NCERT Solutions for Class 10 Maths Chapter 3 : Pair of Linear Equations in Two Variables

Q: Q1. Explain the Graphical Method of Solutions of a Pair of Linear Equations.

Answer: Two lines are represented by the graph of a pair of linear equations with two variables. A system is considered to have a unique solution if the graphs of its two equations intersect at a single point. A system is considered to have no solution if the graphs of two equations are parallel lines. When the graphs of two equations in a system are two coincident lines, the system is said to have an unlimited number of solutions, indicating that it is consistent and dependent.

Q: Q2. How many exercises are there in Chapter 3 of Class 10 Maths?

Answer: Pair of Linear Equations in Two Variables is NCERT Class 10 Maths Chapter 3. This chapter has 7 exercises with 29 well-researched questions, including both graph-based and algebraic sums. Toppr has answers for every exercise. Subject matter specialists developed the NCERT Solutions for Class 10 Maths Chapter 3. You should download them in PDF format so that you can study them whenever and wherever you like.

Q: Q3. Where can I get the Class 10 Maths Chapter 3 solutions?

One of the scoring chapters is Class 10 Maths Chapter 3, Pair of Linear Equation in Two Variables. If you want to receive the solutions for this chapter, go to Toppr's official website. We provide NCERT Solutions for Class 10 Maths in PDF format so that you may utilize them offline. The solutions can also be downloaded via the Toppr App. All you have to do is download the app from the App/Play store, sign in, and get the study material.

Question 1

Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." (Isnt this interesting?) Represent this situation algebraically and graphically.

Medium

Solution

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Answer

Consider Aftab's age as

x

and his daughter's age as

y .

Then , seven years ago,

Aftab's age

= x - 7

His daughter's age

= y - 7

According to the question,

x - 7 = 7 (y - 7)

x - 7 = 7 y - 49

x - 7 y = - 49 + 7

x - 7 y = - 42

…(i)

After three years,
Aftab's age

= x + 3

His daughter's age

= y + 3

According to the question,

x + 3 = 3 (y + 3)

x + 3 = 3 y + 9

x - 3 y = 9 - 3

x - 3 y = 6

…(ii)

Representing equation

(i)

and

(i i)

geometrically, we plot these equations by finding points on the lines representing these two equations

x - 7 y = - 42 ⟹ x = 7 y - 42

$x = 7 y - 42$	$42$	$35$	$49$
$y$	$12$	$11$	$13$

x - 3 y = 6 ⟹ x = 3 y + 6

$x = 3 y + 6$	$42$	$36$	$48$
$y$	$12$	$10$	$14$

From the graph we can see that two lines will intersect at a point.

x - 7 y = - 42

x - 3 y = 6

On subtracting the two equations, we get

4 y = 48

or

y = 12

Substituting value of

y

in (2),

x - 36 = 6

∴ x = 42

Answer

Consider Aftab's age as

x

and his daughter's age as

y .

Then , seven years ago,

Aftab's age

= x - 7

His daughter's age

= y - 7

According to the question,

x - 7 = 7 (y - 7)

x - 7 = 7 y - 49

x - 7 y = - 49 + 7

x - 7 y = - 42

…(i)

After three years,
Aftab's age

= x + 3

His daughter's age

= y + 3

According to the question,

x + 3 = 3 (y + 3)

x + 3 = 3 y + 9

x - 3 y = 9 - 3

x - 3 y = 6

…(ii)

Representing equation

(i)

and

(i i)

geometrically, we plot these equations by finding points on the lines representing these two equations

x - 7 y = - 42 ⟹ x = 7 y - 42

$x = 7 y - 42$	$42$	$35$	$49$
$y$	$12$	$11$	$13$

x - 3 y = 6 ⟹ x = 3 y + 6

$x = 3 y + 6$	$42$	$36$	$48$
$y$	$12$	$10$	$14$

From the graph we can see that two lines will intersect at a point.

x - 7 y = - 42

x - 3 y = 6

On subtracting the two equations, we get

4 y = 48

or

y = 12

Substituting value of

y

in (2),

x - 36 = 6

∴ x = 42

Question 2

The coach of a cricket team buys

3

bats and

6

balls for

c 3900 .

