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NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

Last Updated : 12 Sep, 2023
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NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable is created by a team of professionals at GFG, to help students with any queries they might have as they go through problems from the NCERT textbook. It lessens the frustration of spending a long time working on a problem.

Chapter 2 of Class 8 Maths covers the topic of linear equations in one variable. Including topics such as:

Class 8 Maths NCERT Solutions Chapter 2 Exercises:

Linear Equations in One Variable: Exercise 2.1

Question 1. Solve for x: x – 2 = 7

Solution:

x – 2 = 7

x=7+2  (Adding two on both sides of equation)

x=9

Question 2. Solve for y: y + 3 = 10

Solution:

y + 3 = 10

y = 10 –3  (Subtracting 3 from both sides of equation)

y = 7

Question 3. Solve for z: 6 = z + 2

Solution:

6 = z + 2

z + 2 = 6 (Rearranging to standard form)

z = 6 – 2 (Subtracting two from both sides)

z = 4

Question 4. Solve for x: 3/7 + x = 17/7

Solution:

3/7 + x = 17/7

x = 17/7 – 3/7 (Subtracting 3/7 on both sides)

x = 14/7 (Simplifying the equation)

x = 2

Question 5. Solve for x: 6x = 12

Solution:

6x = 12

x = 12/6 (Dividing both sides by 6)

x = 2

Question 6. Solve for t: t/5 = 10

Solution:

t/5 = 10

t = 10 × 5 (multiplying both sides by 5)

t = 50

Question 7. Solve for x: 2x/3 = 18

Solution:

2x/3 = 18

2x = 18 × 3 (Multiplying both sides by 3)

2x = 54

x = 54/2 (Dividing both sides by 2)

x = 27

Question 8. Solve for y: 1.6 = y/1.5

Solution:

1.6 = y/1.5 

y/1.5 = 1.6 (Rearranging to standard form)

y = 1.6 × 1.5 (Multiplying both sides by 1.5)

y = 2.4

Question 9. Solve for x: 7x – 9 = 16

Solution:

Given that 7x – 9 = 16

7x = 16+9 (Adding 9 on both sides of equation)

7x = 25

x = 25/7 (Dividing both sides by 7)

Question 10. Solve for y: 14y – 8 = 13

Solution:

14y – 8 = 13

14y = 13+8 (Adding 8 on both sides of equation)

14y = 21

y = 21/14 (Dividing both sides by 14)

y = 3/2 (Simplifying the equation)

Question 11. Solve for p: 17 + 6p = 9

Solution:

17 + 6p = 9

6p = 9 – 17 (Subtracting 17 from both sides of equation)

6p = -8

p = -8/6 (Dividing both sides by 6)

p = -4/3(Simplifying the equation)

Question 12. Solve for x: x/3 + 1 = 7/15

Solution:

x/3 + 1 = 7/15

x/3 = 7/15 – 1 (Subtracting both sides by 1)

x/3 = (7 – 15)/15 (Making common denominator)

x/3 = -8/15

x = -8/15 × 3 (Multiplying both sides by 3)

x = -8/5

Linear Equations in One Variable: Exercise 2.2

Question 1. If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8 what is the number?

Solution:

Let the number be ‘a’.

According to the question,

(a – 1/2) × 1/2 = 1/8

a/2 – 1/4 = 1/8

a/2 = 1/8 + 1/4

a/2 = 1/8 + 2/8

a/2 = (1 + 2)/8

a/2 = 3/8

a = (3/8) × 2

So,

a = 3/4

Question 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,

Perimeter of rectangular swimming pool = 154 m 

Let the breadth of rectangle be ‘a’

Length of the rectangle = 2a + 2 We know that,

Perimeter = 2 × (length + breadth)

So, 

2(2a + 2 + a) = 154

2(3a + 2) = 154

3a + 2 = 154/2

3a = 77 – 2

3a = 75

a = 75/3

a = 25

Therefore, Breadth = 25 m

Length = 2a + 2

= (2 × 25) + 2

= 50 + 2

Length = 52 m

Question 3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 62/15 cm. What is the length of either of the remaining equal sides?

Solution:

Base of isosceles triangle = 4/3 cm

Perimeter of triangle = 62/15

Let the length of equal sides of triangle be ‘a’.

