\quad are\quad parallel\quad then\quad the\quad slopes\quad of\quad the\quad two\quad lines\quad \\ should\quad be\quad equal\quad and\quad the\quad intercepts\quad of\quad the\quad two\quad lines\quad \\ should\quad NOT\quad be\quad equal.$

$\\ Writing\quad the\quad equations\quad of\quad lines\quad in\quad slope-intercept\quad form\\ we\quad have,\quad$

$(i)\quad as\quad y=\dfrac { -{ a }_{ 1 } }{ b_{ 1 } } x+\left( \dfrac { -{ c }_{ 1 } }{ { b }_{ 1 } } \right) \quad and\quad \\ (ii)\quad as\quad y=\dfrac { -{ a }_{ 2 } }{ b_{ 2 } } x+\left( \dfrac { -{ c }_{ 2 } }{ { b }_{ 2 } } \right)$

$\\ Now\quad if\quad the\quad lines\quad are\quad parallel\quad then\quad the\quad slopes\quad are\quad equal.$

$\\ i.e\quad \dfrac { -{ a }_{ 1 } }{ b_{ 1 } } =\dfrac { -{ a }_{ 2 } }{ b_{ 2 } } \Longrightarrow \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { b_{ 1 } }{ b_{ 2 } } ........(iii)$

$\\ Also\quad the\quad intercepts\quad are\quad NOT\quad equal.$

$\\ i.e\quad \left( \dfrac { -{ c }_{ 1 } }{ { b }_{ 1 } } \right) \neq \left( \dfrac { -{ c }_{ 2 } }{ { b }_{ 2 } } \right) \Longrightarrow \dfrac { { c }_{ 1 } }{ { c }_{ 2 } } \neq \dfrac { { b }_{ 1 } }{ { b }_{ 2 } } ......(iv)$

$\\ Combining\quad (iii)\quad \& \quad (iv)\quad we\quad have\quad \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { b_{ 1 } }{ b_{ 2 } } \neq \dfrac { { c }_{ 1 } }{ { c }_{ 2 }}$

$\\ Ans-\quad Option\quad B.\\ \\ \\$

STATEMENT - 1 : If the graphs of the two equations are parallel lines, there exists no solution.
STATEMENT - 2 : The system is called an inconsistent system.

0%
Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
0%
Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1
0%
Statement - 1 is True, Statement - 2 is False
0%
Statement - 1 is False, Statement - 2 is True

The solution set of

$x + y -1=0$ and

$3x + 3y -2=0$ is

0%
(1,0)
0%
(0, 1)
0%
An empty set
0%
(1, 1)

Explanation

$x + y -1=0\Rightarrow a_1=1,\,b_1=1,\,c_1=-1$

$3x + 3y -2=0\Rightarrow a_2=3,\,b_2=3,\,c_2=-2$

$\displaystyle \dfrac{1}{3} = \dfrac{1}{3} \neq \dfrac{-1}{-2}$

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

Hence no solution.

Therefore, solution set for

$x + y -1=0$ and

$3x + 3y -2=0$ is an empty set.

Find the solution set of the system of equations:

$\displaystyle \frac{4}{x}+5y=7\:and\:\frac{3}{x}+4y=5.$

0%
$\displaystyle \left ( \frac{1}{3},-1 \right )$
0%
$\displaystyle \left ( \frac{1}{3},-3 \right )$
0%
$\displaystyle \left ( \frac{1}{3},-2 \right )$
0%
$\displaystyle \left ( \frac{1}{3},1 \right )$

Explanation

We put

$\dfrac { 1 }{ x } =p$

Then the system of given equations becomes

$4p+5y=7............(i)$ &

$3p+4y=5........(ii)$

$(i)\times 3\Rightarrow 12p+15y=21........(iii)\quad \& \\ (ii)\times 4\Rightarrow 12p+16y=20.........(iv).\\ (iii)-(iv)\Rightarrow -y=1\\ \Rightarrow y=-1.\quad$

Putting

$y=-1$ in

$(i)$ ,

$4p+5\times (-1)=7\\ \Rightarrow p=\dfrac { 12 }{ 4 } =3.\\ \therefore x=\dfrac { 1 }{ 3 } .\\ \therefore \left( x,y \right) =\left( \dfrac { 1 }{ 3 }, -1 \right) .\quad$

Hence, the answer is

$\left(\dfrac{1}{3},-1\right)$

The system of linear equation

$ax+by=6$ ,

$cx+dy=8$ has no solution if:

0%
$ad-bc>0$
0%
$ad-bc<0$
0%
$ad+bc=0$
0%
$ad-bc=0$

Explanation

The given equations are

$ax+by = 6$ and

$cx+dy = 8$ .

Here,

$a_1=a, b_1=b, c_1=-6\,\,and\,\, a_2=c, b_2=d, c_2=-8$

They have no solution if,

$\dfrac { a_1 }{ a_2 } =\cfrac { b_1 }{ b_2 } \neq \cfrac { c_1 }{ c_2 }$ .

Hence,

$\dfrac { a }{ c } = \cfrac { b }{ d }$

$\Rightarrow ad - bc = 0$

So, option D is correct.