Later, she buys another bat and

3

more balls of the same kind for

c 1300 .

Represent this situation algebraically and geometrically

Easy

Solution

Verified by Toppr

Answer

Reffer imagr.

let rate of bat =

x R s .

and rate of ball =

y R s .

3 x + 6 y = 3900 . . . . . (i)

and

x + 3 y = 1300 . . . . . (i i)

or

x = (1300 - 3 y)

now equation (i) can be written as

3 (1300 - 3 y) + 6 y = 3900

3900 - 9 y + 6 y = 3900

∴ y = 0

Ans:-

x = 1300

Answer

Reffer imagr.

let rate of bat =

x R s .

and rate of ball =

y R s .

3 x + 6 y = 3900 . . . . . (i)

and

x + 3 y = 1300 . . . . . (i i)

or

x = (1300 - 3 y)

now equation (i) can be written as

3 (1300 - 3 y) + 6 y = 3900

3900 - 9 y + 6 y = 3900

∴ y = 0

Ans:-

x = 1300

Question 3

The cost of

2 k g

of apples and

1 k g

of grapes on a day was found to be

R s 160

. After a month, the cost of

4 k g

of apples and

2 k g

of grapes is

R s 300

. Represent the situation algebraically and geometrically.

Medium

Solution

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Answer

$x$	$65$	$55$	$45$	$35$
$y$	$20$	$40$	$60$	$80$

Let the cost of

1

kg of apples be

x

and that of

1

kg be

y

So the algebraic representation can be as follows:

2 x + y = 160

4 x + 2 y = 300 \Rightarrow 2 x + y = 150

The situation can be represented graphically by plotting these two equations.

2 x + y = 160 \Rightarrow y = 160 - 2 x

$x$	$70$	$60$	$50$	$40$
$y = 160 - 2 x$	$20$	$40$	$60$	$80$

2 x + y = 150 \Rightarrow y = 150 - 2 x

$x$	$65$	$55$	$45$	$35$
$y = 150 - 2 x$	$20$	$40$	$60$	$80$

We can see that the lines do not intersect anywhere, i.e. they are parallel. Hence we can not arrive at a solution.

Answer

$x$	$65$	$55$	$45$	$35$
$y$	$20$	$40$	$60$	$80$

Let the cost of

1

kg of apples be

x

and that of

1

kg be

y

So the algebraic representation can be as follows:

2 x + y = 160

4 x + 2 y = 300 \Rightarrow 2 x + y = 150

The situation can be represented graphically by plotting these two equations.

2 x + y = 160 \Rightarrow y = 160 - 2 x

$x$	$70$	$60$	$50$	$40$
$y = 160 - 2 x$	$20$	$40$	$60$	$80$

2 x + y = 150 \Rightarrow y = 150 - 2 x

$x$	$65$	$55$	$45$	$35$
$y = 150 - 2 x$	$20$	$40$	$60$	$80$

We can see that the lines do not intersect anywhere, i.e. they are parallel. Hence we can not arrive at a solution.

Question 4

Form the pair of linear equations in the following problems, and find their solutions graphically.
(i)

10

students of Class X took part in a Mathematics quiz. If the number of girls is

4

more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii)

5

pencils and

7

pens together cost Rs

50,

whereas

7

pencils and

5

pens together cost Rs

46 .

Find the cost of one pencil and that of one pen.

Medium

Solution

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Answer

i) Consider number of boys

= x

and number of girls

= y .

