Class 8th Chapter 2- Linear Equations in One Variable NCERT Solution

Exercise 2.1 | Class 8th Mathematics

1.Solve the equation: x – 2 = 7.
Solution:
Given: x – 2 = 7
⇒ x – 2 + 2 = 7 + 2 (adding 2 on both sides)
⇒ x = 9 (Required solution)

2.Solve the equation: y + 3 = 10.
Given: y + 3 = 10
⇒ y + 3 – 3 = 10 – 3 (subtracting 3 from each side)
⇒ y = 7 (Required solution)

3.Solve the equation: 6 = z + 2
Solution:
We have 6 = z + 2
⇒ 6 – 2 = z + 2 – 2 (subtracting 2 from each side)
⇒ 4 = z
Thus, z = 4 is the required solution.

4.Solve the equation 6x = 12.
Solution:
We have 6x = 12
⇒ 6x ÷ 6 = 12 ÷ 6 (dividing each side by 6)
⇒ x = 2
Thus, x = 2 is the required solution.

5.Solve the equation t5 = 10.
Solution:
Given t5 = 10
⇒ t5 × 5 = 10 × 5 (multiplying both sides by 5)
⇒ t = 50
Thus, t = 50 is the required solution.

6.Solve the equation 2×3 = 18.
Solution:
We have 2×3 = 18
⇒ 2×3 × 3 = 18 × 3 (multiplying both sides by 3)
⇒ 2x = 54
⇒ 2x ÷ 2 = 54 ÷ 2 (dividing both sides by 2)
⇒ x = 27
Thus, x = 27 is the required solution.

7.Solve the equation 1.6 = y1.5
Solution:
Given: 1.6 = y1.5
⇒ 1.6 × 1.5 = y1.5 × 1.5 (multiplying both sides by 1.5)
⇒ 2.40 = y
Thus, y = 2.40 is the required solution.

8.Solve the equation 7x – 9 = 16.
Solution:
We have 7x – 9 = 16
⇒ 7x – 9 + 9 = 16 + 9 (adding 9 to both sides)
⇒ 7x = 25
⇒ 7x ÷ 7 = 25 ÷ 7 (dividing both sides by 7)
⇒ x = 257
Thus, x = 257 is the required solution.

9.Solve the equation 14y – 8 = 13.
Solution:
We have 14y – 8 = 13
⇒ 14y – 8 + 8 = 13 + 8 (adding 8 to both sides)
⇒ 14y = 21
⇒ 14y ÷ 14 = 21 ÷ 14 (dividing both sides by 14)
⇒ y = 2114
⇒ y = 32
Thus, y = 32 is the required solution.

10.Solve the equation 17 + 6p = 9.
Solution:
We have, 17 + 6p = 9
⇒ 17 – 17 + 6p = 9 – 17 (subtracting 17 from both sides)
⇒ 6p = -8
⇒ 6p ÷ 6 = -8 ÷ 6 (dividing both sides by 6)
⇒ p = −86
⇒ p = −43
Thus, p = −43 is the required solution.

 Exercise 2.2 | Class 8th Mathematics

1.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 the breadth of the pool?
Solution:
Let the breadth of the pool be x m.
Condition I: Length = (2x + 2) m.
Condition II: Perimeter = 154 m.
We know that Perimeter of rectangle = 2 × [length + breadth]
2 × [2x + 2 + x] = 154
⇒ 2 × [3x + 2] = 154
⇒ 6x + 4 = 154 (solving the bracket)
⇒ 6x = 154 – 4 [Transposing 4 from (+) to (-)]

⇒ 6x = 150
⇒ x = 150 ÷ 6 [Transposing 6 from (×) to (÷)]
⇒ x = 25
Thus, the required breadth = 25 m

and the length = 2 × 25 + 2 = 50 + 2 = 52 m.

2.Sum of two numbers be 95. If one exceeds the other by 15, find the numbers.
Solution:
Let one number be x
Other number = x + 15
As per the condition of the question, we get
x + (x + 15) = 95
⇒ x + x + 15 = 95
⇒ 2x + 15 = 95
⇒ 2x = 95 – 15 [transposing 15 from (+) to (-)]
⇒ 2x = 80
⇒ x = 802 [transposing 2 from (×) to (÷)]
⇒ x = 40
Other number = 95 – 40 = 55
Thus, the required numbers are 40 and 55.

