Class 8 Maths - Linear Equations in One Variable NCERT Solutions

Chapter 2: Linear Equations in One Variable (NCERT Solutions)

Exercise 2.1

Solve the following equations.

Q1. x - 2 = 7

Transpose -2 to RHS:
x = 7 + 2
x = 9.

Q2. y + 3 = 10

Transpose 3 to RHS:
y = 10 - 3
y = 7.

Q3. 6 = z + 2

Transpose 2 to LHS:
6 - 2 = z
4 = z
z = 4.

Q4. 3/7 + x = 17/7

Transpose 3/7 to RHS:
x = 17/7 - 3/7
x = (17 - 3) / 7
x = 14 / 7
x = 2.

Q5. 6x = 12

Divide both sides by 6:
x = 12 / 6
x = 2.

Q6. t/5 = 10

Multiply both sides by 5:
t = 10 × 5
t = 50.

Q7. 2x/3 = 18

Multiply both sides by 3:
2x = 18 × 3
2x = 54
Divide both sides by 2:
x = 54 / 2
x = 27.

Q8. 1.6 = y/1.5

Multiply both sides by 1.5:
1.6 × 1.5 = y
2.40 = y
y = 2.4.

Q9. 7x - 9 = 16

Transpose -9 to RHS:
7x = 16 + 9
7x = 25
x = 25/7.

Q10. 14y - 8 = 13

Transpose -8 to RHS:
14y = 13 + 8
14y = 21
y = 21 / 14
y = 3/2.

Q11. 17 + 6p = 9

Transpose 17 to RHS:
6p = 9 - 17
6p = -8
p = -8 / 6
p = -4/3.

Q12. x/3 + 1 = 7/15

Transpose 1 to RHS:
x/3 = 7/15 - 1
x/3 = (7 - 15) / 15
x/3 = -8 / 15
Multiply both sides by 3:
x = (-8 / 15) × 3
x = -8/5.

Exercise 2.2

Q1. If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. What is the number?

Let the number be x.
(x - 1/2) × 1/2 = 1/8
Multiply both sides by 2:
x - 1/2 = 1/8 × 2
x - 1/2 = 1/4
Transpose -1/2 to RHS:
x = 1/4 + 1/2 = 1/4 + 2/4
x = 3/4.
The number is 3/4.

Q2. 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?

Let the breadth be b meters.
Then, length = (2b + 2) meters.
Perimeter = 2(length + breadth) = 154
2((2b + 2) + b) = 154
2(3b + 2) = 154
Divide by 2:
3b + 2 = 77
3b = 75
b = 25
Breadth = 25 m.
Length = 2(25) + 2 = 50 + 2 = 52 m.

Q3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 4 & 2/15 cm. What is the length of either of the remaining equal sides?

Let the length of equal sides be x cm.
Perimeter = 4 & 2/15 = 62/15 cm.
Base = 4/3 cm.
x + x + 4/3 = 62/15
2x + 4/3 = 62/15
Transpose 4/3 to RHS:
2x = 62/15 - 4/3 = 62/15 - 20/15
2x = 42/15 = 14/5
Divide by 2:
x = 14/10 = 7/5 = 1 & 2/5 cm.
The length of equal sides is 1 & 2/5 cm.

Q4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Let the first number be x.
The second number will be x + 15.
Sum = x + (x + 15) = 95
2x + 15 = 95
2x = 95 - 15 = 80
x = 80 / 2 = 40.
First number = 40. Second number = 40 + 15 = 55.

Q5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Let the numbers be 5x and 3x.
Their difference is 18: 5x - 3x = 18
2x = 18
x = 9
First number = 5(9) = 45.
Second number = 3(9) = 27.

Q6. Three consecutive integers add up to 51. What are these integers?

Let integers be x, (x+1), (x+2).
x + x + 1 + x + 2 = 51
3x + 3 = 51
3x = 51 - 3 = 48
x = 48 / 3 = 16.
The integers are 16, 17, 18.

Exercise 2.3

Solve the following equations and check your results.