According to the question,

x + y = 10

or,

y = 10 - x

………(i)

y = x + 4

………..(ii)

$x$	$1$	$2$	$3$	$4$
$y$	$9$	$8$	$7$	$6$

$x$	$1$	$2$	$3$	$4$
$y$	$5$	$6$	$7$	$8$

Graphical solution : Girls

= 7,

boys

= 3

(ii) Consider price of a pencil is

x

and price of a pen is

y .

According to the question,

5 x + 7 y = 50

.....(i)

7 x + 5 y = 46

.….(ii)

$x$	$1$	$2$	$3$	$4$
$y$	$7.8$	$6.4$	$5$	$4.2$

$x$	$1$	$2$	$3$	$4$
$y$	$6.4$	$5.7$	$5$	$4.2$

Graphical solution :

Price of a pencil is Rs

3

Price of a pen is Rs

5 .

Answer

i) Consider number of boys

= x

and number of girls

= y .

According to the question,

x + y = 10

or,

y = 10 - x

………(i)

y = x + 4

………..(ii)

$x$	$1$	$2$	$3$	$4$
$y$	$9$	$8$	$7$	$6$

$x$	$1$	$2$	$3$	$4$
$y$	$5$	$6$	$7$	$8$

Graphical solution : Girls

= 7,

boys

= 3

(ii) Consider price of a pencil is

x

and price of a pen is

y .

According to the question,

5 x + 7 y = 50

.....(i)

7 x + 5 y = 46

.….(ii)

$x$	$1$	$2$	$3$	$4$
$y$	$7.8$	$6.4$	$5$	$4.2$

$x$	$1$	$2$	$3$	$4$
$y$	$6.4$	$5.7$	$5$	$4.2$

Graphical solution :

Price of a pencil is Rs

3

Price of a pen is Rs

5 .

Question 5

On comparing the ratios

\frac{a _{1}}{a _{2}}, \frac{b _{1}}{b _{2}}

and

\frac{c _{1}}{c _{2}}

, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.

5 x - 4 y + 8 = 0; 7 x + 6 y - 9 = 0

A

Intersect at a point

B

Parallel

C

Coincident

D

None of these

Medium

Solution

Verified by Toppr

Answer

The given linear equation are

\Rightarrow 5 x - 4 y + 8 = 0 . . . . e q 1

\Rightarrow a_{1} = 5, b_{1} = - 4, c_{1} = 8

\Rightarrow 7 x + 6 y - 9 = 0 . . . e q 2

\Rightarrow a_{2} = 7, b_{2} = 6, c_{2} = - 9

\Rightarrow \frac{a _{1}}{a _{2}} = \frac{5}{7}

\Rightarrow \frac{b _{1}}{b _{2}} = \frac{- 4}{6}

\Rightarrow \frac{c _{1}}{c _{2}} = \frac{8}{9}

comparing

\Rightarrow \frac{a _{1}}{a _{2}}, \frac{b _{1}}{b _{2}}, \frac{c _{1}}{c _{2}}

\Rightarrow \frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

Hence,the line represented by eq1 and eq2 intersect at a point

Answer

The given linear equation are

\Rightarrow 5 x - 4 y + 8 = 0 . . . . e q 1

\Rightarrow a_{1} = 5, b_{1} = - 4, c_{1} = 8

\Rightarrow 7 x + 6 y - 9 = 0 . . . e q 2

\Rightarrow a_{2} = 7, b_{2} = 6, c_{2} = - 9

\Rightarrow \frac{a _{1}}{a _{2}} = \frac{5}{7}

\Rightarrow \frac{b _{1}}{b _{2}} = \frac{- 4}{6}

\Rightarrow \frac{c _{1}}{c _{2}} = \frac{8}{9}

comparing

\Rightarrow \frac{a _{1}}{a _{2}}, \frac{b _{1}}{b _{2}}, \frac{c _{1}}{c _{2}}

\Rightarrow \frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

Hence,the line represented by eq1 and eq2 intersect at a point

Question 6

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.
(i)