3.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.
As per the conditions, we get
5x – 3x = 18
⇒ 2x = 18
⇒ x = 18 ÷ 2 [Transposing 2 from (×) to (÷)]
⇒ x = 9.
Thus, the required numbers are 5 × 9 = 45 and 3 × 9 = 27

4.Three consecutive integers add up to 51. What are these integers?
Solution:
Let the three consecutive integers be x, x + 1 and x + 2.
As per the condition, we get
x + (x + 1) + (x + 2) = 51
⇒ x + x + 1 + x + 2 = 51
⇒ 3x + 3 = 51
⇒ 3x = 51 – 3 [transposing 3 to RHS]
⇒ 3x = 48
⇒ x = 48 ÷ 3 [transposing 3 to RHS]
⇒ x = 16
Thus, the required integers are 16, 16 + 1 = 17 and 16 + 2 = 18, i.e., 16, 17 and 18.

5.The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution:
Let the three consecutive multiples of 8 be 8x, 8x + 8 and 8x + 16.
As per the conditions, we get
8x + (8x + 8) + (8x + 16) = 888
⇒ 8x + 8x + 8 + 8x + 16 = 888
⇒ 24x + 24 = 888
⇒ 24x = 888 – 24 (transposing 24 to RHS)
⇒ 24x = 864
⇒ x = 864 ÷ 24 (transposing 24 to RHS)

⇒ x = 36
Thus, the required multiples are
36 × 8 = 288, 36 × 8 + 8 = 296 and 36 × 8 + 16 = 304,
i.e., 288, 296 and 304.

6.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.
As per the condition, we have
2x + 3(x + 1) + 4(x + 2) = 74
⇒ 2x + 3x + 3 + 4x + 8 = 74
⇒ 9x + 11 = 74
⇒ 9x = 74 – 11 (transposing 11 to RHS)
⇒ 9x = 63
⇒ x = 63 ÷ 9
⇒ x = 7 (transposing 7 to RHS)
Thus, the required numbers are 7, 7 + 1 = 8 and 7 + 2 = 9, i.e., 7, 8 and 9.

7.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 present ages of Rahul and Haroon he 5x years and 7x years respectively.
4 years later, the age of Rahul will be (5x + 4) years.
4 years later, the age of Haroon will be (7x + 4) years.
As per the conditions, we get
(5x + 4) + (7x + 4) = 56
⇒ 5x + 4 + 7x + 4 = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 – 8 (transposing 8 to RHS)
⇒ 12x = 48
⇒ x = 48 ÷ 12 = 4 (transposing 12 to RHS)
Hence, the required age of Rahul = 5 × 4 = 20 years.
and the required age of Haroon = 7 × 4 = 28 years.

8.The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the numbers of girls. What is the total class strength?
Solution:
Let the number of boys be 7x
and the number of girls be 5x
As per the conditions, we get
7x – 5x = 8
⇒ 2x = 8
⇒ x = 8 ÷ 2 = 4 (transposing 2 to RHS)
the required number of boys = 7 × 4 = 28
and the number of girls = 5 × 4 = 20
Hence, total class strength = 28 + 20 = 48

9.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 be x years.
The age of his father = x + 29 years,
and the age of his grandfather = x + 29 + 26 = (x + 55) years.
As per the conditions, we get
x + x + 29 + x + 55 = 135
⇒ 3x + 84 = 135
⇒ 3x = 135 – 84 (transposing 84 to RHS)
⇒ 3x = 51
⇒ x = 51 ÷ 3 (transposing 3 to RHS)
⇒ x = 17
Hence Baichung’s age = 17 years
Baichung’s father’s age = 17 + 29 = 46 years,
and grand father’s age = 46 + 26 = 72 years.

10.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 years.
After 15 years, his age will be = (x + 15) years
As per the conditions, we get
⇒ x + 15 = 4x
⇒ 15 = 4x – x (transposing x to RHS)
⇒ 15 = 3x
⇒ 15 ÷ 3 = x (transposing 3 to LHS)
⇒ x = 5
Hence, the present age of Ravi = 5 years.

