cbse ncert solutions for class 10 mathematics chapter 3

Class- X-CBSE-Mathematics Pair of Linear Equations in Two Variables

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CBSE NCERT Solutions for Class 10 Mathematics Chapter 3 Back of Chapter Questions

1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you werethen. Also, three years from now, I shall be three times as old as you will be.”(Isn’t this interesting?) Represent this situation algebraically and graphically.

Solution:

Let the present age of Aftab be x and present age of daughter be y.

Hence, seven years ago,

Age of Aftab = 𝑥𝑥 − 7

Age of daughter = 𝑦𝑦 − 7

Hence, as per the given condition,(𝑥𝑥 − 7) = 7(𝑦𝑦 − 7)

⟹ x − 7 = 7y − 49

⟹ x − 7y = −42………………………(i)

Three years later,

Age of Aftab = 𝑥𝑥 + 3

Age of daughter = 𝑦𝑦 + 3

Hence, as per the given condition, (𝑥𝑥 + 3) = 3(𝑦𝑦 + 3)

⇒ 𝑥𝑥 + 3 = 3𝑦𝑦 + 9

⇒ 𝑥𝑥 − 3𝑦𝑦 = 6 …………………………(ii)

Hence, equation (i) and (ii) represent given conditions algebraically as:

𝑥𝑥 − 7𝑦𝑦 = −42

𝑥𝑥 − 3𝑦𝑦 = 6

Graphical Representation:

x − 7y = −42

⇒ x = −42 + 7y

Two solutions of this equation are:

𝑥𝑥 −7 0

𝑦𝑦 5 6

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x − 3y = 6

⇒ x = 6 + 3y


𝑥𝑥 6 0

𝑦𝑦 0 −2

The graphical representation is as follows:

2. The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.

Solution:

Let the price of a bat be ₹ x and a ball be ₹ y.

Hence, we can represent algebraically the given conditions as:

3𝑥𝑥 + 6𝑦𝑦 = 3900

𝑥𝑥 + 3𝑦𝑦 = 1300

3𝑥𝑥 + 6𝑦𝑦 = 3900 ⟹ 𝑥𝑥 =3900 − 6𝑦𝑦

3






𝑥𝑥 300 100

𝑦𝑦 500 600

𝑥𝑥 + 3𝑦𝑦 = 1300 ⟹ 𝑥𝑥 = 1300 − 3𝑦𝑦


𝑥𝑥 100 −200

𝑦𝑦 400 500


3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.

Solution:

Let the cost of 1 kg of apples be ₹ 𝑥𝑥 and 1 kg grapes be ₹ y.

The given conditions can be algebraically represented as:





2𝑥𝑥 + 𝑦𝑦 = 160 (1)

4𝑥𝑥 + 2𝑦𝑦 = 300 (2)

2𝑥𝑥 + 𝑦𝑦 = 160 ⟹ 𝑦𝑦 = 160 − 2𝑥𝑥


𝑥𝑥 50 80

𝑦𝑦 60 0

4𝑥𝑥 + 2𝑦𝑦 = 300 ⟹ 𝑦𝑦 =300 − 4𝑥𝑥

2


𝑥𝑥 60 75

𝑦𝑦 30 0


♦ ♦ ♦





EXERCISE 3.2

1. 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 ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Solution:

(i) Let the number of girls in the class be x and number of boys in the class be y.

As per the given conditions

𝑥𝑥 + 𝑦𝑦 = 10 (1)

𝑥𝑥 − 𝑦𝑦 = 4 (2)

𝑥𝑥 + 𝑦𝑦 = 10 ⟹ 𝑥𝑥 = 10 − 𝑦𝑦


𝑥𝑥 0 10

𝑦𝑦 10 0

𝑥𝑥 − 𝑦𝑦 = 4 ⟹ 𝑥𝑥 = 4 + 𝑦𝑦


𝑥𝑥 2 4

𝑦𝑦 −2 0






From the graph, it can be observed that the two lines intersect each other at the point (7, 3).

So,𝑥𝑥 = 7 and y = 3.

Hence, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of one pencil be ₹ 𝑥𝑥 and one pen be ₹ 𝑦𝑦 respectively.

As per the given conditions,

5𝑥𝑥 + 7𝑦𝑦 = 50 (1)

7𝑥𝑥 + 5𝑦𝑦 = 46 (2)

5𝑥𝑥 + 7𝑦𝑦 = 50 ⟹ 𝑥𝑥 =50 − 7𝑦𝑦

5


𝑥𝑥 3 10

𝑦𝑦 5 0





7𝑥𝑥 + 5𝑦𝑦 = 46 ⟹ 𝑥𝑥 =46 − 5𝑦𝑦

7


𝑥𝑥 8 3

𝑦𝑦 −2 5


From the graph, it can be observed that the two lines intersect each other at the point (3, 5).

So,𝑥𝑥 = 3 and 𝑦𝑦 = 5.

Therefore, the cost of one pencil and one pen are ₹3 and ₹5 respectively.





2. On comparing the ratios 𝑎𝑎1𝑎𝑎2

, 𝑏𝑏1𝑏𝑏2

and 𝑐𝑐1𝑐𝑐2

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

(i) 5𝑥𝑥 − 4𝑦𝑦 + 8 = 0 7𝑥𝑥 + 6𝑦𝑦 − 9 = 0

(ii) 9𝑥𝑥 + 3𝑦𝑦 + 12 = 0 18𝑥𝑥 + 6𝑦𝑦 + 24 = 0

(iii) 6𝑥𝑥 − 3𝑦𝑦 + 10 = 0 2𝑥𝑥 − 𝑦𝑦 + 9 = 0

Solution:

(i) 5𝑥𝑥 − 4𝑦𝑦 + 8 = 0

7𝑥𝑥 + 6𝑦𝑦 − 9 = 0

Comparing pair of equations with 𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 + 𝑐𝑐1 = 0 and 𝑎𝑎2𝑥𝑥 + 𝑏𝑏2𝑦𝑦 +𝑐𝑐2 = 0, we get

𝑎𝑎1 = 5, 𝑏𝑏1 = −4, 𝑐𝑐1 = 8

𝑎𝑎2 = 7, 𝑏𝑏2 = 6, 𝑐𝑐2 = −9

𝑎𝑎1𝑎𝑎2

=57

𝑏𝑏1𝑏𝑏2

=−46

=−23

Since, 𝑎𝑎1𝑎𝑎2≠ 𝑏𝑏1

𝑏𝑏2, the given pair of equations intersect at exactly one point.

(ii) 9𝑥𝑥 + 3𝑦𝑦 + 12 = 0

18𝑥𝑥 + 6𝑦𝑦 + 24 = 0

Comparing pair of equations with 𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 + 𝑐𝑐1 = 0 and 𝑎𝑎2𝑥𝑥 + 𝑏𝑏2𝑦𝑦 +𝑐𝑐2 = 0, we get:

𝑎𝑎1 = 9, 𝑏𝑏1 = 3, 𝑐𝑐1 = 12

𝑎𝑎2 = 18, 𝑏𝑏2 = 6, 𝑐𝑐2 = 24

𝑎𝑎1𝑎𝑎2

=9

18=

12

𝑏𝑏1𝑏𝑏2

=36

=12

𝑐𝑐1𝑐𝑐2

= 1224

= 12Since,𝑎𝑎1

𝑎𝑎2= 𝑏𝑏1

𝑏𝑏2= 𝑐𝑐1

𝑐𝑐2 the given pair of equation are coincident.





(iii) 6x − 3y + 10 = 0

2x − y + 9 = 0


𝑎𝑎1 = 6, 𝑏𝑏1 = −3, 𝑐𝑐1 = 10

𝑎𝑎2 = 2, 𝑏𝑏2 = −1, 𝑐𝑐2 = 9

𝑎𝑎1𝑎𝑎2

=62

=31

𝑏𝑏1𝑏𝑏2

=−3−1

=31

𝑐𝑐1𝑐𝑐2

=109

Since, 𝑎𝑎1𝑎𝑎2

= 𝑏𝑏1𝑏𝑏2≠ 𝑐𝑐1

𝑐𝑐2 the given pair of equation are parallel to each other.

3. On comparing the ratios 𝑎𝑎1𝑎𝑎2


and 𝑐𝑐1𝑐𝑐2

, find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3𝑥𝑥 + 2𝑦𝑦 = 5 2𝑥𝑥 − 3𝑦𝑦 = 7

(ii) 2𝑥𝑥 − 3𝑦𝑦 = 8 4𝑥𝑥 − 6𝑦𝑦 = 9

(iii) 32𝑥𝑥 +

53𝑦𝑦 = 7 9𝑥𝑥 − 10𝑦𝑦 = 14

(iv) 5𝑥𝑥 − 3𝑦𝑦 = 11 −10𝑥𝑥 + 6𝑦𝑦 = −22

(v) 43𝑥𝑥 + 2𝑦𝑦 = 8 2𝑥𝑥 + 3𝑦𝑦 = 12

Solution:

(i) 3𝑥𝑥 + 2𝑦𝑦 = 5

2𝑥𝑥 − 3𝑦𝑦 = 7


𝑎𝑎1 = 3, 𝑏𝑏1 = 2, 𝑐𝑐1 = −5

𝑎𝑎2 = 2, 𝑏𝑏2 = −3, 𝑐𝑐2 = −7





𝑎𝑎1𝑎𝑎2

=32

,𝑏𝑏1𝑏𝑏2

=2−3

= −23

,𝑐𝑐1𝑐𝑐2

=−5−7

=57


𝑏𝑏2 the given pair of equation has only one solution.

Hence, the pair of linear equations is consistent.

(ii) 2𝑥𝑥 − 3𝑦𝑦 = 8

4𝑥𝑥 − 6𝑦𝑦 = 9


𝑎𝑎1 = 2, 𝑏𝑏1 = −3, 𝑐𝑐1 = −8

𝑎𝑎2 = 4, 𝑏𝑏2 = −6, 𝑐𝑐2 = −9

𝑎𝑎1𝑎𝑎2

=24

=12

,𝑏𝑏1𝑏𝑏2

=−3−6

=12

,𝑐𝑐1𝑐𝑐2

=−8−9

=89



𝑐𝑐2 the given pair of equation has no solution.

Hence, the pair of linear equations is inconsistent.

