NCERT Solutions for Class 8 Maths Chapter 2 help students learn how to find the value of an unknown variable in simple equations. This chapter focuses on linear equations in one variable, where the highest power of the variable is one. Students learn step-by-step methods to balance equations, simplify expressions, and solve different types of sums. These solutions also cover real-life word problems, making concepts easy to understand and apply. With regular practice, students build strong problem-solving skills, improve logical thinking, and gain confidence in handling algebraic questions for exams and daily mathematical situations.

Class 8 Maths Chapter 2 Linear Equations in One Variable Questions Answers

Exercise 2.1

Solve the following equations.

1. x – 2 = 7

Solution: x – 2 = 7 x=7+2 x = 9

2. y + 3 = 10

Solution: y + 3 = 10 y = 10 –3 y = 7

3. 6 = z + 2

Solution: 6 = z + 2 z + 2 = 6 z = 6-2 z = 4

4. 3/7 + x = 17/7

Solution: 3/7 + x = 17/7 x = 17/7 – 3/7 x = 14/7 x = 2

5. 6x = 12

Solution: 6x = 12 x = 12/6 x = 2

6. t/5 = 10

Solution: t/5 = 10 t = 10 × 5 t = 50

7. 2x/3 = 18

Solution: 2x/3 = 18 2x = 18 × 3 2x = 54 x = 54/2 x = 27

8. 1.6 = y/15

Solution: 1.6 = y/1.5 y/1.5 = 1.6 y = 1.6 × 1.5 y = 2.4

9. 7x – 9 = 16

Solution: 7x – 9 = 16 7x = 16+9 7x = 25 x = 25/7

10. 14y – 8 = 13

Solution: 14y – 8 = 13 14y = 13+8 14y = 21 y = 21/14 y = 3/2

11. 17 + 6p = 9

Solution: 17 + 6p = 9 6p = 9 – 17 6p = -8 p = -8/6 p = -4/3

12. x/3 + 1 = 7/15

Solution: x/3 + 1 = 7/15 x/3 = 7/15 – 1 x/3 = (7 -15)/15 x/3 = -8/15 x = -8/15 × 3 x = -8/5

Read More - NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.2

1. If you subtract ½ from a number and multiply the result by ½, you get 1/8. What is the number?

Solution: Let the number be x. According to the question, (x – 1/2) × ½ = 1/8 x/2 – ¼ = 1/8 x/2 = 1/8 + ¼ x/2 = 1/8 + 2/8 x/2 = (1+ 2)/8 x/2 = 3/8 x = (3/8) × 2 x = ¾

2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m, more than twice its breadth. What are the length and breadth of the pool?

Solution: Given that, The perimeter of the rectangular swimming pool = 154 m. Let the breadth of the rectangle be = x According to the question, Length of the rectangle = 2x + 2 We know that, Perimeter = 2(length + breadth) ⇒ 2(2x + 2 + x) = 154 m ⇒ 2(3x + 2) = 154 ⇒ 3x +2 = 154/2 ⇒ 3x = 77 – 2 ⇒ 3x = 75 ⇒ x = 75/3 ⇒ x = 25 m Therefore, Breadth = x = 25 cm Length = 2x + 2 = (2 × 25) + 2 = 50 + 2 = 52 m

3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle iscm. What is the length of either of the remaining equal sides?

Solution: Base of isosceles triangle = 4/3 cm Perimeter of triangle =image cm = 62/15 Let the length of equal sides of the triangle be x. According to the question, 4/3 + x + x = 62/15 cm ⇒ 2x = (62/15 – 4/3) cm ⇒ 2x = (62 – 20)/15 cm ⇒ 2x = 42/15 cm ⇒ x = (42/30) × (½) ⇒ x = 42/30 cm ⇒ x = 7/5 cm The length of either of the remaining equal sides is 7/5 cm.

4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Solution: Let one of the numbers be = x. Then, the other number becomes x + 15. According to the question, x + x + 15 = 95 ⇒ 2x + 15 = 95 ⇒ 2x = 95 – 15 ⇒ 2x = 80 ⇒ x = 80/2 ⇒ x = 40 First number = x = 40 And, other number = x + 15 = 40 + 15 = 55

5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Solution: Let the two numbers be 5x and 3x. According to the question, 5x – 3x = 18 ⇒ 2x = 18 ⇒ x = 18/2 ⇒ x = 9 Thus, The numbers are 5x = 5 × 9 = 45 And 3x = 3 × 9 = 27.