Thus, the number of winners = 19

 Exercise 2.3 | Class 8th Mathematics

1.3x = 2x + 18
Solution:
We have 3x = 2x + 18
⇒ 3x – 2x = 18 (Transposing 2x to LHS)
⇒ x = 18
Hence, x = 18 is the required solution.
Check: 3x = 2x + 18
Putting x = 18, we have
LHS = 3 × 18 = 54
RHS = 2 × 18 + 18 = 36 + 18 = 54
LHS = RHS
Hence verified.

2.5t – 3 = 3t – 5
Solution:
We have 5t – 3 = 3t – 5
⇒ 5t – 3t – 3 = -5 (Transposing 3t to LHS)
⇒ 2t = -5 + 3 (Transposing -3 to RHS)
⇒ 2t = -2
⇒ t = -2 ÷ 2
⇒ t = -1
Hence t = -1 is the required solution.
Check: 5t – 3 = 3t – 5
Putting t = -1, we have
LHS = 5t – 3 = 5 × (-1)-3 = -5 – 3 = -8
RHS = 3t – 5 = 3 × (-1) – 5 = -3 – 5 = -8
LHS = RHS
Hence verified.

3. 5x + 9 = 5 + 3x
Solution:
We have 5x + 9 = 5 + 3x
⇒ 5x – 3x + 9 = 5 (Transposing 3x to LHS) => 2x + 9 = 5
⇒ 2x = 5 – 9 (Transposing 9 to RHS)
⇒ 2x = -4
⇒ x = -4 ÷ 2 = -2
Hence x = -2 is the required solution.
Check: 5x + 9 = 5 + 3x
Putting x = -2, we have
LHS = 5 × (-2) + 9 = -10 + 9 = -1
RHS = 5 + 3 × (-2) = 5 – 6 = -1
LHS = RHS
Hence verified.

4. 4z + 3 = 6 + 2z
Solution:
We have 4z + 3 = 6 + 2z
⇒ 4z – 2z + 3 = 6 (Transposing 2z to LHS)
⇒ 2z + 3 = 6
⇒ 2z = 6 – 3 (Transposing 3 to RHS)
⇒ 2z = 3
⇒ z = 32
Hence z = 32 is the required solution.
Check: 4z + 3 = 6 + 2z
Putting z = 32, we have
LHS = 4z + 3 = 4 × 32 + 3 = 6 + 3 = 9
RHS = 6 + 2z = 6 + 2 × 32 = 6 + 3 = 9
LHS = RHS
Hence verified.

5. 2x – 1 = 14 – x
Solution:
We have 2x – 1 = 14 – x
⇒ 2x + x = 14 + 1 (Transposing x to LHS and 1 to RHS)
⇒ 3x = 15
⇒ x = 15 ÷ 3 = 5
Hence x = 5 is the required solution.
Check: 2x – 1 = 14 – x
Putting x = 5
LHS we have 2x – 1 = 2 × 5 – 1 = 10 – 1 = 9
RHS = 14 – x = 14 – 5 = 9
LHS = RHS
Hence verified.

6. 8x + 4 = 3(x – 1) + 7
Solution:
We have 8x + 4 = 3(x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7 (Solving the bracket)
⇒ 8x + 4 = 3x + 4
⇒ 8x – 3x = 4 – 4 [Transposing 3x to LHS and 4 to RHS]
⇒ 5x = 0
⇒ x = 0 ÷ 5 [Transposing 5 to RHS]
or x = 0
Thus x = 0 is the required solution.
Check: 8x + 4 = 3(x – 1) + 7
Putting x = 0, we have
8 × 0 + 4 = 3(0 – 1) + 7
⇒ 0 + 4 = -3 + 7
⇒ 4 = 4
LHS = RHS
Hence verified.

7. x = 45 (x + 10)
Solution:
We have x = 45 (x + 10)
⇒ 5 × x = 4(x + 10) (Transposing 5 to LHS)
⇒ 5x = 4x + 40 (Solving the bracket)
⇒ 5x – 4x = 40 (Transposing 4x to LHS)
⇒ x = 40
Thus x = 40 is the required solution.
Check: x = 45 (x + 10)
Putting x = 40, we have
40 = 45 (40 + 10)
⇒ 40 = 45 × 50
⇒ 40 = 4 × 10
⇒ 40 = 40
LHS = RHS
Hence verified.