(iii) 32𝑥𝑥 + 5

3𝑦𝑦 = 7

9𝑥𝑥 − 10𝑦𝑦 = 14


𝑎𝑎1 =32

, 𝑏𝑏1 =53

, 𝑐𝑐1 = −7

𝑎𝑎2 = 9, 𝑏𝑏2 = −10, 𝑐𝑐2 = −14

𝑎𝑎1𝑎𝑎2

=329

=16

,𝑏𝑏1𝑏𝑏2

=53

−10= −

16

,𝑐𝑐1𝑐𝑐2

=−7−14

=12


𝑏𝑏2 the given pair of equation has only one solution






(iv) 5𝑥𝑥 − 3𝑦𝑦 = 11

−10𝑥𝑥 + 6𝑦𝑦 = −22


𝑎𝑎1 = 5, 𝑏𝑏1 = −3, 𝑐𝑐1 = −11

𝑎𝑎2 = −10, 𝑏𝑏2 = 6, 𝑐𝑐2 = 22

𝑎𝑎1𝑎𝑎2

=5

−10= −

12

,𝑏𝑏1𝑏𝑏2

=−36

= −12

,𝑐𝑐1𝑐𝑐2

=−1122

= −12


= 𝑏𝑏1𝑏𝑏2

= 𝑐𝑐1𝑐𝑐2

the given pair of equation has infinite number of solution.


(iv) 43𝑥𝑥 + 2𝑦𝑦 = 8

2𝑥𝑥 + 3𝑦𝑦 = 12


𝑎𝑎1 = 43

, 𝑏𝑏1 = 2, 𝑐𝑐1 = −8

𝑎𝑎2 = 2, 𝑏𝑏2 = 3, 𝑐𝑐2 = −12

𝑎𝑎1𝑎𝑎2

=432

=23

,𝑏𝑏1𝑏𝑏2

=23

,𝑐𝑐1𝑐𝑐2

=−8−12

=23




the given pair of equation has infinite number of Solution.


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

(i) 𝑥𝑥 + 𝑦𝑦 = 5 2𝑥𝑥 + 2𝑦𝑦 = 10





(ii) 𝑥𝑥 − 𝑦𝑦 = 8 3𝑥𝑥 − 3𝑦𝑦 = 16

(iii) 2𝑥𝑥 + 𝑦𝑦 − 6 = 0 4𝑥𝑥 − 2𝑦𝑦 − 4 = 0

(iv) 2𝑥𝑥 − 2𝑦𝑦 − 2 = 0 4𝑥𝑥 − 4𝑦𝑦 − 5 = 0

Solution:

(i) 𝑥𝑥 + 𝑦𝑦 = 5

2𝑥𝑥 + 2𝑦𝑦 = 10


𝑎𝑎1 = 1, 𝑏𝑏1 = 1, 𝑐𝑐1 = −5

𝑎𝑎2 = 2, 𝑏𝑏2 = 2, 𝑐𝑐2 = −10

𝑎𝑎1𝑎𝑎2

=12

,𝑏𝑏1𝑏𝑏2

=12

,𝑐𝑐1𝑐𝑐2

=−5−10

=12




the given pair of linear equation has infinite number of Solution.


Now, x + y = 5 ⟹ x = 5 − y


𝑥𝑥 5 0

𝑦𝑦 0 5

2𝑥𝑥 + 2𝑦𝑦 = 10 ⇒ 𝑥𝑥 =10 − 2𝑦𝑦

2


𝑥𝑥 4 3

𝑦𝑦 1 2

Thus, the graphical representation is as follows:





In the graph, it is observed that the two lines coincide. Hence, the given pair of equations has infinite solutions.

Let x = t, then y = 5 − t. So, the ordered pair (t, 5 − t), where t is a constant, is the solution of the given pair of linear equations.

(ii) 𝑥𝑥 − 𝑦𝑦 = 8

3𝑥𝑥 − 3𝑦𝑦 = 16


𝑎𝑎1 = 1, 𝑏𝑏1 = −1, 𝑐𝑐1 = −8

𝑎𝑎2 = 3, 𝑏𝑏2 = −3, 𝑐𝑐2 = −16

𝑎𝑎1𝑎𝑎2

=13

,𝑏𝑏1𝑏𝑏2

=−1−3

=13

,𝑐𝑐1𝑐𝑐2

=−8−16

=12



𝑐𝑐2 the given pair of linear equation has no solution.






(iii) 2𝑥𝑥 + 𝑦𝑦 − 6 = 0

4𝑥𝑥 − 2𝑦𝑦 − 4 = 0


𝑎𝑎1 = 2, 𝑏𝑏1 = 1, 𝑐𝑐1 = −6

𝑎𝑎2 = 4, 𝑏𝑏2 = −2, 𝑐𝑐2 = −4

𝑎𝑎1𝑎𝑎2

=24

=12

,𝑏𝑏1𝑏𝑏2

=1−2

= −12

,𝑐𝑐1𝑐𝑐2

=−6−4

=32

Since 𝑎𝑎1𝑎𝑎2≠ 𝑏𝑏1

𝑏𝑏2, the given pair of linear equation has only one solution.


Now, 2𝑥𝑥 + 𝑦𝑦 − 6 = 0 ⇒ 𝑦𝑦 = 6 − 2𝑥𝑥


𝑥𝑥 0 3

𝑦𝑦 6 0

4𝑥𝑥 − 2𝑦𝑦 − 4 = 0 ⇒ 𝑦𝑦 =4𝑥𝑥 − 4

2


𝑥𝑥 1 0

𝑦𝑦 0 −2

Hence, the graphical representation is as follows:





In the graph, it is observed that the two lines intersect each other at the point (2, 2). Hence, the solution of the given pair of equations is (2, 2).

(iv) 2𝑥𝑥 − 2𝑦𝑦 − 2 = 0

4𝑥𝑥 − 4𝑦𝑦 − 5 = 0


𝑎𝑎1 = 2, 𝑏𝑏1 = −2, 𝑐𝑐1 = −2

𝑎𝑎2 = 4, 𝑏𝑏2 = −4, 𝑐𝑐2 = −5

𝑎𝑎1𝑎𝑎2

=24

=12

,𝑏𝑏1𝑏𝑏2

=−2−4

=12

,𝑐𝑐1𝑐𝑐2

=−2−5

=25





Since 𝑎𝑎1𝑎𝑎2


𝑐𝑐2, the given pair of linear equation has no solution


5. 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.

Solution:

Let the length of the rectangular garden be 𝑥𝑥 and width be 𝑦𝑦.

As per the given conditions,

𝑥𝑥 − 𝑦𝑦 = 4

𝑥𝑥 + 𝑦𝑦 = 36

𝑥𝑥 − 𝑦𝑦 = 4 ⇒ 𝑥𝑥 = 𝑦𝑦 + 4


𝑥𝑥 0 4

𝑦𝑦 −4 0

𝑥𝑥 + 𝑦𝑦 = 36


𝑥𝑥 0 36

𝑦𝑦 36 0

Hence, the graphical representation is as follows:





From the graph, it can be observed that the two lines intersect each other at the point (20, 16). So, 𝑥𝑥 = 20 and 𝑦𝑦 = 16.

Thus, the length and width of the rectangular garden is 20 m and 16 m respectively.

6. Given the linear equation 2𝑥𝑥 + 3𝑦𝑦 − 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

Solution:

(i) For the two lines 𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 + 𝑐𝑐1 = 0 and a2x + b2y + c2 = 0, to be intersecting, we must have 𝑎𝑎1

𝑎𝑎2≠ 𝑏𝑏1

𝑏𝑏2

Hence, the other linear equation can be 3x + 7y − 15 = 0

𝑎𝑎1𝑎𝑎2

=23

,𝑏𝑏1𝑏𝑏2

=37

,𝑐𝑐1𝑐𝑐2

=−8−15

=8

15





(ii) For the two lines 𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 + 𝑐𝑐1 = 0 and 𝑎𝑎2𝑥𝑥 + 𝑏𝑏2𝑦𝑦 + 𝑐𝑐2 = 0, to be parallel, we must have 𝑎𝑎1


𝑏𝑏2≠ 𝑐𝑐1

𝑐𝑐2

Hence, the other linear equation can be 4𝑥𝑥 + 6𝑦𝑦 + 9 = 0, as

𝑎𝑎1𝑎𝑎2

=24

=12

,𝑏𝑏1𝑏𝑏2

=36

=12

,𝑐𝑐1𝑐𝑐2

=−89

(iii) For the two lines 𝑎𝑎1𝑥𝑥 + 𝑏𝑏1𝑦𝑦 + 𝑐𝑐1 = 0 and 𝑎𝑎2𝑥𝑥 + 𝑏𝑏2𝑦𝑦 + 𝑐𝑐2 = 0, to be parallel, we must have 𝑎𝑎1


𝑏𝑏2= 𝑐𝑐1

𝑐𝑐2

So, the other linear equation can be 6𝑥𝑥 + 9𝑦𝑦 − 24 = 0,

as 𝑎𝑎1𝑎𝑎2

= 26

= 13


= 39

= 13

, 𝑐𝑐1𝑐𝑐2

= −8−24

= 13

7. Draw the graphs of the equations 𝑥𝑥 − 𝑦𝑦 + 1 = 0 and 3𝑥𝑥 + 2𝑦𝑦 − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis and shade the triangular region.

