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# Class 10 NCERT Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.3

• Last Updated : 04 Dec, 2020

### (i) x + y = 14 and x – y = 4

Solution:

x + y = 14 ……….. (1)

x – y = 4 ………….. (2)

x = 14 – y

Substitute x in (2)

(14 – y) – y = 4

14 – 2y = 4

2y = 10

Transposing 2

y = 10/2

y = 5

x = 14 – y

x = 9

Therefore, x = 9 and y = 5.

### (ii) s – t = 3 and (s/3) + (t/2) = 26

Solution:

s – t = 3 …….. (1)

(s/3) + (t/2) = 6 …………. (2)

s = 3 + t

Now, substitute the value of s in (2)

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

Taking 6 as LCM

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

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

(6 + 5t) = 36

5t = 30

t = 6

s = 3 + 6 = 9

Therefore, s = 9 and t = 6.

### (iii) 3x – y = 3 and 9x – 3y = 9

Solution:

3x – y = 3 ……….. (1)

9x – 3y = 9 ……….(2)

From (1)

x = (3 + y) / 3

Substitute x in (2)

9(3 + y) / 3 – 3y = 9

9 + 3y – 3y = 9

0 = 0

Therefore, y has infinite values and x = (3 + y)/3 also has infinite values.

### (iv) 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3

Solution:

0.2x + 0.3y = 1.3 ……… (1)

0.4x + 0.5y = 2.3 …….. (2)

From (1)

x = (1.3 – 0.3y) / 0.2

Putting x in (2)

0.4(1.3 – 0.3y) / 0.2 + 0.5y = 2.3

2(1.3 – 0.3y) + 0.5y = 2.3

2.6 – 0.6y + 0.5y = 2.3

2.6 – 0.1 y = 2.3

0.1 y = 0.3

y = 3

Substitute y in (1)

x = (1.3 – 0.3(3)) / 0.2 = (1.3 – 0.9) / 0.2 = 0.4/0.2 = 2

Therefore, x = 2 and y = 3.

### (v) √2x + √3y = 0 and √3x – √8y = 0

Solution:

√2 x + √3 y = 0 …………… (1)

√3 x – √8 y = 0 …………..  (2)

From (1)

x = – (√3/√2)y

Putting x in (2)

√3(-√3/√2)y – √8y = 0

(-3/√2)y –  √8y = 0

-3y – 4y = 0

-7y = 0

y = 0

Therefore

x = 0

Therefore, x = 0 and y = 0.

### (vi) (3x/2) – (5y/3) = -2 and (x/3) + (y/2) = (13/6)

Solution:

(3x/2) – (5y/3) = -2 ……………. (1)

(x/3) + (y/2) = 13/6 ………. (2)

From (1)

(3/2)x = -2 + (5y/3)

(3/2)x = (-6 + 5y) / 3

x = ((-6 + 5y) / 3) * 2/3

⇒ x = 2(-6 + 5y) / 9 = (-12 + 10y) / 9

Putting x in (2)

((-12 +10y)/9)/3 + y/2 = 13/6

(-12 + 10y)/27 + y/2 = 13/6

Taking 54 as LCM

-24 + 20y + 27y = 117

47y = 117 + 24

47y = 141

y = 3

x = (-12 + 30) / 9

x = 18/9

x = 2

Therefore, x = 2 and y = 3.

### Question 2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution:

2x + 3y = 11…………………………..(1)

2x – 4y = -24………………………… (2)

From (1)

x = (11 – 3y) / 2

Substituting x in equation (2)

2(11 – 3y) / 2 – 4y =- 24

11 – 3y – 4y = -24

-7y = -24 – 11

-7y = -35

y = 5

Putting y in (1)

x = (11 – 3 × 5) / 2 = -4/2 = -2

x = -2, y = 5

y = mx + 3

5 = -2m +3

-2m = 2

m = -1

Therefore, the value of m is -1.

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

Solution:

Let the two numbers be x and y

y = 3x ……………… (1)

y – x = 26 …………..(2)

Substituting the value of y

3x – x = 26

2x = 26

x = 13

y = 39

Therefore, the numbers are 13 and 39.

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

Solution:

Let the larger angle by xo and smaller angle be yo.

Sum of two supplementary pair of angles is 180o.

x + y = 180o……………. (1)

x – y = 18……………..(2)

From (1)

x = 180 – y

Substituting in (2)

180 – y – y = 18

-2y = -162

162 = 2y

y = 81o

x = 180 – y

x = 180 – 81

= 99

Therefore, the angles are 99o and 81o.

### (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.

Solution:

Let the cost of a bat be Rs. x and cost of a ball be Rs. y.

7x + 6y = 3800 ………………. (i)

3x + 5y = 1750 ………………. (ii)

From (i)

y = (3800 – 7x) / 6………………..(iii)

Substituting (iii) in (ii)

3x + 5(3800 – 7x) / 6 =1750

Taking 6 as LCM

18x + 19000 – 7x = 10500

11x = 10500 – 19000

⇒ -17x = -8500

x = 500 ……………………….. (IV)

Substituting the value of x in (III), we get

y = (3800 – 7 × 500)/6 = 300/6 = 50

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

### (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 traveling a distance of 25 km?

Solution:

Let the fixed charge be Rs x and per km charge be Rs y.

x + 10y = 105 …………….. (1)

x + 15y = 155 …………….. (2)

From (1)

x = 105 – 10y ………………. (3)

Substituting the value of x in (2)

105 – 10y + 15y = 155

5y = 50

y = 10

Putting the value of y in (3)

x = 105 – 10 × 10

x = 105 – 100

x = 5

Hence, fixed charge is Rs 5 and per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255

### (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.

Solution:

Let the fraction be x/y.

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

By cross multiplication

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)

By cross multiplication

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

6x + 18 = 5y + 15

6x – 5y = -3 ………………. (2)

From (1)

x = (-4 + 9y)/11 …………….. (3)

Substituting the value of x in (2)

6(-4 + 9y)/11 -5y = -3

Taking 11 as the LCM

-24 + 54y – 55y = -33

-24 – y = -33

-y = -33 + 24

-y = -9

y = 9

Substituting the value of y in (3)

x = (-4 + 9 × 9)/11

x = (-4 + 81)/11

x = 77/11

x = 7

Hence, the fraction is 7/9.

### (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?

Solutions:

Let the age of Jacob = x years and that of son be y years.

(x + 5) = 3(y + 5) ………………… (i)

(x – 5) = 7(y – 5) ………………….. (ii)

From (i)

x + 5 = 3y + 15

x – 3y = 10……………. (iii)

From (ii)

x – 5 = 7y – 35

x – 7y = -30……………..(iv)

Subtracting (iv) from (iii)

-3y + 7y = 40

4y = 40

Transposing 4

y = 40/4

y = 10

Putting y = 10 in (iii)

x – 30 = 10

x = 40

Therefore, present age of Jacob is 40 years and that of son is 10 years

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