Class 8 RD Sharma Solutions - Chapter 9 Linear Equation In One Variable - Exercise 9.1

Chapter 9 of RD Sharma's Class 8 mathematics textbook focuses on Linear Equations in One Variable. This chapter introduces students to the concept of linear equations and their solutions. Linear equations are algebraic equations where the variable appears only to the first power.

The chapter covers various methods to solve these equations, including algebraic manipulation and real-world application problems.

Solve each of the following equations and verify your solution:

Question 1: 9 (1/4) = y - 1 (1/3)

Solution: 

(37 / 4) = y - (4/3)

(37 / 4) + (4 / 3) = y

Calculate LCM of 4 and 3 which is 12.

(((37 * 3) * (4 * 4)) / 12) = y

(111 + 16) / 12 = y

127 / 12 = y

Verification:

Putting y = 127 / 12

RHS = y - (4 / 3)

= 127 / 12 - 4 / 3

Calculate LCM of 12 and 3 which is 12

= (127 - (4 * 4)) / 12

= (127 - 16) / 12

= 111 / 12

Divide it by 3

= 37 / 4

= 9 (1/4)

LHS = RHS

Hence, verified.

Question 2: (5x / 3) + (2 / 5) = 1

Solution:  

(5x / 3) = 1 - (2 / 5) 

Calculate LCM of 1 and 5 which is 5.

(5x / 3) = ((5 * 1) - 2) / 5

(5x / 3) = 3 / 5

Using cross multiplication we get,

25x = 9

x = 9 / 25

Verification:

Putting x = 9 / 25 in LHS we get,

= ((5 * (9 / 25)) / 3) + (2 / 5)

= (3 / 5) + (2/ 5) 

= ((3 + 2) / 5)

= 1

LHS = RHS

Hence, verified.

Question 3: x / 2 + x / 3 + x /4 = 13

Solution: 

Calculate LCM of 2, 3 and 4 which is 12.

(6 * x + 4 * x + 3 * x) / 12 = 13

(13x) / 12 = 13

x = 12

Verification:

Putting x = 12 in LHS we get,

12 / 2 + 12 / 3 + 12 / 4 

= (12 * 6 + 12 * 4 + 12 * 3) / 12

= (72 + 48 + 36) / 12  

= 156 / 12

= 13

LHS = RHS

Hence, verified.

Question 4: x / 2 + x / 8 = 1 / 8

Solution: 

Calculate LCM of 2 and 8  which is 8.

((4 * x + x) / 8 = 1 / 8

5x / 8 = 1 / 8

Using cross multiplication  we get,

x = (1 * 8) / (8 * 5)

x = 1 / 5

Verification:

Putting x = 1 / 5 in LHS we get.

= ((1 / 5) / 2) + ((1 / 5) / 8)

= (1 / 10) + (1/ 40)

Calculate LCM of 10 and 40 which is 40.

= ((4 * 1) + 1) / 40

= 5 / 40

= 1 / 8

LHS = RHS

Hence, Verified

Question 5: 2x / 3 - 3x / 8 = 7 / 12

Solution: 

Calculate LCM of 3 and 8 which is 24.

((2x * 8) - (3x * 3)) / 24 = 7 / 12

(16x - 9x) / 24 = 7 / 12

7x / 24 = 7 / 12

Using cross multiplication we get,

x = (7 * 24) / (12 * 7)

x = 2 

Verification:

Putting x = 2 in LHS we get,

= (2 * 2) / 3 - (3 * 2) / 8

= 4 / 3 - 3 / 4

Calculate LCM of 3 and 4 which is 12.

= (4 * 4) - (3 * 3)/ 12

= (16 - 9) / 12

= 7 / 12

LHS = RHS

Hence, Verified.

Question 6: (x + 2) (x + 3) + (x - 3) (x - 2) - 2x (x + 1) = 0

Solution: 

x (x + 3) + 2 (x + 3) + x (x - 2) -3 (x - 2) - 2x² - 2x = 0

x² + 3x + 2x + 6 + x² - 2x - 3x + 6 - 2x² - 2x = 0

x² + x² - 2x² + 3x + 2x - 2x - 3x - 2x + 6 + 6 = 0

-2x + 12 = 0

12 = 2x

x = 6

Verification:

Putting x = 6 in LHS in we get,

= (6 + 2) (6 + 3) + (6 - 3) (6 - 2) - 2 * 6 (6 + 1)

= 8 * 9 + 3 * 4 - 12 * 7

= 72 + 12 - 84

= 0

LHS = RHS

Hence, Verified.  