Solution:

𝑥𝑥 − 𝑦𝑦 + 1 = 0 ⇒ 𝑥𝑥 = 𝑦𝑦 − 1


𝑥𝑥 0 −1

𝑦𝑦 1 0

3x + 2y − 12 = 0 ⇒ x =12 − 2y

3


𝑥𝑥 4 0

𝑦𝑦 0 6





Now, we have drawn these two equations on graph. Shaded part represents the triangle formed by given two lines and the x −axis:

In the graph, we can observe that the coordinates of the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

♦ ♦ ♦

EXERCISE 3.3

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

(i) 𝑥𝑥 + 𝑦𝑦 = 14 𝑥𝑥 − 𝑦𝑦 = 4

(ii) 𝑠𝑠 − 𝑡𝑡 = 3 𝑠𝑠3

+𝑡𝑡2

= 6

(iii) 3𝑥𝑥 − 𝑦𝑦 = 3 9𝑥𝑥 − 3𝑦𝑦 = 9

(iv) 0.2𝑥𝑥 + 0.3𝑦𝑦 = 1.3 0.4𝑥𝑥 + 0.5𝑦𝑦 = 2.3

(v) √2 𝑥𝑥 + √3 𝑦𝑦 = 0 √3 𝑥𝑥 − √8 𝑦𝑦 = 0

(vi) 3𝑥𝑥2−

5𝑦𝑦3

= −2 𝑥𝑥3

+𝑦𝑦2

=136





Solution;

(i) 𝑥𝑥 + 𝑦𝑦 = 14 … (i)

𝑥𝑥 − 𝑦𝑦 = 4 … (ii)

From (i), we get:

𝑥𝑥 = 14 − 𝑦𝑦… (iii)

Substituting the value of 𝑥𝑥 in equation (ii), we get:

(14 − 𝑦𝑦) − 𝑦𝑦 = 4

⇒ 14 − 2𝑦𝑦 = 4

⇒ 10 = 2𝑦𝑦

⇒ 𝑦𝑦 = 5

Substituting the value of 𝑦𝑦 in equation (iii), we get:

𝑥𝑥 = 9 ∴ 𝑥𝑥 = 9,𝑦𝑦 = 5

(ii) 𝑠𝑠 − 𝑡𝑡 = 3 … (i) 𝑠𝑠3

+𝑡𝑡2

= 6 … (ii)

From (i), we get:

(iii) 𝑠𝑠 = 𝑡𝑡 + 3 … (iii)

Substituting the value of 𝑠𝑠 in equation (iii), we get:𝑡𝑡+33

+ 𝑡𝑡2

= 6

⇒ 2𝑡𝑡 + 6 + 3𝑡𝑡 = 36

⇒ 5𝑡𝑡 = 30

⇒ 𝑡𝑡 = 6

Substituting the value of t in equation (iii), we get:

𝑠𝑠 = 9

∴ 𝑠𝑠 = 9, 𝑡𝑡 = 6

(iv) 3𝑥𝑥 − 𝑦𝑦 = 3 … (i)

9𝑥𝑥 − 3𝑦𝑦 = 9 … (ii)

From (I), we get





𝑦𝑦 = 3𝑥𝑥 − 3 … (iii)

Substituting the value of 𝑦𝑦 in equation (ii), we get:

(v) 9𝑥𝑥 − 3(3𝑥𝑥 − 3) = 9

⇒ 9𝑥𝑥 − 9𝑥𝑥 = 9

⇒9 = 9

This is always true.

Hence, the given pair of equations has infinitely many solutions and variables are related as

𝑦𝑦 = 3𝑥𝑥 − 3

(vi) 0.2𝑥𝑥 + 0.3𝑦𝑦 = 1.3 … (i)

0.4𝑥𝑥 + 0.5𝑦𝑦 = 2.3 … (ii)

From (i), we get:

𝑥𝑥 =1.3 − 0.3𝑦𝑦

0.2 … (iii)


0.4 �1.3 − 0.3𝑦𝑦

0.2� + 0.5𝑦𝑦 = 2.3

⇒ 2.6 − 0.6𝑦𝑦 + 0.5𝑦𝑦 = 2.3

⇒ 2.6 − 2.3 = 0.1𝑦𝑦

⇒ 0.3 = 0.1𝑦𝑦

⇒ 𝑦𝑦 = 3


𝑥𝑥 =1.3 − 0.3 × 3

0.2=

0.40.2

= 2

∴ 𝑥𝑥 = 2,𝑦𝑦 = 3

(v) √2𝑥𝑥 + √3𝑦𝑦 = 0. . . (i)

√3𝑥𝑥 − √8𝑦𝑦 = 0. . . (ii)

From equation (i), we get:





𝑥𝑥 =−√3𝑦𝑦√2

. . . (iii)


√3�−√3𝑦𝑦√2

� − √8𝑦𝑦 = 0

⇒ −3𝑦𝑦√2

− 2√2𝑦𝑦 = 0

⇒ 𝑦𝑦 �−3√2

− 2√2� = 0

⇒ 𝑦𝑦 = 0

Substituting the value of 𝑦𝑦 in equation (iii),we get:

𝑥𝑥 = 0

∴ 𝑥𝑥 = 0,𝑦𝑦 = 0

(vi) 32𝑥𝑥 − 5

3𝑦𝑦 = −2. . . (i)

𝑥𝑥3

+𝑦𝑦2

=136

. . . (ii)

From equation (i), we get:

9𝑥𝑥 − 10𝑦𝑦 = −12

⇒ 𝑥𝑥 =−12 + 10𝑦𝑦

9 … (iii)

Substituting the value of 𝑥𝑥 in equation (ii), we get: −12+10𝑦𝑦

93

+𝑦𝑦2

=136

⇒−24 + 20𝑦𝑦 + 27y

54=

136

⇒ −24 + 47𝑦𝑦 = 13 × 9

⇒ 47𝑦𝑦 = 117 + 24

⇒ 47𝑦𝑦 = 141

⇒ 𝑦𝑦 = 3

Substituting the value of y in equation (iii), we get:





𝑥𝑥 =−12 + 10 × 3

9= 2

∴ 𝑥𝑥 = 2,𝑦𝑦 = 3

2. Solve 2𝑥𝑥 + 3𝑦𝑦 = 11 and 2𝑥𝑥 − 4𝑦𝑦 = −24 and hence find the value of ‘𝑚𝑚’ for which 𝑦𝑦 = 𝑚𝑚𝑥𝑥 + 3.

Solution:

2𝑥𝑥 + 3𝑦𝑦 = 11 . . . (i)

2𝑥𝑥 − 4𝑦𝑦 = −24. . . (ii)

Form equation (i), we get:

𝑥𝑥 =11 − 3𝑦𝑦

2 … (iii)


2 �11 − 3𝑦𝑦

2� − 4𝑦𝑦 = −24

⇒ 11 − 3𝑦𝑦 − 4𝑦𝑦 = −24

⇒ −7𝑦𝑦 = −35

⇒ 𝑦𝑦 = 5


𝑥𝑥 =11 − 3 × 5

2= −2

∴ 𝑥𝑥 = −2,𝑦𝑦 = 5

Given, 𝑦𝑦 = 𝑚𝑚𝑥𝑥 + 3

⇒ 5 = −2𝑚𝑚 + 3

⇒ −2𝑚𝑚 = 2

⇒ 𝑚𝑚 = −1

Hence, 𝑚𝑚 = −1.

3. Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.





(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(ii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(vii) A fraction becomes 911

, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5

6. Find the fraction.

(viii) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution:

(i) Let two numbers are x and y such that y > x.

As per the question:

𝑦𝑦 = 3𝑥𝑥 . . . . (1)

𝑦𝑦 − 𝑥𝑥 = 26 . . . . (2)

On substituting the value of 𝑦𝑦 from equation (1) into equation (2), we get

26 + 𝑥𝑥 = 3𝑥𝑥

⇒ 3𝑥𝑥 − 𝑥𝑥 = 26

⇒ x = 13 . . . (3)

Substituting the value of 𝑥𝑥 in equation (1), we get

y = 39

Hence, two numbers are 13 and 39.

(ii) Let larger angle be x and smaller angle be 𝑦𝑦

As we know, the sum of supplementary angles is always 180°.

Hence, as per the question,

𝑥𝑥 + 𝑦𝑦 = 180 … (1)

𝑥𝑥 − 𝑦𝑦 = 18 … (2)





From (1), we get

𝑥𝑥 = 180° − 𝑦𝑦… (3)

Substituting 𝑥𝑥 in equation (2), we obtain

180o − 𝑦𝑦 − 𝑦𝑦 = 18°

⇒ 162o = 2𝑦𝑦

⇒ 81o = 𝑦𝑦 … (4)

Putting this in equation (3), we obtain

𝑥𝑥 = 180𝑜𝑜 − 81𝑜𝑜 = 99𝑜𝑜

Hence, the angles are 99and 81°.

(iii) Let the cost of a bat be 𝑥𝑥 and a ball be 𝑦𝑦

As per the given information,

7𝑥𝑥 + 6𝑦𝑦 = 3800 … (1)

3𝑥𝑥 + 5𝑦𝑦 = 1750 … (2)

From (1), we get

𝑦𝑦 =3800 − 7𝑥𝑥

6 … (3)

Substituting 𝑥𝑥 in equation (2), we get

3𝑥𝑥 + 5 �3800 − 7𝑥𝑥

6� = 1750

⇒ 3𝑥𝑥 +9500

3−

35𝑥𝑥6

= 1750

⇒ 3𝑥𝑥 −35𝑥𝑥

6= 1750 −

95003

⇒18𝑥𝑥 − 35𝑥𝑥

6=

5250 − 95003

⇒ −17𝑥𝑥

6=−4250

3

⇒ −17𝑥𝑥 = −8500

⇒ 𝑥𝑥 = 500 … (4)

Substituting this in equation (3), we get

𝑦𝑦 =3800 − 7 × 500

6= 50





Hence, the cost of a bat is ₹ 500 and that of a ball is ₹ 50.

(iv) Let the fixed charge be ₹ x and charge per Km be ₹ y.

𝑥𝑥 + 10𝑦𝑦 = 105 … (1)

𝑥𝑥 + 15𝑦𝑦 = 155 … (2)

From (1), we get

𝑥𝑥 = 105 − 10𝑦𝑦 … (3)

Substituting this in equation (2), we get

105 − 10𝑦𝑦 + 15𝑦𝑦 = 155

⇒ 5𝑦𝑦 = 50

⇒ 𝑦𝑦 = 10 … (4)

Putting 𝑦𝑦 in equation (3), we get

𝑥𝑥 = 105 − 10 × 10

𝑥𝑥 = 5

Hence, fixed charge = ₹ 5

And per Km charge = ₹ 10

Hence, charge for 25 km = 𝑥𝑥 + 25𝑦𝑦

= 5 + 250 = ₹ 255

(v) Let the fraction be 𝑥𝑥𝑦𝑦

As per the given information, 𝑥𝑥+2𝑦𝑦+2

= 911

⇒ 11𝑥𝑥 + 22 = 9𝑦𝑦 + 18

⇒ 11𝑥𝑥 − 9𝑦𝑦 = −4 … (1)

𝑥𝑥 + 3𝑦𝑦 + 3

=56

⇒ 6𝑥𝑥 + 18 = 5𝑦𝑦 + 15

⇒ 6𝑥𝑥 − 5𝑦𝑦 = −3 … (2)

From equation (1), we get 𝑥𝑥 = −4+9𝑦𝑦11

… (3)


6 �−4 + 9𝑦𝑦

11� − 5𝑦𝑦 = −3





⇒ −24 + 54𝑦𝑦 − 55𝑦𝑦 = −33

⇒ −𝑦𝑦 = −9

⇒ 𝑦𝑦 = 9 … (4)

Substituting 𝑦𝑦 in equation (3), we obtain

𝑥𝑥 =−4 + 81

11= 7

Hence, the fraction is 79.

