**NCERT Solutions Maths Class 10 Exercise 3.3**

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

**(i) x + y =
14**

* x ***– y = 4**

**(ii) s – t =
3**

** s/3 + t/2 = 6**

**(iii) 3x – y = 3**

** 9 x − 3y
= 9**

**(iv)
0.2 x + 0.3y = 1.3**

** 0.4 x
+ 0.5y = 2.3**

**(v)
√2x + √3y = 0 **

**√3x
+ √8y = 0**

**(vi)
(3/2)x – (5/3)y = −2**

**
x/3 + y/2 = 13/6**

**Solution: (i)** *x *+ *y *= 14 ……(1)

* x ***– y = 4 ….… (2)**

From
equation (2), we get *x *= 4 + *y*

Putting this value of x in
equation (1), we get

4
+ *y *+ *y *= 14

⇒ 2*y *= 10

⇒ *y *= 5

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

*x *+ 5 = 14

⇒ *x *= 14 – 5

⇒
*x *= 9

Therefore, *x *= 9 and *y *= 5.

**(ii) s – t =
3 ...... (1)**

** s/3 + t/2 = 6**

**...... (2)**

From
equation (1), we get *s *= 3 + *t*

Putting this value of s in
equation (2), we get

(3 +
*t*)/3 + *t*/2 = 6

⇒
(6 + 2*t* + 3*t*)/6 = 6

⇒ 5*t *+ 6 = 36

⇒ 5*t *= 30

⇒ *t *= 6

Putting the value of *t* in equation (1), we get

*s *– 6 = 3

⇒ *s *= 3 + 6

⇒ *s *= 9

Therefore, *t *= 6 and *s *= 9.

**(iii)** **3 x – y = 3 … (1)**

** 9 x − 3y = 9 … (2)**

Comparing equation 3*x *– *y *= 3
with a_{1}x + b_{1}y
+ c_{1} = 0 and equation 9*x *− 3*y *= 9
with a_{2}x + b_{2}y
+ c_{2 }= 0, we get

a_{1} = 3, b_{1}
= –1, c_{1} = –3, a_{2} = 9, b_{2} = –3 and c_{2}
= –9

Here, a_{1}/a_{2 }= b_{1}/b_{2
}= c_{1}/c_{2}

Therefore, we have infinitely
many solutions for x and y.

**(iv)** **0.2 x + 0.3y = 1.3 … (1)**

** 0.4 x + 0.5y = 2.3 … (2)**

From equation (1), we have

0.2*x *= 1.3 − 0.3*y*

⇒ *x *= (1.3 − 0.3*y*)/0.2

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

0.4(1.3
− 0.3*y*)/0.2 + 0.5*y *= 2.3

⇒ 2.6
− 0.6*y *+ 0.5*y *= 2.3

⇒
−0.1*y *= 2.3 – 2.6

⇒
−0.1*y *=−0.3

⇒ *y *= 3

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

0.2*x *+ 0.3(3) = 1.3

⇒ 0.2*x *+ 0.9 = 1.3

⇒ 0.2*x* = 1.3 − 0.9

⇒ 0.2*x *= 0.4

⇒ *x *= 2

Therefore, *x *= 2 and *y *= 3.

**(v)
√2x + √3y = 0 **

**……….(1)**

**
√3x − √8y = 0 ……….(2) **

From equation (1), we get

*x *= −√3y/√2

Putting this value of *x* in equation (2), we get

√3(−√3y/√2)
− √8*y*
= 0

⇒ −3*y*/√2 −√8*y* = 0

⇒ *y *= 0

Putting
the value of *y* in equation (1), we
get *x *= 0.

Therefore, *x *= 0 and *y *= 0.

**(vi)
(3/2)x – (5/3)y = −2**

**……….(1)**

**
x/3 + y/2 = 13/6**

**……….(2)**

From equation (2), we have

*x *= (13/6 *– *y/2) ×
3

⇒ *x *= 13/2 *–*
3y/2

Putting this value of *x* in equation (1), we get

⇒ *y *= 12/4

⇒ *y *= 3

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

⇒ *x*/3 = 2/3

⇒ *x* = 2

Therefore, *x *= 2 and *y *= 3.

**2. Solve 2 x + 3y = 11 and 2x −
4y = −24 and hence find the value of ‘m’ for which**

*y ***= mx + 3.**

**Solution: ****2 x +
3y = 11 … (1)**

**2 x − 4y = −24 … (2)**

From equation (2), we get

2*x *= −24 + 4*y*

⇒ *x *= −12 + 2*y*

Putting this value of *x* in equation (1), we get

2(−12
+ 2*y*) + 3*y *= 11

⇒ −24
+ 4*y *+ 3*y *= 11

⇒ 4*y *+ 3*y *= 11 + 24

⇒ 7*y *= 35

⇒ *y *= 5

Putting this value of *y* in equation (1), we get

2*x *+ 3(5) = 11

⇒ 2*x *+ 15 = 11

⇒ 2*x *= 11 – 15 = −4

⇒ *x *= −2

Therefore, *x *= −2 and *y *= 5.