Q1. 3x = 2x + 18

3x - 2x = 18
x = 18.

Check: LHS = 3(18) = 54. RHS = 2(18) + 18 = 36 + 18 = 54. LHS = RHS.

Q2. 5t - 3 = 3t - 5

5t - 3t = -5 + 3
2t = -2
t = -1.

Check: LHS = 5(-1) - 3 = -8. RHS = 3(-1) - 5 = -8. LHS = RHS.

Q3. 5x + 9 = 5 + 3x

5x - 3x = 5 - 9
2x = -4
x = -2.

Check: LHS = 5(-2) + 9 = -10 + 9 = -1. RHS = 5 + 3(-2) = 5 - 6 = -1. LHS = RHS.

Q4. 4z + 3 = 6 + 2z

4z - 2z = 6 - 3
2z = 3
z = 3/2.

Check: LHS = 4(3/2) + 3 = 6 + 3 = 9. RHS = 6 + 2(3/2) = 6 + 3 = 9. LHS = RHS.

Q5. 2x - 1 = 14 - x

2x + x = 14 + 1
3x = 15
x = 5.

Check: LHS = 2(5) - 1 = 9. RHS = 14 - 5 = 9. LHS = RHS.

Q6. 8x + 4 = 3(x - 1) + 7

8x + 4 = 3x - 3 + 7
8x + 4 = 3x + 4
8x - 3x = 4 - 4
5x = 0
x = 0.

Check: LHS = 8(0) + 4 = 4. RHS = 3(0 - 1) + 7 = -3 + 7 = 4. LHS = RHS.

Q7. x = 4/5 (x + 10)

Multiply both sides by 5:
5x = 4(x + 10)
5x = 4x + 40
5x - 4x = 40
x = 40.

Check: LHS = 40. RHS = (4/5)(40 + 10) = (4/5)(50) = 4(10) = 40. LHS = RHS.

Exercise 2.4

Q1. 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?

Let the number be x.
8(x - 5/2) = 3x
8x - (8 × 5)/2 = 3x
8x - 20 = 3x
8x - 3x = 20
5x = 20
x = 20 / 5 = 4.
The number is 4.

Q2. 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?

Let the numbers be x and 5x.
After adding 21, the numbers become x + 21 and 5x + 21.
Obviously, 5x + 21 is larger.
5x + 21 = 2(x + 21)
5x + 21 = 2x + 42
5x - 2x = 42 - 21
3x = 21
x = 7.
The numbers are 7 and 5(7) = 35.

Q3. 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?

Let unit digit be x. Tens digit will be 9 - x.
Original number = 10(9 - x) + x = 90 - 10x + x = 90 - 9x.
Reversed number = 10x + (9 - x) = 9x + 9.
New number = Original number + 27
9x + 9 = 90 - 9x + 27
9x + 9x = 117 - 9
18x = 108
x = 108 / 18 = 6 (unit digit).
Tens digit = 9 - 6 = 3.
Original number = 36.

Exercise 2.5

Q1. Solve the following linear equations.
x/2 - 1/5 = x/3 + 1/4

Transpose x/3 to LHS:
x/2 - x/3 = 1/4 + 1/5
LCM of 2 and 3 is 6. LCM of 4 and 5 is 20.
(3x - 2x) / 6 = (5 + 4) / 20
x / 6 = 9 / 20
x = (9 × 6) / 20 = 54 / 20 = 27 / 10 = 2.7.

Q2. n/2 - 3n/4 + 5n/6 = 21

LCM of denominators 2, 4, 6 is 12.
(6n - 9n + 10n) / 12 = 21
7n / 12 = 21
7n = 21 × 12
n = (21 × 12) / 7 = 3 × 12 = 36.

Q3. x + 7 - 8x/3 = 17/6 - 5x/2

Transpose x terms to LHS:
x - 8x/3 + 5x/2 = 17/6 - 7
LCM of 3, 2 is 6.
(6x - 16x + 15x) / 6 = (17 - 42) / 6
5x / 6 = -25 / 6
Multiply both sides by 6:
5x = -25
x = -5.