(vi) Let the age of Jacob be 𝑥𝑥 and the age of his son be 𝑦𝑦

As per the given information,

(𝑥𝑥 + 5) = 3(𝑦𝑦 + 5)

⇒ 𝑥𝑥 − 3𝑦𝑦 = 10 … (1)

(𝑥𝑥 − 5) = 7(𝑦𝑦 − 5)

⇒ 𝑥𝑥 − 7𝑦𝑦 = −30 … (2)

From (1), we get

𝑥𝑥 = 3𝑦𝑦 + 10 … (3)


3𝑦𝑦 + 10 − 7𝑦𝑦 = −30

⇒ −4𝑦𝑦 = −40

⇒ 𝑦𝑦 = 10 … (4)

Substituting 𝑦𝑦 in equation (3), we get

𝑥𝑥 = 3 × 10 + 10 = 40

Hence, the present age of Jacob is 40 years and the present age of his son is 10 years.

♦ ♦ ♦

EXERCISE 3.4

1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) 𝑥𝑥 + 𝑦𝑦 = 5 and 2𝑥𝑥 − 3𝑦𝑦 = 4

(ii) 3𝑥𝑥 + 4𝑦𝑦 = 10 and 2𝑥𝑥 − 2𝑦𝑦 = 2





(iii) 3𝑥𝑥 − 5𝑦𝑦 − 4 = 0 and 9𝑥𝑥 = 2𝑦𝑦 + 7

(iv) 𝑥𝑥2

+ 2𝑦𝑦3

= −1 and 𝑥𝑥 − 𝑦𝑦3

= 3

Solution:

(i) Elimination method:

𝑥𝑥 + 𝑦𝑦 = 5 … (1)

2𝑥𝑥 − 3𝑦𝑦 = 4 … (2)

Multiplying equation (1) by 3, we get:

3𝑥𝑥 + 3𝑦𝑦 = 15 … (3)

Adding equation (2) and (3), we get:

5𝑥𝑥 = 19

⇒ 𝑥𝑥 =195

Substituting 𝑥𝑥 in equation (1), we get:

𝑦𝑦 = 5 −195

=65

∴ 𝑥𝑥 =195

,𝑦𝑦 =65

Substitution method:

From equation (1), we get:

𝑥𝑥 = 5 − 𝑦𝑦 … (4)

Putting 𝑥𝑥 in equation (2), we get:

2(5 − 𝑦𝑦) − 3𝑦𝑦 = 4

−5𝑦𝑦 = −6

𝑦𝑦 =65

Putting 𝑦𝑦 in equation (4), we get:

𝑥𝑥 = 5 −65

=195

Hence, 𝑥𝑥 = 65

,𝑦𝑦 = 195

(ii) Elimination method:





3𝑥𝑥 + 4𝑦𝑦 = 10 … (1)

2𝑥𝑥 − 2𝑦𝑦 = 2 … (2)


4𝑥𝑥 − 4𝑦𝑦 = 4 . . . (3)


7𝑥𝑥 = 14

⇒ 𝑥𝑥 = 2

Putting x in equation (1), we get:

6 + 4𝑦𝑦 = 10

⇒ 4𝑦𝑦 = 4

⇒ 𝑦𝑦 = 1

Hence, 𝑥𝑥 = 2,𝑦𝑦 = 1



𝑥𝑥 = 𝑦𝑦 + 1 … (4)


3(𝑦𝑦 + 1) + 4𝑦𝑦 = 10

⇒ 7𝑦𝑦 = 7

⇒ 𝑦𝑦 = 1

Putting y in equation (4), we get:

𝑥𝑥 = 1 + 1 = 2

∴ 𝑥𝑥 = 2,𝑦𝑦 = 1

(iii) Elimination method:

3𝑥𝑥 − 5𝑦𝑦 − 4 = 0 … (1)

9𝑥𝑥 = 2𝑦𝑦 + 7

⇒ 9𝑥𝑥 − 2𝑦𝑦 − 7 = 0 … (2)






9𝑥𝑥 − 15𝑦𝑦 − 12 = 0 … (3)

Subtracting equation (2) from equation (3), we get:

13𝑦𝑦 = −5

⇒ 𝑦𝑦 = −5

13


3𝑥𝑥 +2512

− 4 = 0

⇒ 𝑥𝑥 =9

13

∴ 𝑥𝑥 =9

13,𝑦𝑦 = −

513



𝑥𝑥 =5𝑦𝑦 + 4

3 … (4)


9 �5𝑦𝑦 + 4

3� − 2𝑦𝑦 − 7 = 0

13𝑦𝑦 = −5

𝑦𝑦 = −5

13


𝑥𝑥 =5 �− 5

13� + 4

3

𝑥𝑥 =9

13

∴ 𝑥𝑥 =9

13,𝑦𝑦 =

−513

(iv) Elimination method:





𝑥𝑥2

+2𝑦𝑦3

= −1 … (1)

𝑥𝑥 −𝑦𝑦3

= 3 … (2)


2𝑥𝑥 −2𝑦𝑦3

= 6 … (3)


5𝑥𝑥2

= 5

⇒ 𝑥𝑥 = 2


1 +2𝑦𝑦3

= −1

⇒ 𝑦𝑦 = −3

∴ 𝑥𝑥 = 2,𝑦𝑦 = −3



𝑦𝑦 = 3𝑥𝑥 − 9 … (3)


𝑥𝑥2

+2(3𝑥𝑥 − 9)

3= −1

⇒ 3𝑥𝑥 + 4(3𝑥𝑥 − 9) = −6

⇒ 15𝑥𝑥 = 30

⇒ 𝑥𝑥 = 2


𝑦𝑦 = 6 − 9 = −3

∴ 𝑥𝑥 = 2,𝑦𝑦 = −3

Note: Solution must be same in both the cases.





2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1

2 if we only add 1 to the denominator.

What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iv) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

(iv) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

(i) Let the fraction be 𝑥𝑥𝑦𝑦

𝑥𝑥 + 1𝑦𝑦 − 1

= 1

⇒ 𝑥𝑥 − 𝑦𝑦 = −2 … (1)

𝑥𝑥𝑦𝑦 + 1

=12

⇒ 2𝑥𝑥 − 𝑦𝑦 = 1 … (2)

As per the question, subtracting equation (1) from equation (2), we get:𝑥𝑥 = 3


6 − 𝑦𝑦 = 1

⇒ −𝑦𝑦 = −5

⇒ 𝑦𝑦 = 5

Hence, the fraction is 35.





(ii) Let present age of Nuri be x and Sonu be y.

As per the question,

(𝑥𝑥 − 5) = 3(𝑦𝑦 − 5)

⇒ 𝑥𝑥 − 3𝑦𝑦 = −10. . . (1)

(𝑥𝑥 + 10) = 2(𝑦𝑦 + 10)

⇒ 𝑥𝑥 − 2𝑦𝑦 = 10 … (2)


y = 20

Putting the value of y in equation (2), we get:

𝑥𝑥 − 40 = 10

⇒ 𝑥𝑥 = 50

Hence, the age of Nuri is 50 years and the age of Sonu is 20 years.

(iii) Let the unit digit of the number be 𝑥𝑥 and tens digit of the number be 𝑦𝑦 respectively.

Hence, number = 10𝑦𝑦 + 𝑥𝑥

After reversing the digits, number = 10𝑥𝑥 + 𝑦𝑦


𝑥𝑥 + 𝑦𝑦 = 9 … (1)

9(10𝑦𝑦 + 𝑥𝑥) = 2(10𝑥𝑥 + 𝑦𝑦)

⇒ 88𝑦𝑦 − 11𝑥𝑥 = 0

⇒ − 𝑥𝑥 + 8𝑦𝑦 = 0 … (2)

Adding equations (1) and (2), we get:

9𝑦𝑦 = 9

⇒ 𝑦𝑦 = 1


𝑥𝑥 = 8

Hence, the number is 10𝑦𝑦 + 𝑥𝑥 = 10 × 1 + 8 = 18

(v) Let the number of ₹ 50 notes be 𝑥𝑥 and number of ₹ 100 notes be y.

(vi) As per the question,





𝑥𝑥 + 𝑦𝑦 = 25 … (1)

50𝑥𝑥 + 100𝑦𝑦 = 2000

⇒ 𝑥𝑥 + 2𝑦𝑦 = 40 … (2)


𝑦𝑦 = 15


𝑥𝑥 = 10

Hence, Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.

(vii) Let the fixed charge for first three days be ₹ 𝑥𝑥 and each day charge thereafter be ₹ 𝑦𝑦.


𝑥𝑥 + 4𝑦𝑦 = 27 . . . (1)

𝑥𝑥 + 2𝑦𝑦 = 21 . . . (2)


2𝑦𝑦 = 6

⇒ 𝑦𝑦 = 3


𝑥𝑥 + 6 = 21

⇒ 𝑥𝑥 = 15

Hence, the fixed charge is ₹ 15 and each day charge thereafter is ₹ 3.

♦ ♦ ♦

EXERCISE 3.5

1. Which of the following pair of linear equation has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) 𝑥𝑥 − 3𝑦𝑦 − 3 = 0 3𝑥𝑥 − 9𝑦𝑦 − 2 = 0

(ii) 2𝑥𝑥 + 𝑦𝑦 = 5 3𝑥𝑥 + 3𝑦𝑦 = 8





(iii) 3𝑥𝑥 − 5𝑦𝑦 = 20 6𝑥𝑥 − 10𝑦𝑦 = 40

(iv) 𝑥𝑥 − 3𝑦𝑦 − 7 = 0 3𝑥𝑥 − 3𝑦𝑦 − 15 = 0

Solution:

(i) 𝑥𝑥 − 3𝑦𝑦 − 3 = 0

3𝑥𝑥 − 9𝑦𝑦 − 2 = 0

Here, 𝑎𝑎1 = 1, 𝑏𝑏1 = −3, 𝑐𝑐1 = −3

𝑎𝑎2 = 3, 𝑏𝑏2 = −9, 𝑐𝑐2 = −2

𝑎𝑎1𝑎𝑎2

=13

,𝑏𝑏1𝑏𝑏2

=−3−9

=13

,𝑐𝑐1𝑐𝑐2

=−3−2

=32



𝑐𝑐2

Hence, the given pair of equations has no solution.