3 x + 2 y = 5; 2 x - 3 y = 7

(ii)

2 x - 3 y = 8; 4 x - 6 y = 9

(iii)

\frac{3}{2} x + \frac{5}{3} y = 7; 9 x - 10 y = 14

(iv)

5 x - 3 y = 11; - 10 x + 6 y = - 22

(v)

\frac{4}{3} x + 2 y = 8; 2 x + 3 y = 12

Medium

Solution

Verified by Toppr

Answer

(i)

\frac{a _{1}}{a _{2}} = \frac{3}{2}, \frac{b _{1}}{b _{2}} = \frac{2}{- 3}, \frac{c _{1}}{c _{2}} = \frac{5}{7}

∵ \frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

,

the lines intersect and have an unique consistent solution.

(ii)

\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b _{1}}{b _{2}} = \frac{- 3}{- 6} = \frac{1}{2}, \frac{c _{1}}{c _{2}} = \frac{8}{9}

∵ \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

,

the lines are parallel and have no solutions, i.e. the equations are an inconsistent pair

(iii)

\frac{a _{1}}{a _{2}} = \frac{\frac{3}{2}}{9} = \frac{1}{6}, \frac{b _{1}}{b _{2}} = \frac{\frac{5}{3}}{- 1 0} = - \frac{1}{6}, \frac{c _{1}}{c _{2}} = \frac{7}{1 4} = \frac{1}{2}

∵ \frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

,

the lines intersect and have an unique consistent solution.

(iv)

\frac{a _{1}}{a _{2}} = \frac{5}{- 1 0} = - \frac{1}{2}, \frac{b _{1}}{b _{2}} = \frac{- 3}{6} = - \frac{1}{2}, \frac{c _{1}}{c _{2}} = \frac{1 1}{- 2 2} = - \frac{1}{2}

∵ \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

,

the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

(v)

\frac{a _{1}}{a _{2}} = \frac{\frac{4}{3}}{2} = \frac{2}{3}, \frac{b _{1}}{b _{2}} = \frac{2}{3}, \frac{c _{1}}{c _{2}} = \frac{8}{1 2} = \frac{2}{3}

∵ \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

,

the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

Answer

(i)

\frac{a _{1}}{a _{2}} = \frac{3}{2}, \frac{b _{1}}{b _{2}} = \frac{2}{- 3}, \frac{c _{1}}{c _{2}} = \frac{5}{7}

∵ \frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

,

the lines intersect and have an unique consistent solution.

(ii)

\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b _{1}}{b _{2}} = \frac{- 3}{- 6} = \frac{1}{2}, \frac{c _{1}}{c _{2}} = \frac{8}{9}

∵ \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

,

the lines are parallel and have no solutions, i.e. the equations are an inconsistent pair

(iii)

\frac{a _{1}}{a _{2}} = \frac{\frac{3}{2}}{9} = \frac{1}{6}, \frac{b _{1}}{b _{2}} = \frac{\frac{5}{3}}{- 1 0} = - \frac{1}{6}, \frac{c _{1}}{c _{2}} = \frac{7}{1 4} = \frac{1}{2}

∵ \frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

,

the lines intersect and have an unique consistent solution.

(iv)

\frac{a _{1}}{a _{2}} = \frac{5}{- 1 0} = - \frac{1}{2}, \frac{b _{1}}{b _{2}} = \frac{- 3}{6} = - \frac{1}{2}, \frac{c _{1}}{c _{2}} = \frac{1 1}{- 2 2} = - \frac{1}{2}

∵ \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

,

the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

(v)

\frac{a _{1}}{a _{2}} = \frac{\frac{4}{3}}{2} = \frac{2}{3}, \frac{b _{1}}{b _{2}} = \frac{2}{3}, \frac{c _{1}}{c _{2}} = \frac{8}{1 2} = \frac{2}{3}