Q4. (x - 5)/3 = (x - 3)/5

Cross multiply:
5(x - 5) = 3(x - 3)
5x - 25 = 3x - 9
5x - 3x = -9 + 25
2x = 16
x = 8.

Exercise 2.6

Q1. Solve the following equations.
(8x - 3) / 3x = 2

Multiply both sides by 3x:
8x - 3 = 2(3x)
8x - 3 = 6x
8x - 6x = 3
2x = 3
x = 3/2.

Q2. 9x / (7 - 6x) = 15

Multiply both sides by (7 - 6x):
9x = 15(7 - 6x)
9x = 105 - 90x
9x + 90x = 105
99x = 105
x = 105 / 99 = 35/33.

Q3. z / (z + 15) = 4 / 9

Cross multiply:
9z = 4(z + 15)
9z = 4z + 60
9z - 4z = 60
5z = 60
z = 12.

Q4. 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.

Let their present ages be 5x and 7x.
After 4 years: Hari = 5x + 4, Harry = 7x + 4.
Ratio: (5x + 4) / (7x + 4) = 3 / 4
Cross multiply:
4(5x + 4) = 3(7x + 4)
20x + 16 = 21x + 12
20x - 21x = 12 - 16
-x = -4
x = 4
Hari's age = 5(4) = 20 years. Harry's age = 7(4) = 28 years.

Class 8 Maths - Linear Equations in One Variable Practice Questions

Chapter 2: Linear Equations in One Variable (Practice Questions)

RD Sharma / Extra Practice Questions

Q1. Solve for x: 3x - 5 = 13

Transpose -5 to RHS:
3x = 13 + 5
3x = 18
x = 18 / 3
x = 6.

Q2. Solve for y: 2y + 9 = 4

Transpose 9 to RHS:
2y = 4 - 9
2y = -5
y = -5/2.

Q3. Solve for t: t/5 = 10

Multiply both sides by 5:
t = 10 × 5
t = 50.

Q4. Solve for x: 2x/3 = 18

Multiply both sides by 3:
2x = 18 × 3
2x = 54
x = 54 / 2
x = 27.

Q5. Solve for x: 7x - 9 = 16

Transpose -9 to RHS:
7x = 16 + 9
7x = 25
x = 25/7.

Q6. Solve for x: 3x = 2x + 18

Transpose 2x to LHS:
3x - 2x = 18
x = 18.

Q7. Solve for t: 5t - 3 = 3t - 5

Transpose 3t to LHS and -3 to RHS:
5t - 3t = -5 + 3
2t = -2
t = -2 / 2
t = -1.

Q8. Solve for x: 8x + 4 = 3(x - 1) + 7

Simplify RHS:
8x + 4 = 3x - 3 + 7
8x + 4 = 3x + 4
Transpose 3x to LHS and 4 to RHS:
8x - 3x = 4 - 4
5x = 0
x = 0.

Q9. Solve for x: x = (4/5)(x + 10)

Multiply both sides by 5:
5x = 4(x + 10)
5x = 4x + 40
Transpose 4x to LHS:
5x - 4x = 40
x = 40.

Q10. Solve for y: 2y + 5/3 = 26/3 - y

Transpose -y to LHS and 5/3 to RHS:
2y + y = 26/3 - 5/3
3y = (26 - 5) / 3
3y = 21/3
3y = 7
y = 7/3.

Q11. Solve for x: x/2 - 1/5 = x/3 + 1/4

Transpose terms with x to LHS and constants to RHS:
x/2 - x/3 = 1/4 + 1/5
Take LCM for LHS (6) and RHS (20):
(3x - 2x) / 6 = (5 + 4) / 20
x / 6 = 9 / 20
x = (9 × 6) / 20
x = 54 / 20 = 27 / 10
x = 2.7 or 27/10.

Q12. Solve for n: n/2 - 3n/4 + 5n/6 = 21

LCM of 2, 4, 6 is 12.
(6n - 9n + 10n) / 12 = 21
7n / 12 = 21
7n = 21 × 12
n = (21 × 12) / 7
n = 3 × 12
n = 36.