6. Three consecutive integers add up to 51. What are these integers?

Solution: Let the three consecutive integers be x, x+1 and x+2. According to the question, x + (x+1) + (x+2) = 51 ⇒ 3x + 3 = 51 ⇒ 3x = 51 – 3 ⇒ 3x = 48 ⇒ x = 48/3 ⇒ x = 16 Thus, the integers are x = 16 x + 1 = 17 x + 2 = 18

7. The sum of three consecutive multiples of 8 is 888. Find the multiples.

Solution: Let the three consecutive multiples of 8 be 8x, 8(x+1) and 8(x+2). According to the question, 8x + 8(x+1) + 8(x+2) = 888 ⇒ 8 (x + x+1 + x+2) = 888 (Taking 8 as common) ⇒ 8 (3x + 3) = 888 ⇒ 3x + 3 = 888/8 ⇒ 3x + 3 = 111 ⇒ 3x = 111 – 3 ⇒ 3x = 108 ⇒ x = 108/3 ⇒ x = 36 Thus, the three consecutive multiples of 8 are: 8x = 8 × 36 = 288 8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296 8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4, respectively, they add up to 74. Find these numbers.

Solution: Let the three consecutive integers be x, x+1 and x+2. According to the question, 2x + 3(x+1) + 4(x+2) = 74 ⇒ 2x + 3x +3 + 4x + 8 = 74 ⇒ 9x + 11 = 74 ⇒ 9x = 74 – 11 ⇒ 9x = 63 ⇒ x = 63/9 ⇒ x = 7 Thus, the numbers are: x = 7 x + 1 = 8 x + 2 = 9

9. The ages of Rahul and Haroon are in the ratio 5:7. Four years later, the sum of their ages will be 56 years. What are their present ages?

Solution: Let the ages of Rahul and Haroon be 5x and 7x. Four years later, The ages of Rahul and Haroon will be (5x + 4) and (7x + 4), respectively. According to the question, (5x + 4) + (7x + 4) = 56 ⇒ 5x + 4 + 7x + 4 = 56 ⇒ 12x + 8 = 56 ⇒ 12x = 56 – 8 ⇒ 12x = 48 ⇒ x = 48/12 ⇒ x = 4 Therefore, Present age of Rahul = 5x = 5×4 = 20 And, present age of Haroon = 7x = 7×4 = 28

10. The number of boys and girls in a class is in the ratio of 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Solution: Let the number of boys be 7x, and girls be 5x. According to the question, 7x = 5x + 8 ⇒ 7x – 5x = 8 ⇒ 2x = 8 ⇒ x = 8/2 ⇒ x = 4 Therefore, number of boys = 7×4 = 28 And, number of girls = 5×4 = 20 Total number of students = 20+28 = 48

11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Solution: Let the age of Baichung’s father be x. Then, the age of Baichung’s grandfather = (x+26) and, the age of Baichung = (x-29). According to the question, x + (x+26) + (x-29) = 135 ⇒ 3x + 26 – 29 = 135 ⇒ 3x – 3 = 135 ⇒ 3x = 135 + 3 ⇒ 3x = 138 ⇒ x = 138/3 ⇒ x = 46 Age of Baichung’s father = x = 46 Age of Baichung’s grandfather = (x+26) = 46 + 26 = 72 Age of Baichung = (x-29) = 46 – 29 = 17

12. Fifteen years from now, Ravi’s age will be four times his present age. What is Ravi’s present age?

Solution: Let the present age of Ravi be x. Fifteen years later, Ravi’s age will be x+15 years. According to the question, x + 15 = 4x ⇒ 4x – x = 15 ⇒ 3x = 15 ⇒ x = 15/3 ⇒ x = 5 Therefore, the present age of Ravi = 5 years.

13. A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?

Solution: Let the rational be x. According to the question, x × (5/2) + 2/3 = -7/12 ⇒ 5x/2 + 2/3 = -7/12 ⇒ 5x/2 = -7/12 – 2/3 ⇒ 5x/2 = (-7- 8)/12 ⇒ 5x/2 = -15/12 ⇒ 5x/2 = -5/4 ⇒ x = (-5/4) × (2/5) ⇒ x = – 10/20 ⇒ x = -½ Therefore, the rational number is -½.

14. Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, ₹50 and ₹10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹4,00,000. How many notes of each denomination does she have?

Solution: Let the numbers of notes of ₹100, ₹50 and ₹10 be 2x, 3x and 5x, respectively. Value of ₹100 = 2x × 100 = 200x Value of ₹50 = 3x × 50 = 150x Value of ₹10 = 5x × 10 = 50x According to the question, 200x + 150x + 50x = 4,00,000 ⇒ 400x = 4,00,000 ⇒ x = 400000/400 ⇒ x = 1000 Numbers of ₹100 notes = 2x = 2000 Numbers of ₹50 notes = 3x = 3000 Numbers of ₹10 notes = 5x = 5000

15. I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of ₹2 coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Solution: Let the number of ₹5 coins be x. Then, Number ₹2 coins = 3x And, number of ₹1 coins = (160 – 4x) Now, Value of ₹5 coins = x × 5 = 5x Value of ₹2 coins = 3x × 2 = 6x Value of ₹1 coins = (160 – 4x) × 1 = (160 – 4x) According to the question, 5x + 6x + (160 – 4x) = 300 ⇒ 11x + 160 – 4x = 300 ⇒ 7x = 140 ⇒ x = 140/7 ⇒ x = 20 Number of ₹5 coins = x = 20 Number of ₹2 coins = 3x = 60 Number of ₹1 coins = (160 – 4x) = 160 – 80 = 80

16. The organisers of an essay competition decide that a winner in the competition gets a prize of ₹100 and a participant who does not win gets a prize of ₹25. The total prize money distributed is ₹3,000. Find the number of winners, if the total number of participants is 63.

Solution: Let the number of winners be x. Then, the number of participants who didn’t win = 63 – x Total money given to the winner = x × 100 = 100x Total money given to the participant who didn’t win = 25 × (63-x) According to the question, 100x + 25 × (63-x) = 3,000 ⇒ 100x + 1575 – 25x = 3,000 ⇒ 75x = 3,000 – 1575 ⇒ 75x = 1425 ⇒ x = 1425/75 ⇒ x = 19 Therefore, the numbers of winners are 19.

Read More - NCERT Solutions for Class 8 Maths Chapter 4 Data Handling

NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3

Solve the following equations and check your results.

1. 3x = 2x + 18

Solution: 3x = 2x + 18 ⇒ 3x – 2x = 18 ⇒ x = 18 Putting the value of x in RHS and LHS, we get, 3 × 18 = (2 × 18) +18 ⇒ 54 = 54 ⇒ LHS = RHS

2. 5t – 3 = 3t – 5

Solution: 5t – 3 = 3t – 5 ⇒ 5t – 3t = -5 + 3 ⇒ 2t = -2 ⇒ t = -1 Putting the value of t in RHS and LHS, we get, 5× (-1) – 3 = 3× (-1) – 5 ⇒ -5 – 3 = -3 – 5 ⇒ -8 = -8 ⇒ LHS = RHS

3. 5x + 9 = 5 + 3x

Solution: 5x + 9 = 5 + 3x ⇒ 5x – 3x = 5 – 9 ⇒ 2x = -4 ⇒ x = -2 Putting the value of x in RHS and LHS, we get, 5× (-2) + 9 = 5 + 3× (-2) ⇒ -10 + 9 = 5 + (-6) ⇒ -1 = -1 ⇒ LHS = RHS

4. 4z + 3 = 6 + 2z

Solution: 4z + 3 = 6 + 2z ⇒ 4z – 2z = 6 – 3 ⇒ 2z = 3 ⇒ z = 3/2 Putting the value of z in RHS and LHS, we get, (4 × 3/2) + 3 = 6 + (2 × 3/2) ⇒ 6 + 3 = 6 + 3 ⇒ 9 = 9 ⇒ LHS = RHS

5. 2x – 1 = 14 – x

Solution: 2x – 1 = 14 – x ⇒ 2x + x = 14 + 1 ⇒ 3x = 15 ⇒ x = 5 Putting the value of x in RHS and LHS, we get, (2×5) – 1 = 14 – 5 ⇒ 10 – 1 = 9 ⇒ 9 = 9 ⇒ LHS = RHS

6. 8x + 4 = 3 (x – 1) + 7

Solution: 8x + 4 = 3 (x – 1) + 7 ⇒ 8x + 4 = 3x – 3 + 7 ⇒ 8x + 4 = 3x + 4 ⇒ 8x – 3x = 4 – 4 ⇒ 5x = 0 ⇒ x = 0 Putting the value of x in RHS and LHS, we get, (8×0) + 4 = 3 (0 – 1) + 7 ⇒ 0 + 4 = 0 – 3 + 7 ⇒ 4 = 4 ⇒ LHS = RHS

7. x = 4/5 (x + 10)

Solution: x = 4/5 (x + 10) ⇒ x = 4x/5 + 40/5 ⇒ x – (4x/5) = 8 ⇒ (5x – 4x)/5 = 8 ⇒ x = 8 × 5 ⇒ x = 40 Putting the value of x in RHS and LHS, we get, 40 = 4/5 (40 + 10) ⇒ 40 = 4/5 × 50 ⇒ 40 = 200/5 ⇒ 40 = 40 ⇒ LHS = RHS

8. 2x/3 + 1 = 7x/15 + 3

Solution: 2x/3 + 1 = 7x/15 + 3 ⇒ 2x/3 – 7x/15 = 3 – 1 ⇒ (10x – 7x)/15 = 2 ⇒ 3x = 2 × 15 ⇒ 3x = 30 ⇒ x = 30/3 ⇒ x = 10 Putting the value of x in RHS and LHS, we get,

9. 2y + 5/3 = 26/3 – y

Solution: 2y + 5/3 = 26/3 – y ⇒ 2y + y = 26/3 – 5/3 ⇒ 3y = (26 – 5)/3 ⇒ 3y = 21/3 ⇒ 3y = 7 ⇒ y = 7/3 Putting the value of y in RHS and LHS, we get, ⇒ (2 × 7/3) + 5/3 = 26/3 – 7/3 ⇒ 14/3 + 5/3 = 26/3 – 7/3 ⇒ (14 + 5)/3 = (26 – 7)/3 ⇒ 19/3 = 19/3 ⇒ LHS = RHS

10. 3m = 5m – 8/5

Solution: 3m = 5m – 8/5 ⇒ 5m – 3m = 8/5 ⇒ 2m = 8/5 ⇒ 2m × 5 = 8 ⇒ 10m = 8 ⇒ m = 8/10 ⇒ m = 4/5 Putting the value of m in RHS and LHS, we get, ⇒ 3 × (4/5) = (5 × 4/5) – 8/5 ⇒ 12/5 = 4 – (8/5) ⇒ 12/5 = (20 – 8)/5 ⇒ 12/5 = 12/5 ⇒ LHS = RHS

NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.4

1. Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?

Solution: Let the number be x, According to the question, (x – 5/2) × 8 = 3x ⇒ 8x – 40/2 = 3x ⇒ 8x – 3x = 40/2 ⇒ 5x = 20 ⇒ x = 4 Thus, the number is 4.

2. A positive number is 5 times another number. If 21 is added to both numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Solution: Let one of the positive numbers be x, then the other number will be 5x. According to the question, 5x + 21 = 2(x + 21) ⇒ 5x + 21 = 2x + 42 ⇒ 5x – 2x = 42 – 21 ⇒ 3x = 21 ⇒ x = 7 One number = x = 7 Other number = 5x = 5×7 = 35. The two numbers are 7 and 35.

3. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Solution: Let the digit at tens place be x, then the digit at ones place will be (9-x). Original two-digit number = 10x + (9-x) After interchanging the digits, the new number = 10(9-x) + x According to the question, 10x + (9-x) + 27 = 10(9-x) + x ⇒ 10x + 9 – x + 27 = 90 – 10x + x ⇒ 9x + 36 = 90 – 9x ⇒ 9x + 9x = 90 – 36 ⇒ 18x = 54 ⇒ x = 3 Original number = 10x + (9-x) = (10×3) + (9-3) = 30 + 6 = 36 Thus, the number is 36.

4. One of the two digits of a two-digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Solution: Let the digit at tens place be x, then the digit at ones place will be 3x. Original two-digit number = 10x + 3x After interchanging the digits, the new number = 30x + x According to the question, (30x + x) + (10x + 3x) = 88 ⇒ 31x + 13x = 88 ⇒ 44x = 88 ⇒ x = 2 Original number = 10x + 3x = 13x = 13×2 = 26

5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one-third of his mother’s present age. What are their present ages?