∵ \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

,

the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

Question 7

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i)

x + y = 5, 2 x + 2 y = 10

(ii)

x - y = 8, 3 x - 3 y = 16

(iii)

2 x + y - 6 = 0, 4 x - 2 y - 4 = 0

(iv)

2 x - 2 y - 2 = 0, 4 x - 4 y - 5 = 0

Medium

Solution

Verified by Toppr

Answer

(i)

x + y = 5

...(i)

2 x + 2 y = 10

...(ii)

\Rightarrow x + y = 5

\Rightarrow y = 5 - x

x	0	3
y	5	2

Plot

(0, 5)

and

(3, 5)

on graph and join them to get equation

x + y = 5

.

2 x + 2 y = 10

\Rightarrow 2 y = (10 - 2 x)

\Rightarrow y = \frac{1 0 - 2 x}{2} = 5 - x

...(iii)

x	5	2
y	0	3

So, the equation is consistent and has infinitely many solution

(ii)

x - y = 8

....(i)

3 x - 3 y = 16

....(ii)

\Rightarrow x - y = 8

\Rightarrow - x + y = - 8

\Rightarrow y = - 8 + x

\Rightarrow y = x - 8

x	8	0
y	0	-8

3 x - 3 y = 16

...(ii)

\Rightarrow 3 x = 16 + 3 y

\Rightarrow 3 x - 16 = 3 y

\Rightarrow y = \frac{3 x - 1 6}{3} \Rightarrow y = x - \frac{1 6}{3}

$x$	$\frac{1 6}{3}$	$0$
$y$	$0$	$\frac{- 1 6}{3}$

Plotting both the equation in graph, we see that the lines are parallel , so inconsistent.

(iii)

2 x + y - 6 = 0

4 x - 2 y - 4 = 0

2 x + y = 6

....(i)

4 x - 2 y = 4

...(ii)

For equation (i),

2 x + y = 6 \Rightarrow y = 6 - 2 x

$x$	$0$	$3$
$y$	$6$	$0$

Plot point

(0, 6)

and

(3, 0)

on a graph and join then to get equation

3 x + y = 6

For equation (ii),

4 x - 2 y = 4 \Rightarrow \frac{4 x - 4}{2} = y

$x$	$1$	$0$
$y$	$0$	$- 2$

Plot point

(1, 0)

and

(0, - 2)

on a graph and join them to get equation

4 x - 2 y = 0

x = 2, y = 2

is the solution of the given pairs of equation . So. solution is consistent.

(iv)

2 x - 2 y = 2

...(i)

4 x - 4 y = 5

...(ii)

2 x - 2 y = 2 \Rightarrow 2 x - 2 = 2 y

y = x - 1

$x$	$0$	$1$
$y$	$- 1$	$0$

Plot point

(0, 1)

and

(1, 0)

and join to get the equation

2 x - 2 y = 2

on a graph

4 x - 4 y = 5 \Rightarrow 4 x - 5 = 4 y

\Rightarrow \frac{4 x - 5}{4} = y

$x$	$0$	$\frac{5}{4}$
$y$	$- \frac{5}{4}$	$0$

Plot point

(0, - \frac{5}{4})

and

(\frac{5}{4}, 0)

and join them to get the equation

4 x - 4 y = 5

on a graph.

The two lines never intersect, so, the solution is inconsistent.

Answer

(i)

x + y = 5

...(i)

2 x + 2 y = 10

...(ii)

\Rightarrow x + y = 5

\Rightarrow y = 5 - x

x	0	3
y	5	2

Plot

(0, 5)

and

(3, 5)

on graph and join them to get equation

x + y = 5

.