8. 2×3 + 1 = 7×15 + 3
Solution:
We have 2×3 + 1 = 7×15 + 3
15(2×3 + 1) = 15(7×15 + 3)
LCM of 3 and 15 is 15
2×3 × 15 + 1 × 15 = 7×15 × 15 + 3 × 15 [Multiplying both sides by 15]
⇒ 2x × 5 + 15 = 7x + 45
⇒ 10x + 15 = 7x + 45
⇒ 10x – 7x = 45 – 15 (Transposing 7x to LHS and 15 to RHS)
⇒ 3x = 30
⇒ x = 30 ÷ 3 = 10 (Transposing 3 to RHS)
Thus the required solution is x = 10

9. 2y + 53 = 263 – y
Solution:


10. 3m = 5m – 85
Solution:
We have

 Extra Questions Very Short Answer Type | Class 8th Mathematics

Question 1.
Identify the algebraic linear equations from the given expressions.
(a) x² + x = 2
(b) 3x + 5 = 11
(c) 5 + 7 = 12
(d) x + y² = 3
Solution:
(a) x² + x = 2 is not a linear equation.
(b) 3x + 5 = 11 is a linear equation.
(c) 5 + 7 = 12 is not a linear equation as it does not contain variable.
(d) x + y² = 3 is not a linear equation.

Question 2.
Check whether the linear equation 3x + 5 = 11 is true for x = 2.
Solution:
Given that 3x + 5 = 11
For x = 2, we get
LHS = 3 × 2 + 5 = 6 + 5 = 11
LHS = RHS = 11
Hence, the given equation is true for x = 2

Question 3.
Form a linear equation from the given statement: ‘When 5 is added to twice a number, it gives 11.’
Solution:
As per the given statement we have
2x + 5 = 11 which is the required linear equation.

Question 4.
If x = a, then which of the following is not always true for an integer k. (NCERT Exemplar)
(a) kx = ak
(b) xk = ak
(c) x – k = a – k
(d) x + k = a + k
Solution:
Correct answer is (b).

Question 5.
Solve the following linear equations:
(a) 4x + 5 = 9
(b) x + 32 = 2x
Solution:
(a) We have 4x + 5 = 9
⇒ 4x = 9 – 5 (Transposing 5 to RHS)
⇒ 4x = 4
⇒ x = 1 (Transposing 4 to RHS)
(b) We have x + 32 = 2x
⇒ 32 = 2x – x
⇒ x = 32

Question 6.
Solve the given equation 31x × 514 = 1712
Solution:
We have 31x × 514 = 1712

Question 7.
Verify that x = 2 is the solution of the equation 4.4x – 3.8 = 5.
Solution:
We have 4.4x – 3.8 = 5
Putting x = 2, we have
4.4 × 2 – 3.8 = 5
⇒ 8.8 – 3.8 = 5
⇒ 5 = 5
L.H.S. = R.H.S.
Hence verified.

Question 8.

Solution:

⇒ 3x × 3 – (2x + 5) × 4 = 5 × 6
⇒ 9x – 8x – 20 = 30 (Solving the bracket)
⇒ x – 20 = 30
⇒ x = 30 + 20 (Transposing 20 to RHS)
⇒ x = 50
Hence x = 50 is the required solution.

Question 9.
The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.
Solution:
Let the angles of a given triangle be 2x°, 3x° and 4x°.
2x + 3x + 4x = 180 (∵ Sum of the angles of a triangle is 180°)
⇒ 9x = 180
⇒ x = 20 (Transposing 9 to RHS)
Angles of the given triangles are
2 × 20 = 40°
3 × 20 = 60°
4 × 20 = 80°

Question 10.
The sum of two numbers is 11 and their difference is 5. Find the numbers.
Solution:
Let one of the two numbers be x.
Other number = 11 – x.
As per the conditions, we have
x – (11 – x) = 5
⇒ x – 11 + x = 5 (Solving the bracket)
⇒ 2x – 11 = 5
⇒ 2x = 5 + 11 (Transposing 11 to RHS)
⇒ 2x = 16
⇒ x = 8
Hence the required numbers are 8 and 11 – 8 = 3.