So,

2a = (62/15 – 4/3)

2a = (62 – 20)/15

2a = 42/15 

a = (42/30) × (1/2)

a = 42/30

a = 7/5

So, length of either of the remaining equal sides are 7/5 cm each.

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

Solution:

Let one of the numbers be ‘a’.

Then, the other number becomes (a + 15) Given in the question,

Also given that,

a + (a + 15) = 95

2a + 15 = 95

2a = 95 – 15

2a = 80

a = 80/2

a = 40

So, First number = 40

And, other number is = (a + 15) = 40 + 15 = 55

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

Solution:

Let the two numbers be ‘5a’ and ‘3a’.  So, according to the question,

5a – 3a = 18

2a = 18

a = 18/2

a = 19

Thus,

The  first numbers is (5a) = 5 × 9 = 45

And another number (3a) = 3 × 9 = 27.

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

Solution:

Let the three consecutive integers be ‘a’, ‘a + 1’ and ‘a + 2’. So, according to the question,

a + (a + 1) + (a + 2) = 51

3a + 3 = 51

3a = 51 – 3

3a = 48

a = 48/3

a = 16

So, the integers are

First integer will be (a) = 16

Second integer will be (a + 1) = 17

& third integer will be (a + 2) = 18

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

Solution:

Let the three consecutive multiples of 8 be ‘8a’, ‘8(a+1)’ and ‘8(a+2)’. According to the question,

Given,

8a + 8(a + 1) + 8(a + 2) = 888

8 (a + a + 1 + a + 2) = 888 (Taking 8 as common)

8 (3a + 3) = 888

3a + 3 = 888/8

3a + 3 = 111

3a = 111 – 3

3a = 108

a = 108/3

a = 36

Thus, the three consecutive multiples of 8 are:

First no. = 8a = 8 × 36 = 288

Second no. = 8(a + 1) = 8 × (36 + 1) = 8 × 37 = 296

Third No. = 8(a + 2) = 8 × (36 + 2) = 8 × 38 = 304

Question 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 are ‘a’, ‘a+1’ and ‘a+2’. According to the question,

Given,

2a + 3(a + 1) + 4(a + 2) = 74

2a + 3a +3 + 4a + 8 = 74

9a + 11 = 74

9a = 74 – 11

9a = 63

a = 63/9

a = 7

Thus, the numbers are:

First integer. = a = 7

Second integer = a + 1 = 8

and Third integer = a + 2 = 9

Question 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 ‘5a’ and ‘7a’.

Four years later,

The ages of Rahul and Haroon will be (5a + 4) and (7a + 4) respectively. According to the question,

Given, (5a + 4) + (7a + 4) = 56

5a + 4 + 7a + 4 = 56

12a + 8 = 56

12a = 56 – 8

12a = 48

a = 48/12

a = 4

Therefore, Present age of Rahul = 5a = 5 × 4 = 20

And, present age of Haroon = 7a = 7 × 4 = 28

Question 10. The number of boys and girls in a class are in the ratio 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 ‘7a’ and girls be ‘5a’.

According to the question,

Given, 7a = 5a + 8

7a – 5a = 8

2a = 8

a = 8/2

a = 4

Therefore, Number of boys = 7 × 4 = 28

And, Number of girls = 5 × 4 = 20

Total number of students = 20 + 28 = 48

Question 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 age of Baichung’s father be ‘a’.

Then, age of Baichung’s grandfather = (a + 26)

and, Age of Baichung = (a – 29) According to the question,

Given, a + (a + 26) + (a – 29) = 135

3a + 26 – 29 = 135

3a – 3 = 135

3a = 135 + 3

3a = 138

a = 138/3

a = 46

Age of Baichung’s father = a = 46

Age of Baichung’s grandfather = (a + 26) = 46 + 26 = 72

Age of Baichung = (a – 29) = 46 – 29 = 17

Question 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 ‘a’.

Fifteen years later, Ravi age will be (a+15) years. According to the question,

Given, a + 15 = 4a

4a – a = 15

3a = 15

a = 15/3

a = 5

Therefore, Present age of Ravi = 5 years.

Question 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 ‘a’.