2 x + 2 y = 10

\Rightarrow 2 y = (10 - 2 x)

\Rightarrow y = \frac{1 0 - 2 x}{2} = 5 - x

...(iii)

x	5	2
y	0	3

So, the equation is consistent and has infinitely many solution

(ii)

x - y = 8

....(i)

3 x - 3 y = 16

....(ii)

\Rightarrow x - y = 8

\Rightarrow - x + y = - 8

\Rightarrow y = - 8 + x

\Rightarrow y = x - 8

x	8	0
y	0	-8

3 x - 3 y = 16

...(ii)

\Rightarrow 3 x = 16 + 3 y

\Rightarrow 3 x - 16 = 3 y

\Rightarrow y = \frac{3 x - 1 6}{3} \Rightarrow y = x - \frac{1 6}{3}

$x$	$\frac{1 6}{3}$	$0$
$y$	$0$	$\frac{- 1 6}{3}$

Plotting both the equation in graph, we see that the lines are parallel , so inconsistent.

(iii)

2 x + y - 6 = 0

4 x - 2 y - 4 = 0

2 x + y = 6

....(i)

4 x - 2 y = 4

...(ii)

For equation (i),

2 x + y = 6 \Rightarrow y = 6 - 2 x

$x$	$0$	$3$
$y$	$6$	$0$

Plot point

(0, 6)

and

(3, 0)

on a graph and join then to get equation

3 x + y = 6

For equation (ii),

4 x - 2 y = 4 \Rightarrow \frac{4 x - 4}{2} = y

$x$	$1$	$0$
$y$	$0$	$- 2$

Plot point

(1, 0)

and

(0, - 2)

on a graph and join them to get equation

4 x - 2 y = 0

x = 2, y = 2

is the solution of the given pairs of equation . So. solution is consistent.

(iv)

2 x - 2 y = 2

...(i)

4 x - 4 y = 5

...(ii)

2 x - 2 y = 2 \Rightarrow 2 x - 2 = 2 y

y = x - 1

$x$	$0$	$1$
$y$	$- 1$	$0$

Plot point

(0, 1)

and

(1, 0)

and join to get the equation

2 x - 2 y = 2

on a graph

4 x - 4 y = 5 \Rightarrow 4 x - 5 = 4 y

\Rightarrow \frac{4 x - 5}{4} = y

$x$	$0$	$\frac{5}{4}$
$y$	$- \frac{5}{4}$	$0$

Plot point

(0, - \frac{5}{4})

and

(\frac{5}{4}, 0)

and join them to get the equation

4 x - 4 y = 5

on a graph.

The two lines never intersect, so, the solution is inconsistent.

Question 8

Half the perimeter of a rectangular garden, whose length is

4 m

more than its width, is

36 m

. Find the dimensions of the garden.

A

Length is

30 m

and Breadth is

15 m

B

Length is

20 m

and Breadth is

16 m

C

Length is

40 m

and Breadth is

30 m

D

Length is

45 m

and Breadth is

18 m

Medium

Solution

Verified by Toppr

Answer

Let the width of the garden

= x

meter

Then length=

(x + 4)

meter

Half perimeter

= 36 m

So perimeter of garden

= (2 \times 36) = 72

meters

According to the question

\Rightarrow 2 (l + b) = 72

\Rightarrow 2 (x + x + 4) = 72

\Rightarrow 2 x + 2 x + 4 = 74 \Rightarrow 4 x = 64 \Rightarrow x = 16

meters

Hence,the width of the garden

= 16

meters

The length of the garden

= (16 + 4) = 20

meters

Answer

Let the width of the garden

= x

meter

Then length=

(x + 4)

meter

Half perimeter

= 36 m

So perimeter of garden

= (2 \times 36) = 72

meters

According to the question

\Rightarrow 2 (l + b) = 72

\Rightarrow 2 (x + x + 4) = 72

\Rightarrow 2 x + 2 x + 4 = 74 \Rightarrow 4 x = 64 \Rightarrow x = 16

meters

Hence,the width of the garden

= 16

meters

The length of the garden

= (16 + 4) = 20

meters

Question 9

Given the linear equation

2 x + 3 y - 8 = 0

, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines (iii) coincident lines

