NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

Mathematics (Maths) Class 8

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Class 8 : NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

The document NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.
All you need of Class 8 at this link: Class 8

EXERCISE 2.5 
Question 1: Solve the following linear equations.

1. NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

2.NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

3.NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

4. NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

5.  NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

6. NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

Ans: 
1. NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

Transposing NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRevto RHS and x/3  to LHS, we have

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

or

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

or

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

or 

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev(Multiplying both sides by 6) 

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

Check:   NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

∴ LHS = RHS

2.  NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev  

∵  LCM of 2, 4 and 6 = 12 

∴  Multiplying both sides by 12, we have

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

or 

6n – 9n + 10n = 252

or

7n = 252

or  

n = 252/7 = 36 

∴  n = 36

Check: NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

 ∴ LHS = RHS

3. NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

∵ LCM of 3, 6 and 2 is 6. 

∴ Multiplying both sides by 6, we have
 NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

or 6x + 42 – 16x = 17 – 15x 

or (6 – 16)x + 42 = 17 – 15x 

or –10x + 42 = 17 – 15x 

Transposing 42 to RHS and –15x to LHS, we have 

–10x + 15x = 17 – 42 or 5x = –25

or 

5x = –25

or 

x = -25/5 = -5    (Dividing both sides by 5)

∴ x  = –5

Check:   NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev    

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

∴ LHS = RHS

4. NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

∵ LCM of 3 and 5 is 15.

∴ Multiplying both sides by 15, we have

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

or 

5(x – 5) = 3(x – 3)

or

5x – 25 = 3x – 9

Transposing (–25) to RHS and 3x to LHS, we have
5x – 3x = –9 + 25

or

2x = 16

or 

x = 16/2   (Dividing both sides by 2)

∴ x = 8

Check: NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

= 3/3 = 1

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

 LHS = RHS

5. NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

∵  LCM of 4 and 3 is 12.

∴  Multiplying both sides by 12, we have

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

or 

3(3t – 2) – 4(2t + 3) = (4 x2) – 12t
or  

9t – 6 – 8t – 12 = 8 – 12t
or  

(9 – 8)t – (6 + 12) = 8 – 12t

or  

t – 18 = 8 – 12t

Transposing –18 to RHS and –12t to LHS, we have
t + 12t = 8 + 18

or

13t = 26

or

 t = 26/13 

∴ t = 2

Check: NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev
NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev
NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev
 LHS = RHS

6.   NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

Since, LCM of 2 and 3 is 6.

∴   Multiplying both sides by 6, we have

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

or

6m – 3(m – 1) = 6 – 2(m – 2)
or

6m – 3m + 3 = 6 – 2m + 4
or

(6 – 3)m + 3 = (6 + 4) – 2m
or

3m + 3 = 10 – 2m
Transposing 3 to RHS and –2m to LHS, we have
3m + 2m = 10 – 3

or  5m = 7

or 

m = 7/5    (Dividing both sides by 5)

Check: NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev
NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev
NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

 LHS = RHS

Question 2: Simplify and solve the following linear equations.
7 . 3(t – 3) = 5(2t + 1)
8. 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
10. 0.25(4f – 3) = 0.05(10f – 9)

Ans: 7. 3(t – 3) = 5(2t + 1)
or

3t – 9 = 10t + 5

Transposing (–9) to RHS and 10t to LHS, we have
3t – 10t = 5 + 9
or 

–7t = 14

or  
t = 14/-7 = -2               (Dividing both sides by –7)

∴  t = –2

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

Opening the brackets, we have

15y – 60 – 2y + 18 + 5y + 30 = 0

Collecting the like terms,
(15 – 2 + 5)y + (–60 + 18 + 30) = 0
or 

18y + (–12) = 0

Transposing (–12) to RHS, we have

18y = 12

or  

y = 12/18 = 2/3  (Dividing both sides by 18)

∴  y = 2/3

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

Opening the brackets, we have
15z – 21 – 18z + 22 = 32z – 52 – 17
Collecting the like terms on both sides,
(15 – 18)z + (–21 + 22) = 32z + (–52 – 17)

or 

–3z + 1 = 32z – 69
Transposing 1 to RHS and 32z to LHS, we have
–3z – 32z = –69 – 1

or

–35z = –70

or   z = -70/-35 = 2   (Dividing both sides by –35)

∴  z = 2

10. 0.25(4f – 3) = 0.05(10f – 9) 

Opening the brackets, we have

0.25 * 4f – 3 * 0.25 = 0.05 *10f – 0.05 * 9
or  

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

or 

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

Transposing  NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev to RHS and   NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev to LHS, we have

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev
or

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

or

NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev (Dividing both sides by 1/2)

or  NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev


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