According to the question,

Given, a × (5/2) + 2/3 = -7/12

5(a/2) + 2/3 = -7/12

5(a/2) = -7/12 – 2/3

5(a/2) = (-7- 8)/12

5(a/2) = -15/12

5a/2 = -5/4

a = (-5/4) × (2/5)

a = – 10/20

a = -1/2

Therefore, the rational number will be -1/2.

Question 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 ‘2a’ , ‘3a’ and ‘5a’ respectively.

Value of ₹100 = 2a × 100 = 200a

Value of ₹50 = 3a × 50 = 150a

Value of ₹10 = 5a × 10 = 50a According to the question,

Given, 200a + 150a + 50a = 400000

400a = 400000

a = 400000/400

a = 1000

Numbers of ₹100 notes = 2a = 2000

Numbers of ₹50 notes = 3a = 3000

Numbers of ₹10 notes = 5a = 5000

Question 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 number of ₹5 coins be ‘a’.

Then,

Number ₹2 coins = 3a

and, number of ₹1 coins = (160 – 4a) Now,

Value of ₹5 coins= a × 5 = 5a

Value of ₹2 coins = 3a × 2 = 6a

Value of ₹1 coins = (160 – 4a) × 1 = (160 – 4a)

According to the question,

Given, 5a + 6a + (160 – 4a) = 300

11a + 160 – 4a = 300

7a = 140

a = 140/7

a = 20

Number of ₹5 coins =  a = 20

Number of ₹2 coins = 3a = 60

Number of ₹1 coins = (160 – 4a) = 160 – 80 = 80

Question 16. The organizers 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 numbers of winner be ‘a’

Then, the number of participant who didn’t win will be (63 – a)

Total money given to the winner = a × 100 = 100a

Total money given to participant who didn’t win = 25 × (63 – a)

According to the question,

Given, 100a + 25 × (63 – a) = 3000

100a + 1575 – 25a = 3000

75a = 3000 – 1575

75a = 1425

a = 1425/75

a = 19

So, the number of winners are 19.

Linear Equations in One Variable: Exercise 2.3

Question 1. Find the value of x : 3x = 2x + 18

Solution:

3x – 2x =18 (transposing 2x to LHS)

X = 18 (solution) 

Verification — Put the value of x in the equation to verify our solution

3(18) = 2(18) + 18

54 = 36 + 18

54 = 54

 LHS = RHS (so our value of x is correct)

Question 2. Find the value of t : 5t – 3 = 3t – 5

Solution:

 5t – 3 – 3t = -5 (transposing 3t to LHS)

 5t – 3t = -5 + 3 (transposing 3 to RHS)

2t = -2

t = -1 (solution)

Verification — Put the value of t in the equation to verify our solution

5(-1) – 3 = 3(-1) – 5

-5 – 3 = -3 – 5

-8 = -8

LHS = RHS (so our value of t is correct)

Question 3. Find the value of x: 5x + 9 = 5 + 3x

Solution:

5x + 9 – 3x = 5 (transposing 3x to LHS)

5x – 3x = 5 – 9 (transposing 9 to RHS)

2x = -4

x = -2 (solution)

Verification -Put the value of x in the equation to verify our solution

5(-2) + 9 = 5 + 3(-2)

-10 + 9 = 5 -6

-1 = -1

LHS = RHS(so our value of x is correct) 

Question 4. Find the value of z: 4z + 3 = 6 + 2z

Solution: 

 4z +3 – 2z =6 (transposing 2z to LHS)

  4z – 2z = 6 – 3 (transposing 3 to RHS)

2z = 3

z = 3/2 (solution)

Verification — Put the value of z in the equation to verify our solution

4(3/2) + 3 = 6 + 2(3/2)

6 + 3 = 6 + 3

9 = 9

LHS = RHS (so our value of z is correct) 

Question 5. Find the value of x: 2x – 1 = 14 – x 

Solution:

2x – 1 + x = 14 (transposing x to LHS)

2x + x = 14 + 1 (transposing 1 to RHS)

3x = 15

 x = 5 (solution)

Verification — Put the value of x in the equation to verify our solution

2(5) – 1 = 14 – 5

10 – 1 = 14 -5

9 = 9

LHS = RHS  (so our value of x is correct) 

