Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 26 Nov, 2020

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. 

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :