Pair Of Linear Equations In Two Variables Ex 3.2 Solution

Pair of Linear Equations in Two Variables Ex 3.2 introduces the method of solving a pair of linear equations in two variables algebraically. In this exercise, students apply the method of substitution, where one equation is solved for one variable and substituted into the other. This forms the foundation for solving real-life mathematical problems using linear equations.

Pair of Linear Equations in Two Variables Ex 3.2 Textbook solutions

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

(i)

x + y = 14
x − y = 4

Solution:

From (1):
x + y = 14
⇒ x = 14 − y

Substitute into (2):
(14 − y) − y = 4
⇒ 14 − 2y = 4
⇒ 2y = 10
⇒ y = 5

Now,
x = 14 − 5 = 9

Answer:
x = 9, y = 5

(ii)

s − t = 3
s/3 + t/2 = 6

Solution:

From (1):
s = t + 3

Substitute into (2):
(t + 3)/3 + t/2 = 6

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

Now,
s = 6 + 3 = 9

Answer:
s = 9, t = 6

(iii)

3x − y = 3
9x − 3y = 9

Solution:

Divide second equation by 3:
3x − y = 3

Both equations are same ⇒ infinitely many solutions

Answer:
Infinitely many solutions (dependent equations)

(iv)

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

Solution:

Multiply (1) by 10:
2x + 3y = 13

Multiply (2) by 10:
4x + 5y = 23

From (1):
2x = 13 − 3y
⇒ x = (13 − 3y)/2

Substitute into (2):
4[(13 − 3y)/2] + 5y = 23
⇒ 2(13 − 3y) + 5y = 23
⇒ 26 − 6y + 5y = 23
⇒ 26 − y = 23
⇒ y = 3

Now,
x = (13 − 9)/2 = 2

Answer:
x = 2, y = 3

(v)

√2x + √3y = 0
√3x − √8y = 0

Solution:

From (1):
√2x = −√3y
⇒ x = −(√3/√2)y

Substitute into (2):
√3[−(√3/√2)y] − √8y = 0
⇒ −(3/√2)y − √8y = 0

Multiply by √2:
−3y − √16y = 0
⇒ −3y − 4y = 0
⇒ −7y = 0
⇒ y = 0

Then x = 0

Answer:
x = 0, y = 0

(vi)

(3x/2) − (5y/3) = −2
(x/3) + (y/2) = 13/6

Solution:

Multiply (1) by 6:
9x − 10y = −12

Multiply (2) by 6:
2x + 3y = 13

From (2):
2x = 13 − 3y
⇒ x = (13 − 3y)/2

Substitute into (1):
9[(13 − 3y)/2] − 10y = −12
⇒ (117 − 27y)/2 − 10y = −12

Multiply by 2:
117 − 27y − 20y = −24
⇒ 117 − 47y = −24
⇒ 47y = 141
⇒ y = 3

Now,
x = (13 − 9)/2 = 2

Answer:
x = 2, y = 3

Pair of Linear Equations in Two Variables Ex 3.2

2. Solve and find the value of m Given equations:

2x + 3y = 11
2x − 4y = −24

Solution:

Step 1: Solve the system of equations

$\begin{cases}2x + 3y = 11 \\ 2x – 4y = -24\end{cases}$

Subtract the second equation from the first:

(2x + 3y) − (2x − 4y) = 11 − (−24)
⇒ 2x + 3y − 2x + 4y = 35
⇒ 7y = 35
⇒ y = 5

Now substitute y = 5 into the first equation:

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

Step 2: Find the value of m

Given:
y = mx + 3

Substitute x = −2 and y = 5:

5 = m(−2) + 3
⇒ 5 = −2m + 3
⇒ −2m = 2
⇒ m = −1

Pair of Linear Equations in Two Variables Ex 3.2

3. Form the pair of linear equations and solve them by substitution method

(i) Question: 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.

Given:
x − y = 26
x = 3y

Substitute x = 3y into first equation:
3y − y = 26
⇒ 2y = 26
⇒ y = 13

Now,
x = 3 × 13 = 39

Answer:
The numbers are 39 and 13

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

Solution:
Let smaller angle = x
Larger angle = y

Supplementary angles:
x + y = 180

Given:
y = x + 18

Substitute into first equation:
x + (x + 18) = 180
⇒ 2x + 18 = 180
⇒ 2x = 162
⇒ x = 81

Now,
y = 81 + 18 = 99

Answer:
Angles are 81° and 99°

(iii) Question: 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.

Solution:
Let cost of one bat = x
Let cost of one ball = y

Equations:
7x + 6y = 3800
3x + 5y = 1750

From second equation:
3x = 1750 − 5y
⇒ x = (1750 − 5y)/3

Substitute into first:
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

Now,
3x + 5(50) = 1750
⇒ 3x + 250 = 1750
⇒ 3x = 1500
⇒ x = 500

Answer:
Cost of one bat = ₹500
Cost of one ball = ₹50

(iv) Question: The taxi charges in a city consist of a fixed charge and a charge per km. For 10 km, fare is ₹105 and for 15 km, fare is ₹155. Find the fixed charge and per km charge. Also find fare for 25 km.

Solution:
Let fixed charge = x
Charge per km = y

Equations:
x + 10y = 105
x + 15y = 155

From first:
x = 105 − 10y

Substitute into second:
105 − 10y + 15y = 155
⇒ 105 + 5y = 155
⇒ 5y = 50
⇒ y = 10

Now,
x = 105 − 100 = 5

For 25 km:
Fare = x + 25y
= 5 + 250 = 255

Answer:
Fixed charge = ₹5
Charge per km = ₹10
Fare for 25 km = ₹255

Pair of Linear Equations in Two Variables Ex 3.2

(v) Question: A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.

Solution:
Let numerator = x
Denominator = y

Equations:
(x + 2)/(y + 2) = 9/11
(x + 3)/(y + 3) = 5/6

Cross multiply:

11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18
⇒ 11x − 9y = −4

6(x + 3) = 5(y + 3)
⇒ 6x + 18 = 5y + 15
⇒ 6x − 5y = −3

From second:
6x = 5y − 3
⇒ x = (5y − 3)/6

Substitute into first:
11(5y − 3)/6 − 9y = −4

Multiply by 6:
55y − 33 − 54y = −24
⇒ y − 33 = −24
⇒ y = 9

Now,
x = (45 − 3)/6 = 42/6 = 7

Answer:
Fraction = 7/9

Pair of Linear Equations in Two Variables Ex 3.1

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In this exercise Pair of Linear Equations in Two Variables Ex 3.2, we learned to form and solve pairs of linear equations using the substitution method. It helps in solving real-life problems easily by finding unknown values. Regular practice will improve speed and accuracy.