Putting
the values of *x* and *y* in *y *= *mx *+ 3, we get

5
= *m*(−2) + 3

⇒ 5 =
−2*m *+ 3

⇒ −2*m *= 2

⇒ *m *= −1

**3.
Form a 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.**

**(iii)The
coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3
bats and 5 balls for Rs 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 Rs 105
and for a journey of 15 km, the charge paid is Rs 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?**

**(v) A fraction becomes 9/11, 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.**

**(vi)
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
the first number be x and the second number be y. Let x > y.

According to the first
condition, we have

*x *– *y *= 26 … (1)

According to the second condition, we have

*x *= 3*y *… (2)

Putting the value of *x* from equation (2) in equation (1), we
get

3*y *– *y *= 26

⇒ 2*y *= 26

⇒ *y *= 13

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

*x *= 3*y *= 3 × 13 = 39

Therefore, the two numbers
are 13 and 39.

**(ii)** Let the two supplementary angles be x and y. Let y > x.

According to the given conditions, we have

*y *= *x *+ 18 … (1)

Also, x + y = 180° (Sum of supplementary angles is 180°) … (2)

Putting the value of y from
equation (1) in equation (2), we get

*x *+ *x *+ 18 = 180

⇒ 2*x *= 180 – 18 = 162

⇒ x = 81°

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

*y *= *x *+ 18 = 81 + 18 = 99°

Therefore, the two
supplementary angles are 81° and 99°.

**(iii)** Let
the cost of one bat be Rs *x* and the
cost of one ball be Rs y.

According to the first condition,
we have

7*x *+ 6*y *= 3800 … (1)

According to the second condition, we have

3*x *+ 5*y *= 1750 … (2)

From equation (1), we have

7*x *= 3800 − 6*y *

⇒ *x *= (3800 − 6*y**)/7*

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

3(3800
− 6*y**)/7* +
5*y *= 1750

(11400
− 18*y**)/7* +
5*y *= 1750

⇒ 17*y *=
850

⇒ *y *= 50

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

3*x *+ 250 = 1750

⇒ 3*x *=
1500

⇒ *x *= 500

Therefore, the cost of one bat is
Rs 500 and the cost of one ball is Rs 50.

**(iv)** Let
the fixed charge be Rs *x* and the
charge per km be Rs *y*.

According to the given conditions,
we have

*x *+
10*y *=
105… (1)

*x *+
15*y *=
155… (2)

From equation (1), we have

*x *=
105 − 10*y*

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

105
− 10*y *+
15*y *=
155

⇒ 5*y *=
50

⇒ *y *= 10

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

*x *+
10(10) = 105

⇒ *x *= 105 – 100 = 5

⇒ *x *= 5

Therefore, the fixed charge is Rs
5 and the charge per km is Rs 10.

To travel a distance of 25 km, a person
will have to pay = Rs (x + 25y)

= Rs (5 + 25 × 10)

= Rs (5 + 250)

= Rs 255

**(v)** Let
the numerator of the fraction be x and the denominator be y.

According to the first condition,
we have

(*x* + 2)/(*y* + 2) = 9/11

⇒ 11(*x *+
2) = 9(*y *+
2)

⇒ 11*x *+
22 = 9*y *+
18

⇒ 11*x *=
9*y *– 4

⇒ *x *= (9*y*
– 4)/11 …… (1)

According to the second condition,
we have

(*x* + 3)/(*y* + 3) = 5/6

6(*x* + 3) = 5(*y *+ 3)

6*x* + 18 = 5*y* + 15 ……. (2)

Putting
the value of x in equation (2) from equation (1), we get

6(9*y* – 4)/11 + 18 = 5*y* + 15

54*y*/11 – 24/11 + 18 = 5*y* + 15

–24/11
+ 18 – 15 = 5*y* – 54*y*/11

–24/11 + 3 = (55*y* – 54*y*)/11

(–24 + 33)/11 = *y*/11

*y*/11
= 9/11

⇒ *y *= 9

Putting the value of *y* in equation (1), we get

*x *=
(9 × 9 – 4)/11

⇒ *x* = 77/11

⇒ *x *= 7

Therefore, the fraction is 7/9.

**(vi)** Let
the present age of Jacob be *x* years
and the present age of his son be *y*
years.

According to the first condition,
we have

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

x
+ 5 = 3y + 15

x
= 3y + 15 – 5

x
= 3y + 10 ……. (1)

According to the second condition,
we have

(*x *− 5) = 7(*y *− 5)

*x *−
5 = 7*y *–
35

*x *=
7*y *– 35 + 5

*x *=
7*y *– 30 …….. (2)

From equation (1) and equation (2),
we get

3*y *+ 10 = 7*y *– 30

3*y* – 7*y*
= −30 − 10

⇒ −4*y *=
−40

⇒ *y *= 10 years

Putting the value of *y* in equation (1), we get

*X *=
3 × 10 + 10

⇒ *x *= 30 + 10

⇒ *x* = 40 years

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