Solution: Let the present age of Shobo be x, then the age of her mother will be 6x. Shobo’s age after 5 years = x + 5 According to the question, (x + 5) = (1/3) × 6x ⇒ x + 5 = 2x ⇒ 2x – x = 5 ⇒ x = 5 Present age of Shobo = x = 5 years The present age of Shobo’s mother = 6x = 30 years.

6. There is a narrow rectangular plot reserved for a school in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate ₹100 per metre, it will cost the village panchayat ₹75000 to fence the plot. What are the dimensions of the plot?

Solution: Let the length of the rectangular plot be 11x and the breadth be 4x. Rate of fencing per metre = ₹100 Total cost of fencing = ₹75000 Perimeter of the plot = 2(l+b) = 2(11x + 4x) = 2×15x = 30x Total amount of fencing = (30x × 100) According to the question, (30x × 100) = 75000 ⇒ 3000x = 75000 ⇒ x = 75000/3000 ⇒ x = 25 Length of the plot = 11x = 11 × 25 = 275m Breadth of the plot = 4 × 25 = 100m.

7. Hasan buys two kinds of cloth materials for school uniforms; shirt material that costs him ₹50 per metre and trouser material that costs him ₹90 per metre. For every 3 meters of the shirt material, he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit, respectively. His total sale is ₹36,600. How much trouser material did he buy?

Solution: Let 2x m of trouser material and 3x m of shirt material be bought by him Selling price of shirt material per meter = ₹ 50 + 50 ×(12/100) = ₹ 56 Selling price of trouser material per meter = ₹ 90 + 90 × (10/100) = ₹ 99 Total amount of sale = ₹36,600 According to the question, (2x × 99) + (3x × 56) = 36600 ⇒ 198x + 168x = 36600 ⇒ 366x = 36600 ⇒ x = 36600/366 ⇒ x = 100 Total trouser material he bought = 2x = 2 × 100 = 200 m.

8. Half of a herd of deer is grazing in the field, and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Solution: Let the total number of deer be x. Deer grazing in the field = x/2 Deer playing nearby = x/2 × ¾ = 3x/8 Deer drinking water = 9 According to the question, x/2 + 3x/8 + 9 = x (4x + 3x)/8 + 9 = x ⇒ 7x/8 + 9 = x ⇒ x – 7x/8 = 9 ⇒ (8x – 7x)/8 = 9 ⇒ x = 9 × 8 ⇒ x = 72

9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Solution: Let the age of granddaughter be x and grandfather be 10x. Also, he is 54 years older than her. According to the question, 10x = x + 54 ⇒ 10x – x = 54 ⇒ 9x = 54 ⇒ x = 6 Age of grandfather = 10x = 10×6 = 60 years. Age of granddaughter = x = 6 years.

10. Aman’s age is three times his son’s age. Ten years ago, he was five times his son’s age. Find their present ages.

Solution: Let the age of Aman’s son be x, then the age of Aman will be 3x. According to the question, 5(x – 10) = 3x – 10 ⇒ 5x – 50 = 3x – 10 ⇒ 5x – 3x = -10 + 50 ⇒ 2x = 40 ⇒ x = 20 Aman’s son age = x = 20 years Aman age = 3x = 3×20 = 60 years

NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.5

Solve the following linear equations.

1. x/2 – 1/5 = x/3 + ¼

Solution: x/2 – 1/5 = x/3 + ¼ ⇒ x/2 – x/3 = ¼+ 1/5 ⇒ (3x – 2x)/6 = (5 + 4)/20 ⇒ 3x – 2x = 9/20 × 6 ⇒ x = 54/20 ⇒ x = 27/10

2. n/2 – 3n/4 + 5n/6 = 21

Solution: n/2 – 3n/4 + 5n/6 = 21 ⇒ (6n – 9n + 10n)/12 = 21 ⇒ 7n/12 = 21 ⇒ 7n = 21 × 12 ⇒ n = 252/7 ⇒ n = 36

3. x + 7 – 8x/3 = 17/6 – 5x/2

Solution: x + 7 – 8x/3 = 17/6 – 5x/2 ⇒ x – 8x/3 + 5x/2 = 17/6 – 7 ⇒ (6x – 16x + 15x)/6 = (17 – 42)/6 ⇒ 5x/6 = – 25/6 ⇒ 5x = – 25 ⇒ x = – 5

4. (x – 5)/3 = (x – 3)/5

Solution: (x – 5)/3 = (x – 3)/5 ⇒ 5(x-5) = 3(x-3) ⇒ 5x-25 = 3x-9 ⇒ 5x – 3x = -9+25 ⇒ 2x = 16 ⇒ x = 8