(ii) 2𝑥𝑥 + 𝑦𝑦 = 5

⇒ 2𝑥𝑥 + 𝑦𝑦 − 5 = 0 … (1)

3𝑥𝑥 + 2𝑦𝑦 = 8

⇒ 3𝑥𝑥 + 2𝑦𝑦 − 8 = 0 … (2)

Here, 𝑎𝑎1 = 2, 𝑏𝑏1 = 1, 𝑐𝑐1 = −5

𝑎𝑎2 = 3, 𝑏𝑏2 = 2, 𝑐𝑐2 = −8

𝑎𝑎1𝑎𝑎2

=23

,𝑏𝑏1𝑏𝑏2

=12

,𝑐𝑐1𝑐𝑐2

=−5−8

=58


𝑏𝑏2

Hence, the given pair of equations has unique solution.

Now, using cross-multiplication method we know that,

𝑥𝑥𝑏𝑏1𝑐𝑐2 − 𝑏𝑏2𝑐𝑐1

=𝑦𝑦

𝑐𝑐1𝑎𝑎2 − 𝑐𝑐2𝑎𝑎1=

1𝑎𝑎1𝑏𝑏2 − 𝑎𝑎2𝑏𝑏1

⇒𝑥𝑥

−8 − (−10) =𝑦𝑦

−15 − (−16) =1

4 − 3

⇒𝑥𝑥2

=𝑦𝑦1

=11

⇒𝑥𝑥2

= 1,𝑦𝑦1

= 1





∴ 𝑥𝑥 = 2,𝑦𝑦 =1

(iii) 3𝑥𝑥 − 5𝑦𝑦 = 20

⇒ 3𝑥𝑥 − 5𝑦𝑦 − 20 = 0 … (1)

6𝑥𝑥 − 10𝑦𝑦 = 40

⇒ 6𝑥𝑥 − 10𝑦𝑦 − 40 = 0 … (2)

Here, 𝑎𝑎1 = 3, 𝑏𝑏1 = −5, 𝑐𝑐1 = −20

𝑎𝑎2 = 6, 𝑏𝑏2 = −10, 𝑐𝑐2 = −40

𝑎𝑎1𝑎𝑎2

=36

=12

,𝑏𝑏1𝑏𝑏2

=−5−10

=12

,𝑐𝑐1𝑐𝑐2

=−20−40

=12




Hence, the given pair of equations has infinite solutions.

The solution will be found by assuming the value of 𝑥𝑥 to be 𝑘𝑘. Hence the ordered pair �𝑘𝑘, 3𝑘𝑘−20

5�, where 𝑘𝑘 is a constant, are the solutions of the

given pair of equations.

(vi) 𝑥𝑥 − 3𝑦𝑦 − 7 = 0

3𝑥𝑥 − 3𝑦𝑦 − 15 = 0

Here, 𝑎𝑎1 = 1, 𝑏𝑏1 = −3, 𝑐𝑐1 = −7

𝑎𝑎2 = 3, 𝑏𝑏2 = −3, 𝑐𝑐2 = −15

𝑎𝑎1𝑎𝑎2

=13

,𝑏𝑏1𝑏𝑏2

=−3−3

= 1,𝑐𝑐1𝑐𝑐2

=−7−15

=7

15,


𝑏𝑏2

Hence, the given pair of equations has unique solution.

Now, using cross-multiplication method we know that,


=𝑦𝑦



⇒𝑥𝑥

45 − (21) =𝑦𝑦

−21 − (−15) =1

−3 − (−9)

⇒𝑥𝑥

24=

𝑦𝑦−6

=16





⇒ 𝑥𝑥24

= 16 and 𝑦𝑦

−6= 1

6

∴ 𝑥𝑥 = 4,𝑦𝑦 = −1

2. (i) For which values of 𝑎𝑎 and 𝑏𝑏 does the following pair of linear equations have an infinite number of solutions?

2𝑥𝑥 + 3𝑦𝑦 = 7

(𝑎𝑎 − 𝑏𝑏)𝑥𝑥 + (𝑎𝑎 + 𝑏𝑏)𝑦𝑦 = 3𝑎𝑎 + 𝑏𝑏 − 2

(ii) For which value of 𝑘𝑘 will the following pair of linear equations have no solution?

3𝑥𝑥 + 𝑦𝑦 = 1

(2𝑘𝑘 − 1)𝑥𝑥 + (𝑘𝑘 − 1)𝑦𝑦 = 2𝑘𝑘 + 1

Solution:

(i) 2𝑥𝑥 + 3𝑦𝑦 = 7

⇒ 2𝑥𝑥 + 3𝑦𝑦 − 7 = 0 … (1)

(𝑎𝑎 − 𝑏𝑏)𝑥𝑥 + (𝑎𝑎 + 𝑏𝑏)𝑦𝑦 = 3𝑎𝑎 + 𝑏𝑏 − 2

⇒ (𝑎𝑎 − 𝑏𝑏)𝑥𝑥 + (𝑎𝑎 + 𝑏𝑏)𝑦𝑦 − (3𝑎𝑎 + 𝑏𝑏 − 2) = 0 … (2)

Here, 𝑎𝑎1 = 2, 𝑏𝑏1 = 3, 𝑐𝑐1 = −7

𝑎𝑎2 = (𝑎𝑎 − 𝑏𝑏), 𝑏𝑏2 = (𝑎𝑎 + 𝑏𝑏), 𝑐𝑐2 = −(3𝑎𝑎 + 𝑏𝑏 − 2)

𝑎𝑎1𝑎𝑎2

=2

𝑎𝑎 − 𝑏𝑏, 𝑏𝑏1𝑏𝑏2

=3

𝑎𝑎 + 𝑏𝑏,𝑐𝑐1𝑐𝑐2

=−7

−(3𝑎𝑎 + 𝑏𝑏 − 2)=

7(3𝑎𝑎 + 𝑏𝑏 − 2)

As we know that, for the equations to have infinitely many solutions:

𝑎𝑎1𝑎𝑎2

=𝑏𝑏1𝑏𝑏2

=𝑐𝑐1𝑐𝑐2

Hence, 2𝑎𝑎−𝑏𝑏

= 73𝑎𝑎+𝑏𝑏−2

⇒ 6𝑎𝑎 + 2𝑏𝑏 − 4 = 7𝑎𝑎 − 7𝑏𝑏

⇒ 𝑎𝑎 − 9𝑏𝑏 + 4 = 0 … (3)

And 2𝑎𝑎−𝑏𝑏

= 3𝑎𝑎+𝑏𝑏

⇒ 2𝑎𝑎 + 2𝑏𝑏 = 3𝑎𝑎 − 3𝑏𝑏

⇒ 𝑎𝑎 − 5𝑏𝑏 = 0 … (4)

Now, using cross-multiplication method in equation (3) and (4),





𝑎𝑎0 − (−20)

=𝑏𝑏

4 − 0=

1−5 − (−9)

⇒𝑎𝑎

20=𝑏𝑏4

=14

⇒ 𝑎𝑎 = 5, 𝑏𝑏 = 1

Hence, the values of 𝑎𝑎 and 𝑏𝑏 are 5 and 1 respectively.

(ii) 3𝑥𝑥 + 𝑦𝑦 = 1

⇒ 3𝑥𝑥 + 𝑦𝑦 − 1 = 0 … (1)

(2𝑘𝑘 − 1)𝑥𝑥 + (𝑘𝑘 − 1)𝑦𝑦 = (2𝑘𝑘 + 1)

⇒ (2𝑘𝑘 − 1)𝑥𝑥 + (𝑘𝑘 − 1)𝑦𝑦 − (2𝑘𝑘 + 1) = 0 … (2)

Here, 𝑎𝑎1 = 3, 𝑏𝑏1 = 1, 𝑐𝑐1 = −1

𝑎𝑎2 = (2𝑘𝑘 − 1), 𝑏𝑏2 = (𝑘𝑘 − 1), 𝑐𝑐2 = −(2𝑘𝑘 + 1)

𝑎𝑎1𝑎𝑎2

=3

2𝑘𝑘 − 1,𝑏𝑏1𝑏𝑏2

=1

𝑘𝑘 − 1,𝑐𝑐1𝑐𝑐2

=−1

−2𝑘𝑘 − 1=

12𝑘𝑘 + 1

,

As we know that, for the equations to have no solutions:

𝑎𝑎1𝑎𝑎2

=𝑏𝑏1𝑏𝑏2≠𝑐𝑐1𝑐𝑐2

⇒3

2𝑘𝑘 − 1=

1𝑘𝑘 − 1

≠1

2𝑘𝑘 + 1

⇒3

2𝑘𝑘 − 1=

1𝑘𝑘 − 1

⇒ 3𝑘𝑘 − 3 = 2𝑘𝑘 − 1

⇒ 𝑘𝑘 = 2

Again for 𝑘𝑘 = 2, 32𝑘𝑘−1

= 1𝑘𝑘−1

= 1 but 12𝑘𝑘+1

= 15

Since, 1 ≠ 15

Hence, the values of 𝑘𝑘 is 2 for no solution.





3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8𝑥𝑥 + 5𝑦𝑦 = 9

3𝑥𝑥 + 2𝑦𝑦 = 4

Solution:


8𝑥𝑥 + 5𝑦𝑦 = 9 … (1)

3𝑥𝑥 + 2𝑦𝑦 = 4 … (2)


𝑦𝑦 =4 − 3𝑥𝑥

2 …(3)


8𝑥𝑥 + 5 �4 − 3𝑥𝑥

2� = 9

⇒ 16𝑥𝑥 + 20 − 15𝑥𝑥 = 18

So, 𝑥𝑥 = −2

Putting 𝑥𝑥 = −2 in (3), we get

𝑦𝑦 = �4 − 3 × −2

2� = 5

∴ 𝑥𝑥 = −2,𝑦𝑦 = 5

Cross-multiplication method:

8𝑥𝑥 + 5𝑦𝑦 − 9 = 0

3𝑥𝑥 + 2𝑦𝑦 − 4 = 0

Here, 𝑎𝑎1 = 8, 𝑏𝑏1 = 5, 𝑐𝑐1 = −9

𝑎𝑎2 = 3, 𝑏𝑏2 = 2, 𝑐𝑐2 = −4

𝑥𝑥−20 − (−18) =

𝑦𝑦−27 − (−32) =

116 − 15

⇒𝑥𝑥−2

=𝑦𝑦5

=11

⇒ 𝑥𝑥−2

= 1 and 𝑦𝑦5

= 1

⇒ 𝑥𝑥 = −2 and 𝑦𝑦 = 5





4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 13 when 1 is subtracted from the numerator and it

becomes 14 when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(iv) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:

(i) Let the fixed charge of the food be x and the charge for food per day be y.


x + 20y = 1000 . . . (1)

x + 26y = 1180 . . . (2)


6y = 180

⇒ y = 30


x + 26 × 30 = 1180

⇒ x = 1180 − 780

⇒ x = 400





Hence, the fixed charge of the food and the charge per day are ₹ 400 and ₹ 30 respectively.