Q13. Solve for x: x + 7 - 8x/3 = 17/6 - 5x/2

Transpose x terms to LHS, constants to RHS:
x - 8x/3 + 5x/2 = 17/6 - 7
LCM of LHS is 6, LCM of RHS is 6.
(6x - 16x + 15x) / 6 = (17 - 42) / 6
5x / 6 = -25 / 6
5x = -25
x = -25 / 5
x = -5.

Q14. Solve for x: (x - 5) / 3 = (x - 3) / 5

Cross multiply:
5(x - 5) = 3(x - 3)
5x - 25 = 3x - 9
Transpose 3x to LHS and -25 to RHS:
5x - 3x = -9 + 25
2x = 16
x = 8.

Q15. Solve for t: 3(t - 3) = 5(2t + 1)

Expand both sides:
3t - 9 = 10t + 5
Transpose 10t to LHS and -9 to RHS:
3t - 10t = 5 + 9
-7t = 14
t = 14 / -7
t = -2.

Q16. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Let the common ratio multiplier be x.
The numbers are 5x and 3x.
According to the problem: 5x - 3x = 18
2x = 18
x = 9
The numbers are 5(9) = 45 and 3(9) = 27.

Q17. 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?

Let present ages be 5x and 7x.
After 4 years: Rahul = 5x + 4, Haroon = 7x + 4.
Sum in 4 years: (5x + 4) + (7x + 4) = 56
12x + 8 = 56
12x = 48
x = 4
Rahul's age = 5(4) = 20 years. Haroon's age = 7(4) = 28 years.

Q18. Solve for m: m - (m-1)/2 = 1 - (m-2)/3

Multiply the entire equation by LCM(2,3) = 6.
6 × m - 6 × (m-1)/2 = 6 × 1 - 6 × (m-2)/3
6m - 3(m - 1) = 6 - 2(m - 2)
6m - 3m + 3 = 6 - 2m + 4
3m + 3 = 10 - 2m
Transpose -2m to LHS, 3 to RHS:
3m + 2m = 10 - 3
5m = 7
m = 7/5.

Q19. 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?

Let breadth be b meters.
Length = 2b + 2 meters.
Perimeter = 2(Length + Breadth) = 154
2((2b + 2) + b) = 154
2(3b + 2) = 154
Divide both sides by 2:
3b + 2 = 77
3b = 75
b = 25
Breadth = 25 m.
Length = 2(25) + 2 = 50 + 2 = 52 m.

Q20. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Let one number be x.
Then the other number is x + 15.
Sum = x + (x + 15) = 95
2x + 15 = 95
2x = 95 - 15
2x = 80
x = 40
The numbers are 40 and (40+15) = 55.

Class 8 Maths - Linear Equations in One Variable Summary

Chapter 2: Linear Equations in One Variable (Concepts & Formulas)

1. Introduction

An algebraic equation is an equality involving variables. It has an equality sign (=). The expression on the left of the equality sign is the Left Hand Side (LHS), and the expression on the right is the Right Hand Side (RHS).

In a linear equation in one variable, the highest power of the variable appearing in the equation is 1. Operations like addition, subtraction, multiplication, and division are used to solve it.

A value of the variable which makes the equation true (LHS = RHS) is called the solution or root of the equation.

2. Rules for Solving Linear Equations

We can add the same number to both sides of the equation.
We can subtract the same number from both sides of the equation.
We can multiply both sides of the equation by the same non-zero number.
We can divide both sides of the equation by the same non-zero number.
Transposition: Any term of an equation may be taken from one side to the other with a change in its sign.
(+) becomes (-)
(-) becomes (+)
(×) becomes (÷)
(÷) becomes (×)

3. Steps to Solve Applications (Word Problems)

Read the problem carefully and identify the unknown quantity.
Represent the unknown quantity by a variable like x, y, z, etc.
Translate the given conditions or relationships in the problem into mathematical expressions and form an equation.
Solve the equation for the variable.
Check the solution by substituting it back into the original problem to see if it satisfies all conditions.

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