5. (3t – 2)/4 – (2t + 3)/3 = 2/3 – t

Solution: (3t – 2)/4 – (2t + 3)/3 = 2/3 – t ⇒ ((3t – 2)/4) × 12 – ((2t + 3)/3) × 12 ⇒ (3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t ⇒ 9t – 6 – 8t – 12 = 8 – 12t ⇒ 9t – 6 – 8t – 12 = 8 – 12t ⇒ t – 18 = 8 – 12t ⇒ t + 12t = 8 + 18 ⇒ 13t = 26 ⇒ t = 2

6. m – (m – 1)/2 = 1 – (m – 2)/3

Solution: m – (m – 1)/2 = 1 – (m – 2)/3 ⇒ m – m/2 – 1/2 = 1 – (m/3 – 2/3) ⇒ m – m/2 + ½ = 1 – m/3 + 2/3 ⇒ m – m/2 + m/3 = 1 + 2/3 – ½ ⇒ m/2 + m/3 = ½ + 2/3 ⇒ (3m + 2m)/6 = (3 + 4)/6 ⇒ 5m/6 = 7/6 ⇒ m = 7/6 × 6/5 ⇒ m = 7/5

Simplify and solve the following linear equations.

7. 3 (t – 3) = 5(2t + 1)

Solution: 3(t – 3) = 5(2t + 1) ⇒ 3t – 9 = 10t + 5 ⇒ 3t – 10t = 5 + 9 ⇒ -7t = 14 ⇒ t = 14/-7 ⇒ t = -2

8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0

Solution: 15(y – 4) –2(y – 9) + 5(y + 6) = 0 ⇒ 15y – 60 -2y + 18 + 5y + 30 = 0 ⇒ 15y – 2y + 5y = 60 – 18 – 30 ⇒ 18y = 12 ⇒ y = 12/18 ⇒ y = 2/3

9. 3 (5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Solution: 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17 ⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17 ⇒ 15z – 18z – 32z = -52 – 17 + 21 – 22 ⇒ -35z = -70 ⇒ z = -70/-35 ⇒ z = 2

10. 0.25(4f – 3) = 0.05(10f – 9)

Solution: 0.25(4f – 3) = 0.05(10f – 9) ⇒ f – 0.75 = 0.5f – 0.45 ⇒ f – 0.5f = -0.45 + 0.75 ⇒ 0.5f = 0.30 ⇒ f = 0.30/0.5 ⇒ f = 3/5 ⇒ f = 0.6

NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.6

Solve the following equations.

1. (8x – 3)/3x = 2

Solution: (8x – 3)/3x = 2 ⇒ 8x/3x – 3/3x = 2 ⇒ 8/3 – 1/x = 2 ⇒ 8/3 – 2 = 1/x ⇒ (8 – 6)/3 = 1/x ⇒ 2/3 = 1/x ⇒ x = 3/2

2. 9x/(7 – 6x) = 15

Solution: 9x/(7 – 6x) = 15 ⇒ 9x = 15(7 – 6x) ⇒ 9x = 105 – 90x ⇒ 9x + 90x = 105 ⇒ 99x = 105 ⇒ x = 105/99 = 35/33

3. z/(z + 15) = 4/9

Solution: z/(z + 15) = 4/9 ⇒ z = 4/9 (z + 15) ⇒ 9z = 4(z + 15) ⇒ 9z = 4z + 60 ⇒ 9z – 4z = 60 ⇒ 5z = 60 ⇒ z = 12

4. (3y + 4)/(2 – 6y) = -2/5

Solution: (3y + 4)/(2 – 6y) = -2/5 ⇒ 3y + 4 = -2/5 (2 – 6y) ⇒ 5(3y + 4) = -2(2 – 6y) ⇒ 15y + 20 = -4 + 12y ⇒ 15y – 12y = -4 – 20 ⇒ 3y = -24 ⇒ y = -8

5. (7y + 4)/(y + 2) = -4/3

Solution: (7y + 4)/(y + 2) = -4/3 ⇒ 7y + 4 = -4/3 (y + 2) ⇒ 3(7y + 4) = -4(y + 2) ⇒ 21y + 12 = -4y – 8 ⇒ 21y + 4y = -8 – 12 ⇒ 25y = -20 ⇒ y = -20/25 = -4/5

6. The ages of Hari and Harry are in the ratio of 5:7. Four years from now, the ratio of their ages will be 3:4. Find their present ages.