(ii) Let the fraction be xy

As per the question, x−1y

= 13

⇒ 3x − y = 3 . . . (1) x

y+8= 1

4

⇒ 4x − y = 8 . . . (2)


x = 5


20 − y = 8

⇒ y = 12

Hence, the fraction is 512

(iii) Let the number of right answers be x and number of wrong answers be y.


3x − y = 40 … (1)

4x − 2y = 50

⇒ 2𝑥𝑥 − 𝑦𝑦 = 25 . . . . (2)


x = 15


30 − y = 25

y = 5

Hence, the number of right answers and the number of wrong answers is 15 and 5 respectively.

Hence, the total number of questions is 20.





(iv) Let the speed of first car be u km/h and speed of second car be v km/h respectively.

Speed of both cars while they are travelling in same direction = (u −v) km/h

Speed of both cars while they are travelling in opposite directions i.e., when they are travelling towards each other = (u + v)km/h

Distance travelled = speed × Time


5(u − v) = 100

⇒ u − v = 20 . . . (1)

1(u + v) = 100

⇒ u + v = 100 . . . (2)


2u = 120

⇒ u = 60

Putting u in equation (2), we get:

v = 40

Hence, speed of the first car is 60 km/h and speed of the second car is 40 km/h.

(v) Let length of rectangle be x units and breadth of rectangle be y units.

Hence, area = xy


(x − 5)(y + 3) = xy − 9

⇒ 3𝑥𝑥 − 5𝑦𝑦 − 6 = 0 … (1)

(x + 3)(y + 2) = xy + 67

⇒ 2𝑥𝑥 + 3𝑦𝑦 − 61 = 0 … (2)

Using cross-multiplication method,


=𝑦𝑦







we get:

𝑥𝑥305 − (−18)

=𝑦𝑦

−12 − (−183)=

19 − (−10)

⇒𝑥𝑥

323=

𝑦𝑦171

=1

19

⇒ 𝑥𝑥 = 17,𝑦𝑦 = 9

Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.

EXERCISE 3.6

1. Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) 12𝑥𝑥

+1

3𝑦𝑦= 2

13𝑥𝑥

+1

2𝑦𝑦=

136

(ii) 2√𝑥𝑥

+3�𝑦𝑦

= 2 4√𝑥𝑥

−9�𝑦𝑦

= −1

(iii) 4𝑥𝑥

+ 3𝑦𝑦 = 14 3𝑥𝑥− 4𝑦𝑦 = 23

(iv) 5𝑥𝑥 − 1

+1

𝑦𝑦 − 2= 2

6𝑥𝑥 − 1

−3

𝑦𝑦 − 2= 1

(v) 7𝑥𝑥 − 2𝑦𝑦𝑥𝑥𝑦𝑦

= 5 8𝑥𝑥 + 7𝑦𝑦𝑥𝑥𝑦𝑦

= 15

(vi) 6𝑥𝑥 + 3𝑦𝑦 = 6𝑥𝑥𝑦𝑦 2𝑥𝑥 + 4𝑦𝑦 = 5𝑥𝑥𝑦𝑦

(vii) 10𝑥𝑥 + 𝑦𝑦

+2

𝑥𝑥 − 𝑦𝑦= 4

15𝑥𝑥 + 𝑦𝑦

−5

𝑥𝑥 − 𝑦𝑦= −2

(viii) 13𝑥𝑥 + 𝑦𝑦

+1

3𝑥𝑥 − 𝑦𝑦=

34

1

2(3𝑥𝑥 + 𝑦𝑦) −1

2(3𝑥𝑥 − 𝑦𝑦) =−18





Solutions:

(i) 12𝑥𝑥

+ 13𝑦𝑦

= 2

13𝑥𝑥

+ 12𝑦𝑦

= 136

Let 1𝑥𝑥

= 𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 1𝑦𝑦

= 𝑞𝑞

Hence both equations convert to: 𝑝𝑝2

+ 𝑞𝑞3

= 2

⇒ 3p + 2q − 12 = 0. . . (1) 𝑝𝑝3

+ 𝑞𝑞2

= 136

⇒ 2𝑝𝑝 + 3𝑞𝑞 − 13 = 0 . . . (2)

Using cross-multiplication method, we get: 𝑝𝑝

−26−(−36)= 𝑞𝑞

−24−(−39)= 1

9−4

⇒ 𝑝𝑝10

= 𝑞𝑞15

= 15

⇒ 𝑝𝑝 = 2, 𝑞𝑞 = 3

∵ 1𝑥𝑥

= 𝑝𝑝 = 2, 1𝑦𝑦

= 𝑞𝑞 = 3

Hence, 𝑥𝑥 = 12

,𝑦𝑦 = 13

(ii) 2√𝑥𝑥

+ 3

√𝑦𝑦= 2

4√𝑥𝑥− 9

√𝑦𝑦= −1





let 1√𝑥𝑥

= 𝑝𝑝 and 1√𝑦𝑦

= 𝑞𝑞

Hence both equations convert to:

2𝑝𝑝 + 3𝑞𝑞 = 2 . . . (1)

4𝑝𝑝 − 9𝑞𝑞 = −1 . . . (2)


4𝑝𝑝 + 6𝑞𝑞 = 4 . . . (3)


15𝑞𝑞 = 5

⇒ 𝑞𝑞 = 13

Putting the value of 𝑞𝑞 in equation (1), we get:

2𝑝𝑝 + 3 × 13

= 2

⇒ 2p = 1

⇒ 𝑝𝑝 = 12

Since, 𝑝𝑝 = 1√𝑥𝑥

= 12

⇒ √𝑥𝑥 = 2

⇒ x = 4

𝑞𝑞 = 1

√𝑦𝑦= 1

3

⇒ �𝑦𝑦 = 3

⇒ y = 9

∴ 𝑥𝑥 = 4,𝑦𝑦 = 9

(iii) 4𝑥𝑥

+ 3𝑦𝑦 = 14

3𝑥𝑥− 4𝑦𝑦 = 23

Let 1𝑥𝑥

= 𝑝𝑝


4p + 3y = 14





⇒ 4𝑝𝑝 + 3𝑦𝑦 − 14 = 0 . . . (1)

3p − 4y = 23

⇒ 3𝑝𝑝 − 4𝑦𝑦 − 23 = 0 . . . (2)


−69−56= 𝑦𝑦

−42−(−92)= 1

−16−9

⇒ 𝑝𝑝−125

= 𝑦𝑦50

= −125

⇒ 𝑝𝑝−125

= −125

, 𝑦𝑦50

= −125

⇒ p = 5, y = −2

Since, 𝑝𝑝 = 1𝑥𝑥

= 5

⇒ 𝑥𝑥 = 15 and y = −2

(iv) 5𝑥𝑥−1

+ 1𝑦𝑦−2

= 2

6𝑥𝑥−1

− 3𝑦𝑦−2

= 1

Let 1𝑥𝑥−1

= 𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 1𝑦𝑦−2

= 𝑞𝑞


5p + q = 2 … (1)

6p − 3q = 1 … (2)


15p + 3q = 6 … (3)

Adding (2) and (3), we get:

21p = 7

⇒ 𝑝𝑝 = 13

Putting 𝑝𝑝 in equation (1), we get:

5 × 13

+ 𝑞𝑞 = 2

⇒ 𝑞𝑞 = 2 − 53

= 13





Since, 𝑝𝑝 = 1𝑥𝑥−1

= 13

⇒ 𝑥𝑥 − 1 = 3

⇒ 𝑥𝑥 = 4

𝑞𝑞 = 1𝑦𝑦−2

= 13

⇒ y − 2 = 3

⇒ y = 5

∴ 𝑥𝑥 = 4,𝑦𝑦 = 5

(v) 7𝑥𝑥−2𝑦𝑦𝑥𝑥𝑦𝑦

= 5

⇒ 7𝑦𝑦− 2

𝑥𝑥= 5 . . . (1)

8𝑥𝑥+7𝑦𝑦𝑥𝑥𝑦𝑦

= 15

⇒ 8𝑦𝑦

+ 7𝑥𝑥

= 15 . . . (2)

Let 1𝑥𝑥

= 𝑝𝑝 and 1𝑦𝑦

= 𝑞𝑞


−2p + 7q = 5

⇒ −2𝑝𝑝 + 7𝑞𝑞 − 5 = 0. . . (3)

7p + 8q = 15

⇒ 7𝑝𝑝 + 8𝑞𝑞 − 15 = 0 . . . (4)

using cross-multiplication method, we get: 𝑝𝑝

−105−(−40)= 𝑞𝑞

−35−30= 1

−16−49

⇒ 𝑝𝑝−65

= 𝑞𝑞−65

= 1−65

⇒ 𝑝𝑝−65

= 1−65

, 𝑞𝑞−65

= 1−65

⇒ p = 1, q = 1


= 1, 𝑞𝑞 = 1𝑦𝑦

= 1

Hence, x = 1, y = 1





(vi) 6x + 3y = 6xy

⇒ 6𝑦𝑦

+ 3𝑥𝑥

= 6. . . (1)

2x + 4y = 5xy

⇒ 2𝑦𝑦

+ 4𝑥𝑥

= 5 . . . (2)

Let 1𝑥𝑥

= 𝑝𝑝 and 1𝑦𝑦

= 𝑞𝑞


3p + 6q − 6 = 0

4p + 2q − 5 = 0


−30−(−12)= 𝑞𝑞

−24−(−15)= 1

6−24

⇒ 𝑝𝑝−18

= 𝑞𝑞−9

= 1−18

⇒ 𝑝𝑝−18

= 1−18

, 𝑞𝑞−9

= 1−18

⇒ 𝑝𝑝 = 1, 𝑞𝑞 = 12


= 1, 𝑞𝑞 = 1𝑦𝑦

= 12

Hence, x = 1, y = 2

(vii) 10𝑥𝑥+𝑦𝑦

+ 2𝑥𝑥−𝑦𝑦

= 4

15𝑥𝑥+𝑦𝑦

− 5𝑥𝑥−𝑦𝑦

= −2

Let 1𝑥𝑥+𝑦𝑦

= 𝑝𝑝 and 1𝑥𝑥−𝑦𝑦

= 𝑞𝑞


10p + 2q = 4

⇒ 10𝑝𝑝 + 2𝑞𝑞 − 4 = 0 . . . (1)