Question 6. Find the value of x: 8x + 4 = 3 (x – 1) + 7 

Solution:

 8x + 4 = 3x – 3 + 7 (solving RHS)

 8x + 4 – 3x = – 3 + 7 (transposing 3x to LHS)

  8x – 3x = – 3 + 7– 4 (transposing 4 to RHS)

5x = 0

 x = 0 (solution)

Verification — Put the value of x in the equation to verify our solution

8(0) +4 = 3(0-1) + 7

0 + 4 = -3 +7

4 = 4

LHS = RHS (so our value of x is correct) 

Question 7. Find the value of x: x = 4/5 (x + 10)

Solution:

 5x = 4 (x + 10)                  

5x = 4x + 40

5x – 4x = 40 (transposing 4x to LHS)

 x = 40 (solution)

Verification — Put the value of x in the equation to verify our solution

40 = 4/5 ( 40 +10)

40 = 4(50)/5

40 = 40

 LHS = RHS (so our value of x is correct) 

Question 8. Find the value of x: 2x/3 + 1 = 7x/15 + 3

Solution:

 (2x + 3) / 3 = (7x + 45) / 15 (solving LHS and RHS)

15 (2x + 3) = 3 (7x + 45) (transposing 15 and 3 )

30x + 45 = 21x + 135 (solving brackets)

 30x + 45 – 21x = 135 (transposing 21x to LHS)

30x – 21x = 135 – 45 (transposing 45 to RHS)

9x = 90

 x = 10(solution)

Verification — Put the value of x in the equation to verify our solution

2(10)/3 + 1 = 7(10)/15 + 3

20/3 + 1 = 14/3 + 3

23/3 = 23/3

LHS = RHS (so our value of x is correct) 

Question 9. Find the value of y: 2y + 5/3 = 26/3 – y

Solution:

(6y + 5) / 3 = (26 – 3y) / 3(canceling 3 at denominator from both sides)

 6y + 5 = 26 – 3y(solving brackets)

 6y + 5 + 3y = 26 (transposing 3y to LHS)

 9y = 26 – 5 (transposing 5 to RHS)

9y = 21   

 y = 7/3 (solution)

Verification — Put the value of y in the equation to verify our solution

2(7/3) + 5/3 = 26/3 – 7/3

(14 + 5)/3 = (26 – 7)/3

19/3 = 19/3

LHS = RHS (so our value of y is correct)

Question 10. Find the value of m: 3m = 5m – 8/5

Solution:

 3m = 25m -8/5

15m = 25m – 8     

15m – 25m = -8 (transposing 25m to LHS)

-10m = -8

m = 8/10 or m = 4/5 (solution)

Verification — Put the value of m in the equation to verify our solution                                                        

3(4/5) = 5(4/5) – 8/5

12/5 = 20/5 – 8/5

12/5 = 12/5

 LHS = RHS (so our value of m is correct)

Linear Equations in One Variable: Exercise 2.4

Question 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

Solving for x, 

⇒ 8x – 40/2 = 3x

⇒ 8x – 3x = 40/2 (Bring like terms together)

⇒ 5x = 20 (Divide both sides by 5)

⇒ x = 4

Thus, the number is 4.

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

Solution:

According to the question, 

Let one of the positive number be x then other number will be 5x. 

Therefore,

5x + 21 = 2(x + 21)

Solving for x,

⇒ 5x + 21 = 2x + 42

⇒ 5x – 2x = 42 – 21 (Bring like terms together)

⇒ 3x = 21 (Dividing both sides by 3)

⇒ x = 7

1st number = x = 7

2nd number = 5x = 5×7 = 35 

Question 3. The 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 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

Solving for x,

⇒ 10x + 9 – x + 27 = 90 – 10x + x 

⇒ 9x + 36 = 90 – 9x

⇒ 9x + 9x = 90 – 36 (Bring like terms together)

⇒ 18x = 54 ( Divide both sides by 18)

⇒ x = 3

Original number = 10x + (9 – x) = (10 × 3) + (9 – 3) = 30 + 6 = 36

Thus, the number is 36.

