Class 10 Math Ex 3.3 Linear Equation-NCERT Solution

Class 10 Math Ex 3.3 Linear Equation-NCERT Solution focuses on solving pairs of linear equations using the elimination method. The problems involve simplifying and manipulating the given equations to eliminate one variable, allowing students to find the solution for both variables. This exercise strengthens algebraic skills and builds a strong foundation for solving linear systems efficiently.

Class 10 Math Ex 3.3 Linear Equation-NCERT Solution

Question 1.| Class 10 Maths Ex 3.3 Solutions – Linear Equations
Solve the following pairs of linear equations by the substitution method:

SOL:

Class 10 Math Ex 3.3 Linear Equation-NCERT Solution

Question 2.| Class 10 Maths Ex 3.3 Solutions – Linear Equations
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of’m’ for which y = mx +3.
Solution:

Ex 3.3 Class 10 Maths |Class 10 Maths Ex 3.3 Solutions – Linear Equations

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

(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 ₹3800. Later, she buys 3 bats and 5 balls for ₹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 ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charges per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 911, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 56. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five year ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution:

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

Let the numbers be x and y.
Equation 1: x – y = 26
Equation 2: x = 3y

Substitute equation 2 into equation 1:
3y – y = 26
2y = 26 → y = 13
Then x = 3 × 13 = 39

Answer: The numbers are 39 and 13.

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

Let the smaller angle be x and the larger be y.
Equation 1: x + y = 180
Equation 2: y = x + 18

Substitute equation 2 into equation 1:
x + (x + 18) = 180
2x + 18 = 180
2x = 162 → x = 81
Then y = 81 + 18 = 99

Answer: The angles are 81° and 99°.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750.

Let the cost of a bat be x and the cost of a ball be y.
Equation 1: 7x + 6y = 3800
Equation 2: 3x + 5y = 1750

From equation 2:
3x = 1750 – 5y → x = (1750 – 5y) / 3

Substitute into equation 1:
7[(1750 – 5y)/3] + 6y = 3800
(12250 – 35y)/3 + 6y = 3800
(12250 – 35y + 18y)/3 = 3800
(12250 – 17y)/3 = 3800
Multiply both sides by 3:
12250 – 17y = 11400
17y = 850 → y = 50
Then x = (1750 – 5×50) / 3 = 1500 / 3 = 500

Answer: Bat = ₹500, Ball = ₹50

(iv) The taxi charges are ₹105 for 10 km and ₹155 for 15 km.

Let the fixed charge be x and the per km charge be y.
Equation 1: x + 10y = 105
Equation 2: x + 15y = 155

Subtract equation 1 from equation 2:
(x + 15y) – (x + 10y) = 155 – 105
5y = 50 → y = 10
Then from equation 1:
x + 10×10 = 105 → x = 5

To find charge for 25 km:
x + 25y = 5 + 25×10 = 255

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

(v) A fraction becomes 9/11 if 2 is added to both numerator and denominator. It becomes 5/6 if 3 is added to both.

Let the fraction be x/y.
Equation 1: (x + 2)/(y + 2) = 9/11 → 11(x + 2) = 9(y + 2)
Equation 2: (x + 3)/(y + 3) = 5/6 → 6(x + 3) = 5(y + 3)

From equation 1:
11x + 22 = 9y + 18
11x – 9y = -4 … (A)

From equation 2:
6x + 18 = 5y + 15
6x – 5y = -3 … (B)

From (B): 6x = 5y – 3 → x = (5y – 3)/6
Substitute into (A):
11[(5y – 3)/6] – 9y = -4
(55y – 33)/6 – 9y = -4
(55y – 33 – 54y)/6 = -4
(y – 33)/6 = -4
y – 33 = -24 → y = 9
Then x = (5×9 – 3)/6 = (45 – 3)/6 = 42/6 = 7

Answer: The fraction is 7/9

(vi) Five years hence, Jacob’s age will be three times his son’s. Five years ago, Jacob’s age was seven times his son’s.

Let present age of Jacob be x and his son be y.
Equation 1: x + 5 = 3(y + 5)
Equation 2: x – 5 = 7(y – 5)

From equation 1:
x + 5 = 3y + 15 → x = 3y + 10
Substitute into equation 2:
(3y + 10) – 5 = 7(y – 5)
3y + 5 = 7y – 35
4y = 40 → y = 10
Then x = 3×10 + 10 = 40

Answer: Jacob is 40 years old, son is 10 years old