Medium

Solution

Verified by Toppr

Answer

a) Intersecting lines
Solution: For intersecting line, the linear equations should meet following condition:

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

For getting another equation to meet this criterion, multiply the coefficient of x with any number and multiply the coefficient of y with any other number. A possible equation can be as follows:

4 x + 9 y - 8 = 0

(b) Parallel lines
Solution: For parallel lines, the linear equations should meet following condition:

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

For getting another equation to meet this criterion, multiply the coefficients of x and y with the same number and multiply the constant term with any other number. A possible equation can be as follows:

4 x + 6 y - 24 = 0

(c) Coincident lines
Solution: For getting coincident lines, the equations should meet following condition;

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

For getting another equation to meet this criterion, multiply the whole equation with any number. A possible equation can be as follows:

4 x + 6 y - 16 = 0

Answer

a) Intersecting lines
Solution: For intersecting line, the linear equations should meet following condition:

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

For getting another equation to meet this criterion, multiply the coefficient of x with any number and multiply the coefficient of y with any other number. A possible equation can be as follows:

4 x + 9 y - 8 = 0

(b) Parallel lines
Solution: For parallel lines, the linear equations should meet following condition:

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

For getting another equation to meet this criterion, multiply the coefficients of x and y with the same number and multiply the constant term with any other number. A possible equation can be as follows:

4 x + 6 y - 24 = 0

(c) Coincident lines
Solution: For getting coincident lines, the equations should meet following condition;

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

For getting another equation to meet this criterion, multiply the whole equation with any number. A possible equation can be as follows:

4 x + 6 y - 16 = 0

Question 10

Draw the graphs of

x - y + 1 = 0

and

3 x + 2 y - 12 = 0

. Determine the coordinates of the vertices of the triangle formed by these lines and

x

-axis and shade the triangular area.

Medium

Solution

Verified by Toppr

Answer

Given,

x - y + 1 = 0

....(1)

3 x + 2 y - 12 = 0

....(2)

put

x = 0

in (1)

0 - y + 1 = 0 \Rightarrow y = 1

(0, 1)

.....(a)

put

y = 0

in (2)

3 x + 2 (0) - 12 = 0 \to x = 4

(4, 0)

.....(b)

put

x = 0

in (2)

3 (0) + 2 y - 12 = 0 \to y = 6

(0, 6)

......(c)

put

y = 0

in (1)

x - 0 + 1 = 0 \Rightarrow x = - 1

(- 1, 0)

...(d)

From figure, the vertices of triangle are

(- 1, 0), (2, 3), (4, 0)

Answer

Given,

x - y + 1 = 0

....(1)

3 x + 2 y - 12 = 0

....(2)

put

x = 0

in (1)

0 - y + 1 = 0 \Rightarrow y = 1

(0, 1)

.....(a)

put

y = 0

in (2)

3 x + 2 (0) - 12 = 0 \to x = 4

(4, 0)

.....(b)

put

x = 0

in (2)

3 (0) + 2 y - 12 = 0 \to y = 6

(0, 6)

......(c)

put

y = 0

in (1)

x - 0 + 1 = 0 \Rightarrow x = - 1

(- 1, 0)

...(d)

From figure, the vertices of triangle are

(- 1, 0), (2, 3), (4, 0)

NCERT Solutions for Class 10 Maths Chapter 3 : Pair of Linear Equations in Two Variables

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NCERT Solutions for Class 10 Maths Chapter 3 : Pair of Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables – Brief Overview

1. Linear Equations

2. Pair of Linear Equations in Two Variables

3. Method of Solution for the above Pair of Linear Equations in Two Variables

5. Algebraic Method Of Solution Of A Pair Of Linear Equations

Frequently Asked Questions on NCERT Class 10 Maths Chapter 3 : Pair of Linear Equations in Two Variables