Solution: Let the age of Hari be 5x and Harry be 7x. 4 years later, Age of Hari = 5x + 4 Age of Harry = 7x + 4 According to the question, (5x + 4)/(7x + 4) = ¾ ⇒ 4(5x + 4) = 3(7x + 4) ⇒ 20x + 16 = 21x + 12 ⇒ 21x – 20x = 16 – 12 ⇒ x = 4 Hari’s age = 5x = 5 × 4 = 20 years Harry’s age = 7x = 7 × 4 = 28 years

7. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.

Solution: Let the numerator be x, then the denominator will be (x + 8) According to the question, (x + 17)/(x + 8 – 1) = 3/2 ⇒ (x + 17)/(x + 7) = 3/2 ⇒ 2(x + 17) = 3(x + 7) ⇒ 2x + 34 = 3x + 21 ⇒ 34 – 21 = 3x – 2x ⇒ 13 = x The rational number is x/(x + 8) = 13/21

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Understanding the Basics of Linear Equations

When you start looking at class 8 maths chapter 2 solutions, the first thing you notice is the word "Linear." In very simple terms, a linear equation is like a balance scale. On one side, you have some numbers and letters (variables), and on the other side, you have another set of numbers. The "equal to" sign (=) means that both sides must weigh exactly the same.

In this chapter, we only use one letter, like x or y. This is why it is called "in one variable." If you see an equation like 2x - 3 = 7 your goal is to find out what x is. To do this, we move the numbers around until the letter is all by itself. This process is the heart of class 8 maths chapter 2 solutions, which focus on making these steps clearer for every student.

How to Solve Equations Step-by-Step

Solving these math problems is like playing a game of "Opposites." If a number is being added on one side, it becomes subtracted when you move it to the other side. If it is being multiplied, it becomes division.

Step 1: Transposing Numbers

In many class 8 maths chapter 2 solutions, we use a method called "transposing." This just means moving a number from the left side to the right side of the equals sign. Don't forget the golden rule: when you move a number across the "equal" bridge, its sign changes! Plus becomes minus, and minus becomes plus.

Step 2: Simplifying the Equation

Sometimes, an equation looks messy. You might see something like 3x + 5 = 2x + 10. In these cases, we want to bring all the letters to one side and all the plain numbers to the other. This is a key part of class 8 maths chapter 2 solutions ganita prakash style learning, where we keep things organized.

Applications and Word Problems

Math isn't just about numbers on a page; it’s about solving real-life puzzles. Many students search for class 8 maths chapter 2 solutions pdf because the word problems can be a bit tricky. These problems talk about ages, lengths of a rectangle, or even how many coins are in a bag.

Let’s look at a simple example. If the sum of two numbers is 95 and one is 10 more than the other, how do we find them?

1. Let the first number be x.

2. The second number will be x+ 10

3. The equation becomes x + (x + 10) = 95

4. Now, we solve for x just like we practiced!

Using class 8 maths chapter 2 solutions power play techniques, you can break down these big stories into small, easy math sentences. This makes the subject feel much less scary for students in younger classes like class 4th or 7th who are looking ahead.

Why Practice with NCERT Solutions?

Using the class 8 maths chapter 2 solutions is the best way to prepare for exams. These solutions follow the official syllabus, ensuring you don't waste time on topics that aren't important.

• Clear Logic: Every step is explained so you know "why" we did a calculation, not just "how."

• Better Grades: Since most school exams use NCERT patterns, practicing these specific problems helps you score higher.

• Building Foundation: Learning linear equations now will make high school algebra feel much easier later on.

If you ever feel stuck, you can download a class 8 maths chapter 2 solutions pdf to study even when you don't have an internet connection. This is very helpful for quick revision before a class test.

Practice Exercises for Students

To help you get started, try solving these simple problems. You can check the class 8 maths chapter 2 solutions if you get confused.

1. x - 2 = 7 (Hint: Move -2 to the other side)

2. 5t - 3 = 3t - 5(Hint: Get all 't' letters on one side)

3. z / 4 = 9 (Hint: The opposite of division is multiplication)

When you practice these daily, you will find that math is just like a puzzle waiting to be solved. Using class 8 maths chapter 2 solutions ensures you are up to date with the latest school standards.

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