15p − 5q = −2

⇒ 15𝑝𝑝 − 5𝑞𝑞 + 2 = 0. . . (2)

Using cross-multiplication method, we get:





𝑝𝑝4−20

= 𝑞𝑞−60−20

= 1−50−30

⇒ 𝑝𝑝−16

= 1−80

𝑎𝑎𝑎𝑎𝑎𝑎 𝑞𝑞−80

= 1−80

⇒ 𝑝𝑝 = 15

𝑎𝑎𝑎𝑎𝑎𝑎 𝑞𝑞 = 1

Since, 𝑝𝑝 = 1𝑥𝑥+𝑦𝑦

= 15

𝑎𝑎𝑎𝑎𝑎𝑎 𝑞𝑞 = 1𝑥𝑥−𝑦𝑦

= 1

x + y = 5. . . . (3)

x − y = 1. . . . . (4)

Adding both equations (3) and (4), we get:

2x = 6

⇒ x = 3


y = 2

∴ 𝑥𝑥 = 3,𝑦𝑦 = 2

(viii) 13𝑥𝑥+𝑦𝑦

+ 13𝑥𝑥−𝑦𝑦

= 34

12(3𝑥𝑥+𝑦𝑦)

− 12(3𝑥𝑥−𝑦𝑦)

= −18

let 13𝑥𝑥+𝑦𝑦

= 𝑝𝑝 and 13𝑥𝑥−𝑦𝑦

= 𝑞𝑞


𝑝𝑝 + 𝑞𝑞 = 34

… (1)

𝑝𝑝2− 𝑞𝑞

2= −1

8

⇒ 𝑝𝑝 − 𝑞𝑞 = −14

… (2)


2𝑝𝑝 = 34− 1

4

⇒ 2𝑝𝑝 = 12

⇒ 𝑝𝑝 = 14

Putting p in (1), we get:





14

+ 𝑞𝑞 = 34

⇒ 𝑞𝑞 = 34− 1

4= 1

2

Since, 𝑝𝑝 = 13𝑥𝑥+𝑦𝑦

= 14

⇒ 3x + y = 4 … (3)

𝑞𝑞 = 13𝑥𝑥−𝑦𝑦

= 12

⇒ 3x − y = 2 … (4)


6x = 6

⇒ x = 1

Putting the value of x in (3), we get:

3(1) + y = 4

⇒ y = 1

∴ 𝑥𝑥 = 1,𝑦𝑦 = 1

2. Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(v) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution:

(i) Let the speed of Ritu be x km/h in still water and the speed of stream be y km/h.

Hence, speed of Ritu while rowing upstream = (x − y) km/h

And speed of Ritu while rowing downstream = (x + y) km/h

According to the question,





2(x + y) = 20

⇒ x + y = 10 … (1)

2(x − y) = 4

⇒ x − y = 2 … (2)


2x = 12

⇒ x = 6

Putting the value of x in equation (2), we get:

y = 4

Thus, Rita's speed is 6km/h in still water and the speed of the current is 4 km/h.

(ii) Let the number of days taken by a woman to finish the work be 𝑥𝑥 and a man to finish the work be 𝑦𝑦.

Work done by a woman in 1 day = 1𝑥𝑥

Work done by a man in 1 day = 1𝑦𝑦

According to the question,

4 �2𝑥𝑥

+ 5𝑦𝑦� = 1

⇒2𝑥𝑥

+5𝑦𝑦

=14

… (1)

3 �3𝑥𝑥

+6𝑦𝑦� = 1

⇒ 3𝑥𝑥

+ 6𝑦𝑦

= 13

… (2)

Let 1𝑥𝑥

= 𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 1𝑦𝑦

= 𝑞𝑞

Hence given equations convert to:

2𝑝𝑝 + 5𝑞𝑞 = 14

⇒ 8𝑝𝑝 + 20𝑞𝑞 − 1 = 0 … (3)

3𝑝𝑝 + 6𝑞𝑞 = 13





⇒ 9𝑝𝑝 + 18𝑞𝑞 − 1 = 0 … (4)


−20−(−18)= 𝑞𝑞

−9−(−8)= 1

144−180

⇒ 𝑝𝑝−2

= 𝑞𝑞−1

= 1−36

⇒ 𝑝𝑝−2

= 1−36

, 𝑞𝑞−1

= 1−36

⇒ 𝑝𝑝 = 118

, 𝑞𝑞 = 136


= 118

, 𝑞𝑞 = 1𝑦𝑦

= 136


Hence, the number of days taken by a woman and a man to finish the work is 18 and 36.

(iii) Let the speed of train be u km/h and speed of bus be v km/h.

Now as we know, Time taken = Distance  travelledSpeed

Hence, As per the question, 60𝑢𝑢

+ 240𝑣𝑣

= 4. . . (1)

100𝑢𝑢

+ 200𝑣𝑣

= 256

. . . (2)(Since, 10 minutes= 16 hours)

Let 1𝑢𝑢

= 𝑝𝑝 and 1𝑣𝑣

= 𝑞𝑞

Hence given equations convert to:

60p + 240q = 4 . . . (3)

100𝑝𝑝 + 200𝑞𝑞 = 256

⇒ 600𝑝𝑝 + 1200𝑞𝑞 = 25. . . . (4)


600p + 2400q = 40. . . . . (5)


1200q = 15

⇒ 𝑞𝑞 = 151200

= 180





Putting q in equation (3), we get:

60p + 3 = 4

⇒ 60p = 1

⇒ 𝑝𝑝 = 160

Since, 𝑝𝑝 = 1𝑢𝑢

= 160

, 𝑞𝑞 = 1𝑣𝑣

= 180

⇒ u = 60 km/h, v = 80 km/h

Hence, the speed of train and the speed of bus are 60 km/h and 80 km/h respectively.

♦ ♦ ♦

EXERCISE 3.7 (Optional)*

1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Solution:

The difference of ages of Ani and Biju is 3 years. Hence, either Biju is 3 years older than Ani or Ani is 3 years older than Biju.

Let the age of Ani and Biju be x years and y years respectively.

Age of Dharma = 2x years

Age of cathy = 𝑦𝑦2 years

Case I:

Ani is older than Biju by 3 years

𝑥𝑥 − 𝑦𝑦 = 3 … (1)

2x −y2

= 30

⇒ 4x − y = 60 . . . (2)

Subtracting (1) from (2), we get:

3x = 60 − 3 = 57 ⇒ x = 19





Hence, age of Ani = 19 years

And age of Biju = 19 − 3 = 16 years

Case II: Biju is older than Ani by 3 years

y − x = 3 … (3)

2x −y2

= 30

⇒ 4x − y = 60 … (4)


3x = 63

⇒ x = 21

Hence, age of Ani = 21 years

And age of Biju = 21 + 3 = 24 years

2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara 𝐼𝐼𝐼𝐼]

[Hint: 𝑥𝑥 + 100 = 2(𝑦𝑦 − 100),𝑦𝑦 + 10 = 6(𝑥𝑥 − 10)].

Solution:

Let the money with the first person be ₹ x and money with the second person be ₹ y.


x + 100 = 2(y − 100)

⇒ x + 100 = 2y − 200

⇒ x − 2y = −300 … (1)

6(x − 10) = (y + 10)

⇒ 6x − 60 = y + 10

⇒ 6x − y = 70 … (2)


12x − 2y = 140 … (3)






11x = 140 + 300

⇒ 11x = 440

⇒ x = 40


40 − 2y = −300

⇒ 40 + 300 = 2y

⇒ 2y = 340

⇒ y = 170

Hence, the two friends has ₹ 40 and ₹ 170 respectively.

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km h⁄ faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km h⁄ ; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Solution:

Let total distance travel by train be d km, the speed of the train be x km/h and the time taken by train to travel d km be t hours.

Since, Speed= Distance travlledTime taken to tranvel that distance

⇒ 𝑥𝑥 =𝑎𝑎𝑡𝑡

⇒ 𝑎𝑎 = 𝑥𝑥𝑡𝑡. . . (1)

As per the question, (𝑥𝑥 + 10) = 𝑑𝑑(𝑡𝑡−2)

⇒ (𝑥𝑥 + 10)(𝑡𝑡 − 2) = 𝑎𝑎

⇒ 𝑥𝑥𝑡𝑡 + 10𝑡𝑡 − 2𝑥𝑥 − 20 = 𝑎𝑎

By using equation (1), we get:

−2𝑥𝑥 + 10𝑡𝑡 = 20 . . . (2)

(𝑥𝑥 − 10) =𝑎𝑎

(𝑡𝑡 + 3)

⇒ (𝑥𝑥 − 10)(𝑡𝑡 + 3) = 𝑎𝑎

By using equation (1), we get:

3x − 10t = 30 … (3)






𝑥𝑥 = 50

Substituting x in equation (2), we get:

(−2) × (50) + 10t = 20

⇒ −100 + 10t = 20

⇒ 10t = 120

⇒ t = 12


d = xt = 50 x 12 = 600

Hence, the distance covered by the train is 600 km.

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution:

Let the number of rows be x and number of students in a row be y.

Total number of students in the class = Number of rows × Number of students in a row = xy


Total number of students = (x − 1)(y + 3)

⇒ xy = (x − 1)(y + 3)

⇒ 𝑥𝑥𝑦𝑦 = xy − y + 3x − 3

⇒ 3x − y − 3 = 0

⇒ 3x − y = 3 … (1)

Again, Total number of students = (x + 2)(y − 3)

⇒ xy = xy + 2y − 3x − 6

⇒ 3x − 2y = −6 … (2)

Subtracting equation (2) from (1), we get:

y = 9

Substituting the value of y in equation (1), we get:

3x − 9 = 3





⇒ 3x = 9 + 3 = 12

⇒ x = 4

Hence, number of rows, x = 4

And number of students in a row, y = 9

Hence, total number of students in a class = 𝑥𝑥𝑦𝑦 = 4 × 9 = 36

5. In a ∆ABC,∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.

Solution:

Given,

∠C = 3 ∠B = 2(∠A + ∠B)

⇒ 3∠B = 2(∠A + ∠B)

⇒ ∠B = 2 ∠A ⇒ 2 ∠A − ∠B = 0 . . . (1)

We know that the sum of all angles of a triangle is 180°

Hence, ∠A + ∠B + ∠C = 180o

⇒ ∠𝐴𝐴 + ∠𝐵𝐵 + 3∠B = 180o

⇒ ∠A + 4∠B = 180o … (2)


8∠𝐴𝐴 − 4 ∠𝐵𝐵 = 0 … (3)


9 ∠A = 180o

⇒ ∠A = 20o


20o + 4 ∠B = 180o

⇒ 4∠B = 160o

⇒ ∠B = 40o

Hence, ∠C = 3 ∠B = 3 × 40o = 120o

Thus, the value of ∠A,∠B and ∠C are 20o, 40o and 120o respectively.