The cost of 9 chairs and 3 tables is Rs. 306, while the cost of 6 chairs and 3 tables is Rs.Then the cost of 6 chairs and 1 table is

0%
Rs. 161
0%
Rs. 162
0%
Rs. 169
0%
Rs. 175

Explanation

$9x+3y=306$ ---- (1)

$6x+3y=246$ -----(2)

Subtracting eq(2) from eq(1)

$3x=60$

$x=20$

Substituting

$x=20$ in eq(1)

$9(20)+3y = 306$

$180+3y=306$

$3y=126$

$y=42$

Hence,

$6x+y =162$ .

Check whether the pair of equations

$x + 3y = 6$ , and

$2x - 3y = 12$ is consistent. If so, solve graphically

0%
Yes; $x=6, y=0$
0%
No; $x=6, y=0$
0%
Ambiguous
0%
Data insufficient

Explanation

The given pair can be written as

$x+3y-6=0$ and

$2x-3y-12=0$
Here

$a_1=1,b1=3,c_1=-6...........eq1 and a_2=2,b_2=-3,c_2=-12.........eq2$

$\therefore \frac{a_1}{a_2}=\frac{1}{2},\frac{b_1}{b_2}=\frac{3}{-6}=\frac{1}{-2}$
Hence

$\frac{a_1}{a_2}\neq\frac {b_1}{b_2}$
Thus the given pair of equation is consistent.
To solve them graphically we have
Consider

$x+3y-6=0$

$x=0 \implies y =2$

$x=3 \implies y=1$
Join the points by drawing line.
Consider

$2x-3y-12=0$

$x=0 \implies y =-4$

$x=3 \implies y=-2$
Join the two points by drawing line.
The given pair of linear equation has unique solution

$x=6$ and

$y=0$ .

The value of

$k$ for which the following system of equations has infinitely many solutions

$5x+2y=k$

$10x+4y=3$

0%
$\cfrac{1}{2}$
0%
$3$
0%
$\cfrac{3}{2}$
0%
$1$

Explanation

For infinitely many solutions, both equations need to be identical in the ratios
i.e.

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$
where,

$a_1,b_1,c_1$ are the coefficients of a equation
and

$a_2,b_2,c_2$ are the coefficients of another equation

Thus from

$5x+2y=k$

$10x+4y=3$

To have infinitely many solutions

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

$\dfrac{5}{10}=\dfrac{2}{4}=\dfrac{k}{3}$

Taking

$\dfrac{2}{4}=\dfrac{k}{3}$

$\Rightarrow \dfrac{1}{2}=\dfrac{k}{3}$

Thus,

$2k=3$

$\Rightarrow k=\cfrac{3}{2}$ is the required answer.

If a pair of linear equations is inconsistent. In the below graph, the lines are _________.

0%
Parallel
0%
Coincidence
0%
Intersecting
0%
Divergence

If a pair of linear equations is consistent and dependent. In the below graph, the lines are __________.

0%
Parallel
0%
Coincident
0%
Intersecting
0%
none of these

Explanation

If a pair of linear equations is consistent, then the number of solutions can be either one or infinite.

Here, the lines are dependent as well and hence, have an infinite number of solutions.

$\therefore$ Lines having an infinite number of solutions are coincident

Answer: B

Which of the following condition is true if the system of equations below is shown to be consistent and dependent?

$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$

0%
$\displaystyle \frac{a_1}{a_2}= \frac{b_1}{b_2}\neq \frac{c_1}{c_2}$
0%
$\displaystyle \frac{a_1}{a_2}= \frac{b_1}{b_2}= \frac{c_1}{c_2}$
0%
$\displaystyle \frac{a_1}{a_2}\neq \frac{b_1}{b_2}\neq \frac{c_1}{c_2}$
0%
$\displaystyle \frac{a_1}{a_2}\neq \frac{b_1}{b_2}= \frac{c_1}{c_2}$

Explanation

If the system of equations are consistent, then there can be exactly one solution or infinite solutions.

The solutions are infinite when they are dependent.

We know that, for two linear equations

$a_{1}x+b_{1}=c_{1}$ and

$a_{2}x+b_{2}y=c_{2}$

$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}$ , the system is consistent and has infinitely many solutions.

Hence, a system of equations will be consistent and dependent if

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

The value of

$k$ for which the system

$kx+3y=7$ and

$2x-5y=3$ has no solution is:

0%
$7$ and $k=-\cfrac{3}{14}$
0%
$4$ and $k=\cfrac{3}{14}$
0%
$\cfrac{6}{5}$ and $k\ne \cfrac{14}{3}$
0%
$-\cfrac{6}{5}$ and $k\ne \cfrac{14}{3}$

Explanation

$kx+3y=7$

$2x-5y=3$
Here,

${a}_{1}={k}_{1}, {b}_{1}=3, {c}_{1}=-7$

${a}_{2}=2, {b}_{2}=-5,{c}_{2}=-3$
For no solution,

$\cfrac{k}{2}=\cfrac{3}{-5}\ne \cfrac{-7}{-3}$

$\Rightarrow k=\cfrac{-6}{5}$ and

$k\ne \cfrac{14}{3}$

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 1 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 1 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.