Question 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 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

Solving for x,

⇒ 31x + 13x = 88

⇒ 44x = 88 (Divide both sides by 44)

⇒ x = 2

Original number = 10x + 3x = 13x = 13×2 = 26

Question 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 age of her mother will be 6x.

Shobo’s age after 5 years = x + 5

According to the question,

(x + 5) = (1/3) × 6x

Solving for x, 

⇒ x + 5 = 2x

⇒ 2x – x = 5 (Bring like terms together)

⇒ x = 5

Present age of Shobo = x = 5 years

Present age of Shobo’s mother = 6x = 30 years.

Question 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 meter 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 breadth be 4x.

Rate of fencing per meter = ₹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.

Question 7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹50 per meter and trouser material that costs him ₹90 per meter. For every 3 meters of the shirt material he buys 2 meters 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.

Question 8. Half of a herd of deer are 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

Question 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.

Question 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 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

Linear Equations in One Variable: Exercise 2.5

Solve the following linear equations.

Question 1.  x/2 – 1/5 = x/3 + 1/4

Solution:

(5x – 2)/10 = (4x + 3)/12   …(Taking LCM on both the sides)

12(5x – 2) = 10 (4x + 3)  …(Cross multiplying)

60x – 24 = 40x + 30   …(Solving the brackets)

60x – 40x = 30 + 24  …(Transposing terms of x to LHS and others to RHS)

20x = 54

x = 54/20 or 27/10 … (Solution)

Verification:

Putting value of “x” in the equation to check if our answer is correct

27/20 – 1/5 = 27/30 + 1/4

(27 – 4)/20 = (108 + 30)/120

23/20 = 138/120

23/20 = 23/20

LHS = RHS (Hence Proved that solution is correct) 

Question 2.  n/2 – 3n/4 + 5n/6 = 21

Solution:

(6n – 9n + 10n)/12 = 21   …(Taking LCM and solving LHS)

7n/12 = 21 (Solving LHS)

7n = 21 × 12

n = 36  …(Solution)

Verification:

Putting value of “n” in the equation to check if our answer is correct

36/2 – 108/4 + 180/6 = 21

18 – 27 + 30 = 21

21 = 21

LHS = RHS (Hence Proved that solution is correct)

Question 3.  x + 7 – 8x/3 = 17/6 – 5x/2

Solution:

x – 8x/3 + 5x/2 = 17/6 – 7  …(Transposing terms of x to LHS and others to RHS)

(6x – 16x + 15x)/6 = (17 – 42)/6  …(Taking LCM and solving)

5x/6 = -25/6

x = -5  …(Solution)

Verification –

Putting value of “x” in the equation to check if our answer is correct

-5 + 7 – (-40)/3 = 17/6 – (-25)/2

2 + 40/3 = 17/6 + 25/2

46/3 = (17 + 75)/6

46/3 = 92/6

46/3 = 46/3

LHS = RHS  (Hence Proved that solution is correct)

Question 4.  (x – 5)/3 = (x – 3)/5

Solution:

5(x – 5) = 3(x – 3)  …(Cross multiply)

5x – 25 = 3x – 9  

2x = 16

x = 8  …(Solution)

Verification – 

Putting value of “x” in the equation to check if our answer is correct

(8 – 5)/3 = (8 – 3)/5

3/3 = 5/5

1 = 1

LHS = RHS (Hence Proved that solution is correct)

Question 5.  (3t – 2)/4 – (2t + 3)/3 = 2/3 – t

Solution:

3t/4 – 1/2 – 2t/3 -1 = 2/3 – t  …(Solving brackets)

3t/4 – 2t/3 + t = 2/3 + 1 + 1/2  …(Transposing terms of x to LHS and others to RHS)

(9t – 8t + 12t)/12 = (4 + 6 + 3)/6  …(Taking LCM both sides)

13t/12 = 13/6

t = 2   …(Solution)

Verification –

Putting value of “t” in the equation to check if our answer is correct

(3 × 2 – 2)/4 – (2 × 2 + 3)/3 = 2/3 – 2

4/4 – 7/3 = 2/3 – 2

(12 – 28)/12 = (2 – 6)/3

-16/12 = -4/3

-4/3 = -4/3

LHS = RHS (Hence Proved that solution is correct)

