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

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

**(i) x + y = 14 and x â€“ y = 4**

**Solution:**

x + y = 14 ……….. (1)

x â€“ y = 4 ………….. (2)

x = 14 â€“ y

Substitute x in (2)

(14 â€“ y) â€“ y = 4

14 â€“ 2y = 4

2y = 10

Transposing 2

y = 10/2

y = 5

x = 14 â€“ y

x = 9

Therefore, x = 9 and y = 5.

**(ii) s â€“ t = 3 and (s/3) + (t/2) = 26**

**Solution:**

s â€“ 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 â€“ y = 3 and 9x â€“ 3y = 9**

**Solution:**

3x â€“ y = 3 ……….. (1)

9x â€“ 3y = 9 ……….(2)

From (1)

x = (3 + y) / 3

Substitute x in (2)

9(3 + y) / 3 â€“ 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 â€“ 0.3y) + 0.5y = 2.3

2.6 â€“ 0.6y + 0.5y = 2.3

2.6 â€“ 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) âˆš2x + âˆš3y = 0 and âˆš3x – âˆš8y = 0**

**Solution:**

âˆš2 x + âˆš3 y = 0 …………… (1)

âˆš3 x â€“ âˆš8 y = 0 ………….. (2)

From (1)

x = â€“ (âˆš3/âˆš2)y

Putting x in (2)

âˆš3(-âˆš3/âˆš2)y â€“ âˆš8y = 0

(-3/âˆš2)y – âˆš8y = 0

-3y – 4y = 0

-7y = 0

y = 0

Therefore

x = 0

Therefore, x = 0 and y = 0.

**(vi) (3x/2) â€“ (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

â‡’ 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 â€“ 4y = â€“ 24 and hence find the value of â€˜mâ€™ for which y = mx + 3.**

**Solution:**

2x + 3y = 11â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(1)

2x â€“ 4y = -24â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (2)

From (1)

x = (11 – 3y) / 2

Substituting x in equation (2)

2(11 – 3y) / 2 â€“ 4y =- 24

11 â€“ 3y â€“ 4y = -24

-7y = -24 – 11

-7y = -35

y = 5

Putting y in (1)

x = (11 – 3 Ã— 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.

**Question 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.**

**Solution:**

Let the two numbers be x and y

y = 3x â€¦â€¦â€¦â€¦â€¦â€¦ (1)

y â€“ x = 26 â€¦â€¦â€¦â€¦..(2)

Substituting the value of y

3x â€“ 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 x

^{o}and smaller angle be y^{o}.Sum of two supplementary pair of angles is 180

^{o}.x + y = 180

^{o}â€¦â€¦â€¦â€¦â€¦. (1)x â€“ y = 18

^{o }â€¦â€¦â€¦â€¦â€¦..(2)From (1)

x = 180 – y

Substituting in (2)

180

^{ }â€“ y â€“ y = 18-2y = -162

162 = 2y

y = 81

^{o}x = 180 – y

x = 180 – 81

= 99

Therefore, the angles are 99

^{o}and 81^{o}.

**(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 â€¦â€¦â€¦â€¦â€¦â€¦. (i)

3x + 5y = 1750 â€¦â€¦â€¦â€¦â€¦â€¦. (ii)

From (i)

y = (3800 – 7x) / 6â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

Substituting (iii) in (ii)

3x + 5(3800 – 7x) / 6 =1750

Taking 6 as LCM

18x + 19000 – 7x = 10500

11x = 10500 – 19000

â‡’ -17x = -8500

x = 500 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (IV)

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

y = (3800 – 7 Ã— 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 â€¦â€¦â€¦â€¦â€¦.. (1)

x + 15y = 155 â€¦â€¦â€¦â€¦â€¦.. (2)

From (1)

x = 105 â€“ 10y â€¦â€¦â€¦â€¦â€¦â€¦. (3)

Substituting the value of x in (2)

105 â€“ 10y + 15y = 155

5y = 50

y = 10

Putting the value of y in (3)

x = 105 â€“ 10 Ã— 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 â€“ 9y = -4 â€¦â€¦â€¦â€¦â€¦.. (1)

By cross multiplication

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

6x + 18 = 5y + 15

6x â€“ 5y = -3 â€¦â€¦â€¦â€¦â€¦â€¦. (2)

From (1)

x = (-4 + 9y)/11 â€¦â€¦â€¦â€¦â€¦.. (3)

Substituting the value of x in (2)

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

Taking 11 as the LCM

-24 + 54y â€“ 55y = -33

-24 – y = -33

-y = -33 + 24

-y = -9

y = 9

Substituting the value of y in (3)

x = (-4 + 9 Ã— 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, Jacobâ€™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 â€“ 5) = 7(y â€“ 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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