Pair of Linear Equations in Two Variables Class 10 Exercise 3.2 Solution.

- June 09, 2026

Pair of Linear Equations in Two Variables Class 10 Exercise 3.2 Solution.

Introduction:

दो चरों वाले रैखिक समीकरणों (Pair of Linear Equations in Two Variables) के युग्म ऐसे समीकरण होते हैं जिनमें दो अज्ञात राशियाँ होती हैं। इन समीकरणों का हल वह मान होता है जो दोनों समीकरणों को एक साथ संतुष्ट करता है। इस अभ्यास (Exercise 3.2) में हमने प्रतिस्थापन विधि (Substitution Method) का उपयोग करके विभिन्न प्रश्नों को हल करना सीखा तथा दो चरों के मान ज्ञात किए।

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

i) x + y = 14
x − y = 4

Solution:

Step 1:

From first equation:

$x = 14 - y$

Step 2: Substitute

(14 − y) − y = 4
⇒ 14 − 2y = 4
⇒ 2y = 10
⇒ y = 5

Step 3:

x = 14 − 5 = 9

Answer:

x = 9, y = 5

ii) s − t = 3

$\frac{s}{3} + \frac{t}{2} = 6$

$Solution:$

Step 1:

s = t + 3

Step 2: Substitute

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

Multiply by 6:
2(t + 3) + 3t = 36
⇒ 2t + 6 + 3t = 36
⇒ 5t = 30
⇒ t = 6

s = t + 3 = 9

Answer:

(s, t) = (9, 6)

iii) 3x − y = 3
9x − 3y = 9

Solution:

Second equation = 3 × first

Same equation

Answer:

Infinite solutions (coincident lines)

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

Solution:

From first:

$x = \frac{1.3 - 0.3y}{0.2}$

Substitute:
⇒ y = 3
⇒ x = 2

Answer:

(x, y) = (2, 3)

$\sqrt{3}x - \sqrt{8}y = 0$

Step 1:

From first:
x = −(√3/√2)y

Step 2: Substitute

Solve → y = 0
⇒ x = 0

Answer:

(x, y) = (0, 0)

2. Solve 2x + 3y = 11 and 2x - 4y = - 24 and hence find the value of 'm' for which y = mx + 3

Solution:

Given Equations

2x + 3y = 11
2x − 4y = −24

Step 1: Eliminate x

Subtract second equation from first:

(2x + 3y) − (2x − 4y) = 11 − (−24)

⇒ 2x + 3y − 2x + 4y = 35
⇒ 7y = 35
⇒ y = 5

Step 2: Find x

Substitute y = 5 in first equation:

2x + 3(5) = 11
⇒ 2x + 15 = 11
⇒ 2x = −4
⇒ x = −2

Solution of Equations

(x, y) = (−2, 5)

Step 3: Find m

Substitute x = −2, y = 5 in y = mx + 3

5 = m(−2) + 3
5 = −2m + 3
2 = −2m
m = −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.

Solution:

Step 1: Let Variables

Let:

First number = x
Second number = y

Step 2: Form Equations

1st Condition (Difference = 26):

$x - y = 26$

2nd Condition (One is three times the other):

$x = 3y$

Step 3: Substitution Method

Substitute x = 3y in first equation:

3y − y = 26
⇒ 2y = 26
⇒ y = 13

Step 4: Find x

x = 3y = 3 × 13 = 39

Final Answer

First number = 39
Second number = 13

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

Solution:

Step 1: Let Variables

Let:

Smaller angle = x°
Larger angle = y°

Step 2: Form Equations

1st Condition (Supplementary angles):

Sum = 180°

$x + y = 180$

$2nd Condition (Larger exceeds smaller by 18°):$

$y = x + 18$

Step 3: Substitution Method

Substitute y = x + 18 in first equation:

x + (x + 18) = 180
2x + 18 = 180
2x = 162
x = 81

Step 4: Find y

y = x + 18 = 81 + 18 = 99

Final Answer

Smaller angle = 81°
Larger angle = 99°

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:

Step 1: Let Variables

Let:

Cost of one bat = x
Cost of one ball = y

Step 2: Form Equations

1st Condition:

$7x + 6y = 3800$

2nd Condition:

$3x + 5y = 1750$

Step 3: Substitution Method

From second equation:

$3x = 1750 - 5y$

$x = \frac{1750 - 5y}{3}$

Step 4: Substitute in First Equation

7(1750 − 5y)/3 + 6y = 3800

Multiply by 3:

7(1750 − 5y) + 18y = 11400

12250 − 35y + 18y = 11400
12250 − 17y = 11400
17y = 850
y = 50

Step 5: Find x

x = (1750 − 5×50)/3
x = (1750 − 250)/3
x = 1500/3
x = 500

Final Answer

Cost of one bat = Rs. 500
Cost of one ball = Rs. 50

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

$Solution:$

Step 1: Let Variables

Let:

Fixed charge = x
Charge per km = y

Step 2: Form Equations

For 10 km:

$x + 10y = 105$

For 15 km:

$x + 15y = 155$

Step 3: Subtract Equations

(x + 15y) − (x + 10y) = 155 − 105

5y = 50
y = 10

Step 4: Find Fixed Charge

Substitute y = 10:

x + 10(10) = 105
x + 100 = 105
x = 5

Charges

Fixed charge = Rs. 5
Charge per km = Rs. 10

Step 5: Fare for 25 km

$x + 25y = 5 + 25 \times 10$

= 5 + 250 = Rs. 255

Final Answer

Fixed charge = Rs. 5
Charge per km = Rs. 10
Fare for 25 km = 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:

Step 1: Let the Fraction

Let numerator = x
Let denominator = y

Original fraction = x/y

Step 2: Form Equations

Condition 1:

$\frac{x+2}{y+2} = \frac{9}{11}$

Cross multiply:
11(x + 2) = 9(y + 2)
11x + 22 = 9y + 18
11x − 9y = −4 …… (1)

Condition 2:

$\frac{x+3}{y+3} = \frac{5}{6}$

Cross multiply:
6(x + 3) = 5(y + 3)
6x + 18 = 5y + 15
6x − 5y = −3 …… (2)

Step 3: Solve Equations

Multiply (2) by 11:
66x − 55y = −33

Multiply (1) by 6:
66x − 54y = −24

Subtract:
(66x − 55y) − (66x − 54y) = −33 − (−24)
−y = −9
y = 9

Step 4: Find x

Substitute in (2):
6x − 5(9) = −3
⇒ 6x − 45 = −3
⇒ 6x = 42
⇒ x = 7

Final Answer

Original fraction = 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?

Solution:

Step 1: Let Variables

Let:

Present age of Jacob = x years
Present age of son = y years

Step 2: Form Equations

Condition 1 (After 5 years):

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

⇒ x + 5 = 3y + 15
⇒ x − 3y = 10 …… (1)

Condition 2 (5 years ago):

$x - 5 = 7(y - 5)$

x − 5 = 7y − 35
x − 7y = −30 …… (2)

Step 3: Solve Equations

Subtract (2) from (1):

(x − 3y) − (x − 7y) = 10 − (−30)
x − 3y − x + 7y = 40
4y = 40
y = 10

Step 4: Find x

Substitute in (1):

x − 3(10) = 10
x − 30 = 10
x = 40

Final Answer

Jacob’s present age = 40 years
Son’s present age = 10 years

$Conclusion:$

इस अभ्यास से हमें दो चरों वाले रैखिक समीकरणों (Pair of Linear Equations in Two Variables) के युग्म को प्रतिस्थापन विधि द्वारा हल करना आया। हमने सीखा कि सही गणना और तार्किक चरणों का पालन करके अज्ञात राशियों के मान आसानी से निकाले जा सकते हैं। यह अध्याय दैनिक जीवन की अनेक समस्याओं को गणितीय रूप से हल करने में भी सहायक है।

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