Question 7: x/2 - 4/5 + x / 5 + 3x/10 = 1/5

Solution: 

x / 2 + x / 5 + 3x / 10 = 1 / 5 + 4 / 5

Calculate LCM of 2, 5, 10 which is 10.

(5x + 2x + 3x) / 10 = (1 + 4) / 5

10x / 10 = 5 / 5

x = 1

Verification:

Putting x = 1 in LHS we get,

= 1 / 2 - 4 / 5 + 1 / 5 + 3 / 10

= (1 * 5 - 4 * 2 + 1 * 2 + 3) / 10

= (5 - 8 + 2 + 3) / 10

= 2 / 10

= 1 / 5

LHS = RHS

Hence, Verified.  

Question 8: 7 / x + 35 = 1 / 10

Solution: 

7 / x  = 1 / 10 - 35

Calculate LCM of 1 , 10 which is 10.

7 / x = (1 - (35 * 10)) / 10

7 / x = - 349 / 10

Using cross multiplication we get,

(7 * 10 / -349) = x

x = -70 / 349

Verification:

Putting x = -70 / 349 in LHS we get,

= 7 / (-70 / 349) + 35

= (7 * 349) / (- 70) + 35

= -349 / 10 + 35

= (-349 + 350) / 10

= 1/10

LHS = RHS

Hence, Verified. 

Question 9: (2x - 1) / 3 - (6x - 2) / 5 = 1/3

Solution: 

Calculate LCM of 3 and 5 which is 15.

(5 (2x - 1) - 3 (6x - 2)) / 15 = 1 / 3

(10x - 5 - 18x + 6) / 15 = 1 / 3

(-8x + 1) / 15 = 1 / 3

Using cross multiplication we get,

3 (-8x + 1) = 15

-24x + 3 = 15

-24x = 12

x = 12 / -24 

x = - 1 /2

Verification:

Putting x = -1 / 2 in LHS we get,

(1* (-1 / 2) - 1) / 3 - (6 * (-1 / 2) - 2) / 5

(-1 - 1) / 3 - (-3 - 2) / 5

(-2 / 3) - (-5 / 5)

(-2 / 3) + 1

Calculate LCM of 1 and 3 which is 3.

(-2 + 1 * 3) / 3

1 / 3

LHS = RHS

Hence, Verified.

Question 10: 13 (y - 4) - 3 (y - 9) - 5 (y + 4) = 0

Solution: 

13y - 52 - 3y + 27 - 5y - 20 = 0

13y - 3y -5y - 52 + 27 - 20 = 0

5y - 45 = 0

5y = 45

y = 9

Verification:

Putting y = 9 in LHS we get,

= 13 (9 - 4) - 3 (9 - 9) - 5 (9 + 4)

= 13 * 5 - 0 - 5 * 13

= 65 - 65 

= 0

LHS = RHS

Hence, Verified.

Question 11: (2 / 3) (x - 5) - (1 / 4) (x - 2) = 9 / 2

Solution: 

2x / 3 - 10 / 3 - x / 4  + 1 / 2 = 9 / 2

2x / 3 -x / 4 = 9 /2 + 10 / 3 - 1 / 2

Calculate LCM of 3 , 4 which is 12 and calculate LCM of  2, 3 which is 6.

(8x - 3x) / 12 = (9 * 3 +  10 * 2 - 1 * 3) / 6

5x / 12 = (27 + 20 - 3) / 6

5x / 12 = 44 / 6

x = (44 * 12) / (6 * 5)

x = 88 / 5

Verification:

Putting x = 88 / 5 in LHS we get,

(2 / 3) (88 / 5 - 5) - (1 / 4) (88 / 5 - 2) 

(2 / 3) * ((88 - 25) / 5) - (1 / 4) * ((88 - 10) / 5)

(2 / 3) * (63 / 5) - (1 / 4) * (78 / 5)

42 / 5 - 39  / 10

Calculate LCM of 5 and 10 which is 10.

(42 * 2 - 39 * 1) / 10

(84 - 39) / 10

45 / 10

9 / 2

LHS = RHS

Hence, Verified. 

Summary

Chapter 9 of RD Sharma's Class 8 mathematics textbook delves into Linear Equations in One Variable, a fundamental concept in algebra. Students learn to identify, formulate, and solve linear equations using various methods such as the balance method, cross-multiplication, and transposition. The chapter emphasizes the importance of maintaining equation balance while performing operations on both sides. It introduces real-world applications, demonstrating how linear equations can model and solve practical problems in areas like geometry, age calculation, and basic physics. Through numerous examples and exercises, students develop problem-solving skills, logical thinking, and the ability to translate word problems into mathematical equations. The chapter builds a strong foundation for more advanced algebraic concepts and prepares students for complex equation solving in higher grades.