Question 6.  m – (m – 1)/2 = 1 – (m – 2)/3

Solution:

(2m – m + 1)/2 = (3 – m + 2)/3  …(Taking LCM both sides)

(m + 1)/2 = (5 – m)/3

3(m + 1) = 2(5 – m)  …(Cross multiplying)

3m + 3 = 10 – 2m

5m = 7

m = 7/5  …(Solution)

Verification –

Putting value of “m” in the equation to check if our answer is correct

7/5 – (7/5 – 1)/2 = 1 – (7/5 – 2)/3

7/5 – 1/5 = 1 – (-3)/15

6/5 = 1 + 1/5

6/5 = 6/5

LHS = RHS (Hence Proved that solution is correct)

Question 7. 3(t – 3) = 5(2t + 1)

Solution:

3t – 9 = 10t + 5  …(Opening brackets)

3t – 10t = 9 + 5

-7t = 14

t = -2  …(Solution)

Verification –

Putting value of “t” in the equation to check if our answer is correct 

3(-2 – 3) = 5(2(-2) + 1)

3(-5) = 5(-4 +1)

-15 = -15

LHS = RHS (Hence Proved that solution is correct)

Question 8. 15(y – 4) – 2(y – 9) + 5(y + 6) = 0

Solution:

15y – 60 – 2y + 18 + 5y + 30 = 0

18y – 12 = 0

y = 12/18 or 2/3  …(Solution)

Verification –

Putting value of “y” in the equation to check if our answer is correct 

15(2/3 – 4) – 2(2/3 – 9) + 5(2/3 + 6) = 0

10 – 60 – 4/3 +18 + 10/3 + 30 = 0

-50 -4/3 + 48 + 10/3 = 0

-2 + 6/3 = 0

-2 + 2 = 0

0 = 0

LHS = RHS (Hence Proved that solution is correct)

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

Solution:

15z – 21 – 18z + 22 = 32z – 52 – 17  …(Solving the brackets)

-3z + 1 = 32z – 69

-35z = -70

z = 2  …(Solution)

Verification –

Putting value of “z” in the equation to check if our answer is correct 

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

3(3) – 2(7) = 4(3) – 17

9 – 14 = 12 – 17

-5 = -5

LHS = RHS (Hence Proved that solution is correct)

Question 10. 0.25(4f – 3) = 0.05(10f – 9)

Solution:

f – 0.25(3) = 0.5f – 0.05(9)

f – 0.75 = 0.5f – 0.45

0.5f = 0.75 – 0.45

f = 3/5 or 0.6 (Solution)

Verification – 

Putting value of “f” in the equation to check if our answer is correct

0.25(4(0.6) – 3) = 0.05(10(0.6) – 9)

0.25(2.4 – 3) = 0.05(6 – 9)

0.25 × (-0.6) = 0.05 × (-3)

-0.15 = -0.15

LHS = RHS (Hence Proved that solution is correct)

Linear Equations in One Variable: Exercise 2.6

We use cross multiplication in this exercise a lot of times, so it is explained here before.

Let, a/b = c/d Now if we multiply both sides by the denominators of left side and right side, we get, (a/b) X (b X d) = (c/d) X (b X d) => a X d = b X c This is called cross multiplication.

Question.1 Solve the following equations.

We can solve the problems 1 to 5 by trying to bring all the unknown variables to the left side.

1. (8x-3) / 3x = 2

Solution:

By multiplying on both sides by 3x we get, => (8x-3) X (3x) / 3x = 2 X (3x) => 8x-3 = 6x => 8x-6x-3 = 0 => 2x-3 = 0 => 2x = 3 => x = 3/2

2. 9x / (7-6x) = 15

Solution:

By multiplying both sides by (7-6x) we get, => (9x) X (7-6x) / (7-6x) = 15 X (7-6x) => 9x = (15 X 7) – (15 X 6)x => 9x = 105 – 90x => 9x + 90x = 105 => 99x = 105 => x = 105/99 => x = 35/33

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

Solution:

By cross multiplication, => z X 9 = (z+15) X 4 => 9z = 4z + 4 X 15 => 9z – 4z = 60 => 5z = 60 => z = 60/5 => z = 12