6. Draw the graphs of the equations 5𝑥𝑥 − 𝑦𝑦 = 5 and 3𝑥𝑥 − 𝑦𝑦 = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the 𝑦𝑦 axis.

Solution:

5x − y = 5

⇒ y = 5x − 5


𝑥𝑥 0 1

𝑦𝑦 −5 0

3x − y = 3

⇒ y = 3x − 3


𝑥𝑥 0 1

𝑦𝑦 −3 0

The graphical representation of the two lines will be as follows:





We have shown required triangle △ ABC in the graph. The coordinates of its vertices are A (1, 0), B (0,−3), C (0,−5).

7. Solve the following pair of linear equations:

Solution:

(i) Both equations can be written as:

𝑝𝑝𝑥𝑥 + 𝑞𝑞𝑦𝑦 − (𝑝𝑝 − 𝑞𝑞) = 0 … (1)

𝑞𝑞𝑥𝑥 − 𝑝𝑝𝑦𝑦 − (𝑝𝑝 + 𝑞𝑞) = 0 … (2)

By using cross multiplication method, we get:

𝑥𝑥−𝑞𝑞(𝑝𝑝 + 𝑞𝑞) − 𝑝𝑝(𝑝𝑝 − 𝑞𝑞)

=𝑦𝑦

−𝑞𝑞(𝑝𝑝 − 𝑞𝑞) + 𝑝𝑝(𝑝𝑝 + 𝑞𝑞)=

1−𝑝𝑝2 − 𝑞𝑞2

⇒𝑥𝑥

−𝑝𝑝2 − 𝑞𝑞2=

𝑦𝑦𝑝𝑝2 + 𝑞𝑞2

=1

−𝑝𝑝2 − 𝑞𝑞2

⇒ 𝑥𝑥 = 1,𝑦𝑦 = −1

(ii) Both equations can be written as:

𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 − 𝑐𝑐 = 0 … (1)

𝑏𝑏𝑥𝑥 + 𝑎𝑎𝑦𝑦 − (1 + 𝑐𝑐) = 0 . . . (2)

By using cross multiplication method, we get:

𝑥𝑥−𝑏𝑏(1 + 𝑐𝑐) + 𝑎𝑎𝑐𝑐

=𝑦𝑦

−𝑏𝑏𝑐𝑐 + 𝑎𝑎(1 + 𝑐𝑐)=

1𝑎𝑎2 − 𝑏𝑏2

(i) 𝑝𝑝𝑥𝑥 + 𝑞𝑞𝑦𝑦 = 𝑝𝑝 − 𝑞𝑞 𝑞𝑞𝑥𝑥 − 𝑝𝑝𝑦𝑦 = 𝑝𝑝 + 𝑞𝑞

(ii) 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑐𝑐 𝑏𝑏𝑥𝑥 + 𝑎𝑎𝑦𝑦 = 1 + 𝑐𝑐

(iii) 𝑥𝑥𝑎𝑎−𝑦𝑦𝑏𝑏

= 0 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑎𝑎2 + 𝑏𝑏2

(iv) (𝑎𝑎 − 𝑏𝑏)𝑥𝑥 + (𝑎𝑎 + 𝑏𝑏)𝑦𝑦= 𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 − 𝑏𝑏2 (𝑎𝑎 + 𝑏𝑏)(𝑥𝑥 + 𝑦𝑦) = 𝑎𝑎2 + 𝑏𝑏2

(v) 152𝑥𝑥 − 378𝑦𝑦 = −74 −378𝑥𝑥 + 152𝑦𝑦 = −604





⇒𝑥𝑥

c(a − b) − b=

𝑦𝑦𝑐𝑐(𝑎𝑎 − 𝑏𝑏) + 𝑎𝑎

=1

𝑎𝑎2 − 𝑏𝑏2

⇒ 𝑥𝑥 =𝑐𝑐(𝑎𝑎 − 𝑏𝑏) − 𝑏𝑏𝑎𝑎2 − 𝑏𝑏2

, 𝑦𝑦 =𝑐𝑐(𝑎𝑎 − 𝑏𝑏) + 𝑎𝑎𝑎𝑎2 − 𝑏𝑏2

(iii) 𝑥𝑥𝑎𝑎− 𝑦𝑦

𝑏𝑏= 0

𝑏𝑏𝑥𝑥 − 𝑎𝑎𝑦𝑦 = 0 … (1)

𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 = 𝑎𝑎2 + 𝑏𝑏2 … (2)

Multiplying equation (1) and (2) by b and a respectively, we get:

𝑏𝑏2𝑥𝑥 − 𝑎𝑎𝑏𝑏𝑦𝑦 = 0 … (3)

𝑎𝑎2𝑥𝑥 + 𝑎𝑎𝑏𝑏𝑦𝑦 = 𝑎𝑎3 + 𝑎𝑎𝑏𝑏2 … (4)


𝑏𝑏2𝑥𝑥 + 𝑎𝑎2𝑥𝑥 = 𝑎𝑎3 + 𝑎𝑎𝑏𝑏2

⇒ 𝑥𝑥(𝑏𝑏2 + 𝑎𝑎2) = 𝑎𝑎(𝑎𝑎2 + 𝑏𝑏2)

⇒ 𝑥𝑥 = 𝑎𝑎


𝑏𝑏(𝑎𝑎) − 𝑎𝑎𝑦𝑦 = 0

𝑎𝑎𝑏𝑏 − 𝑎𝑎𝑦𝑦 = 0

𝑎𝑎𝑦𝑦 = 𝑎𝑎𝑏𝑏

𝑦𝑦 = 𝑏𝑏

Hence, 𝑥𝑥 = 𝑎𝑎,𝑦𝑦 = 𝑏𝑏

(iv) (𝑎𝑎 − 𝑏𝑏)𝑥𝑥 + (𝑎𝑎 + 𝑏𝑏)𝑦𝑦 = 𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 − 𝑏𝑏2 … (1)

(𝑎𝑎 + 𝑏𝑏)(𝑥𝑥 + 𝑦𝑦) = 𝑎𝑎2 + 𝑏𝑏2

⇒ (𝑎𝑎 + 𝑏𝑏)𝑥𝑥 + (𝑎𝑎 + 𝑏𝑏)𝑦𝑦 = 𝑎𝑎2 + 𝑏𝑏2. . . (2)

Subtracting equation (2) from (1), we get:

(𝑎𝑎 − 𝑏𝑏)𝑥𝑥 − (𝑎𝑎 + 𝑏𝑏)𝑥𝑥 = (𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 − 𝑏𝑏2) − (𝑎𝑎2 + 𝑏𝑏2)

(𝑎𝑎 − 𝑏𝑏 − 𝑎𝑎 − 𝑏𝑏)𝑥𝑥 = −2𝑎𝑎𝑏𝑏 − 2𝑏𝑏2 − 2𝑏𝑏𝑥𝑥 = −2𝑏𝑏(𝑎𝑎 + 𝑏𝑏)

𝑥𝑥 = 𝑎𝑎 + 𝑏𝑏






(𝑎𝑎 − 𝑏𝑏)(𝑎𝑎 + 𝑏𝑏) + (𝑎𝑎 + 𝑏𝑏)𝑦𝑦 = 𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 − 𝑏𝑏2

𝑎𝑎2 − 𝑏𝑏2 + (𝑎𝑎 + 𝑏𝑏)𝑦𝑦 = 𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 − 𝑏𝑏2

(v) (𝑎𝑎 + 𝑏𝑏)𝑦𝑦 = −2𝑎𝑎𝑏𝑏

𝑦𝑦 =−2𝑎𝑎𝑏𝑏𝑎𝑎 + 𝑏𝑏

Hence, 𝑥𝑥 = 𝑎𝑎 + 𝑏𝑏,𝑦𝑦 = −2𝑎𝑎𝑏𝑏𝑎𝑎+𝑏𝑏

(vi) 152𝑥𝑥 − 378𝑦𝑦 = −74. . (1)

−378𝑥𝑥 + 152𝑦𝑦 = −604 … (2)

Adding the equations (1) and (2), we get:

−226𝑥𝑥 − 226𝑦𝑦 = −678

⇒ 𝑥𝑥 + 𝑦𝑦 = 3 … (3)

Subtracting the equation (2) from equation (1), we get:

530𝑥𝑥 − 530𝑦𝑦 = 530

⇒ 𝑥𝑥 − 𝑦𝑦 = 1 … (4)


2𝑥𝑥 = 4

⇒ 𝑥𝑥 = 2


𝑦𝑦 = 1


8. ABCD is a cyclic quadrilateral (see Fig.) Find the angles of the cyclic quadrilateral.





Solution:

As we know, the sum of opposite angles in a cyclic quadrilateral is 180°.

∠A + ∠C = 180

⇒ 4𝑦𝑦 + 20° − 4𝑥𝑥 = 180

⇒ −4𝑥𝑥 + 4𝑦𝑦 = 160°

⇒ 𝑥𝑥 − 𝑦𝑦 = −40 … (1)

Also, ∠B + ∠D = 180°

⇒ 3𝑦𝑦 − 5° − 7𝑥𝑥 + 5° = 180°

⇒ −7𝑥𝑥 + 3𝑦𝑦 = 180 … (2)


3𝑥𝑥 − 3𝑦𝑦 = −120 … (3)


−4𝑥𝑥 = 60°

⇒ 𝑥𝑥 = −15°


−15° − 𝑦𝑦 = −40°

⇒ 𝑦𝑦 = −15° + 40° = 25

Hence,

∠A = 4𝑦𝑦 + 20° = 4(25°) + 20° = 120𝑜𝑜

∠B = 3𝑦𝑦 − 5° = 3(25°) − 5° = 70𝑜𝑜

∠C = −4𝑥𝑥 = −4(−15°) = 60𝑜𝑜

∠D = −7𝑥𝑥 + 5° = −7(−15°) + 5° = 110𝑜𝑜

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cbse ncert solutions for class 10 mathematics chapter 3

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