4. (3y+4) / (2-6y) = 2 / 5

Solution:

By cross multiplication, => (3y+4) X 5 = (-2) X (2-6y) => (5 X 3)y + (4 X 5) = (-2 X 2) + (-2 X -6)y => 15y + 20 = -4 + 12y => 15y -12y = (-4) + (-20) => 3y = -24 => y = -24/3 => y = -8

5. (7y+4) / (y+2) = – 4 / 3

Solution:

By cross multiplication, => (7y+4) X 3 = -4 X (y+2) => (7 X 3)y + (4 X 3) = -4y + (-4 X 2) => 21y + 12 = (-4y) + (-8) => 21y + 4y = (-8) + (-12) => 25y = -20 => y = -20/25 => y=-4/5

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

Solution:

Let the present age of Hari be x , Let the present age of Harry be y . Presently their ages are in ratio 5:7 , So we get => x : y = 5 : 7 We get, => x/y = 5/7 By cross multiplication, => 7x = 5y => x = (5y) / 7 ……….. (1) After 4 years, Hari’s age will be x+4, Harry’s age will be y+4. The ratio between their ages after four years is 3:4. So we get, => (x+4) : (y+4) = 3 : 4 => (x+4)/(y+4) = 3/4 By cross multiplication, => (x+4) X 4 = 3 X (y+4) => 4x +16 = 3y + 12 => 4x – 3y = -4 ……….. (2) Now , we got two euations. x = (5y) / 7 ……….. (1) 4x – 3y = -4 ……….. (2) If we substitute this x value from (1) in equation (2) we get => 4 X (5y/7) – 3y = -4 => 20y/7 – 3y = -4 => 20y/7 – (7X3)y/7 = -4 => (20y -21y) / 7 = -4 => -y/7 = -4 => y = (-4) X (-7) => y = 28 By substituting y=28 value in (1) we get => x = (5 X 28) / 7 => x = (5 X 4) => x =20 So here Hari’s present age is 20 years and Harry’s present age is 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, and the denominator be y. From the first part of the question we get, => denominator = numerator + 8 => y = x + 8 ……….. (1) Now, from the second part of the question , => (x+17) / (y-1) = 3/2 By cross multiplication, => (x+17) X 2 = 3 X (y-1) => 2x + 34 = 3y – 3 => 2x – 3y = -34 – 3 => 2x – 3y = -37 ……….. (2) We got two equations. Substituting (1) in (2), we get => 2x − 3 X (x+8) = −37 => 2x − 3x − 24 = −37 => 37 − 24 = x => x = 13 By substituting x=13 in (1) we get, => y = 13 +8 => y = 21 We got x=13 and y=21 Hence the original rational fraction will be 13/21

Important Points to Remember:

  • These NCERT solutions are developed by the GfG team, with a focus on students’ benefit.
  • These solutions are entirely accurate and can be used by students to prepare for their board exams. 
  • Each solution is presented in a step-by-step format with comprehensive explanations of the intermediate steps.

FAQs on NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable

1. Why is it important to learn Linear Equations in One Variable?

Learning about linear equations in one variable is important for class 8 students because it helps develop problem-solving skills that are applicable in real-life situations. Linear equations are widely used in various fields, such as economics, physics, and engineering. Understanding how to solve these equations enables students to analyze and solve problems involving unknown quantities. Additionally, learning linear equations sets a strong foundation for advanced mathematical concepts and prepares students for higher-level mathematics courses.

2. What topics are covered in NCERT Solutions for Chapter 2 – Linear Equations in One Variable?

NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable covers topics such as introduction to linear equations and their solutions. Students learn various methods for solving linear equations, including the use of inverse operations and cross-multiplication, practical applications of linear equations in real-life situations, such as finding the cost of an item or determining the number of years.

3. How can NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable help me?

NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable can help you solve the NCERT exercise without any limitations. If you are stuck on a problem, you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

4. How many exercises are there in Class 8 Maths Chapter 2 – Linear Equations in One Variable?

There are 6 exercises in the Class 8 Maths Chapter 2 – Linear Equations in One Variable which covers all the important topics and sub-topics.

5. Where can I find NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable?

You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.



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