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

Mathematics (Maths) Class 8

Created by: Full Circle

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

REDUCING EQUATIONS TO SIMPLER FORM

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

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

Multiplying both sides by 6, we have   (∵ LCM of 3 and 6 is 6)

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

or 

x + 3 + 6 = 2(6x – 1) 

or 

x + 9 = 12x – 2 

9 to RHS and 12x to LHS, we have 

x – 12x = –2 – 9 

or 

–11x = –11 

or  

x = -11/-11 = 1 (Dividing both sides by (–11))

∴ x = 1

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

Ques 2: Solve: 2(3x + 1) – 5x NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev

Ans: First opening the brackets, 

LHS = 2(3x + 1) – 5x 

= 6x + 2 – 5x 

= (6 – 5)x + 2 = x + 2

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

Now, the equation is

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

Transposing 2 to RHS and –4x 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 5)

∴ x = -5/2

Check: LHS = 2(3x + 1) – 5x

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

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

EQUATIONS REDUCIBLE TO THE LINEAR FORM

REMEMBER
When we multiply the numerator on LHS with the denominator on RHS and vice-versa and equate the products so obtained, it is known as cross-multiplication.

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

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

By cross multiplication, 

8(x – 1) = 5(2x + 3)
or

8x – 8 = 10x + 15
or

8x = 10x + 15 + 8   (Transposing (–8) to RHS)
or

8x – 10x = 23   (Transposing 10x to LHS)

or

–2x = 23

Dividing both sides by (–2), we have  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

= LHS
i.e. LHS = RHS

Ques 2: The denominator of a rational number greater than its numerator by 3. If the numerator is increased by 16 and the denominator is decreased by 1, then the number obtained is 5/3. Find the rational number.
Ans: Let the numerator = x

∴ Denominator = x + 3

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

New numerator = (x + 16)
and New denominator = (x + 3) – 1
= (x + 2)

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

But the new number = 5/3

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

By cross multiplication, we have

3(x + 16) = 5(x + 2)

or

3x + 48 = 5x + 10

Transposing 48 to RHS and 10 to LHS, we have

3x – 5x = 10 – 48
or

–2x = –38

Dividing both sides by –2, we have

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

or 

x = 19 ⇒ Numerator
and x + 3 = 19 + 3 = 22 ⇒ Denominator

∴  The rational number = 19/22

Check: (i) 22 – 19 = 3, i.e. numerator is 3 less than the denominator.

(ii) New number = NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev which is true.

Ques 3: Present ages of Shaloo and Preeti are in the ratio 5 : 4. 5 years from now the ratio of their ages will be 6 : 5. Find their ages.
Ans: Let the present age of Shaloo = 5x years
Let the present age of Preeti = 4x years
After 5 years, Age of Shaloo = (5x + 5) years
Age of Preeti = (4x + 5) years

According to the condition, we have
NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev
By cross multiplication, we have
5(5x + 5) = 6(4x + 5)
or

25x + 25 = 24x + 30

Transposing 25x to RHS and 24x to LHS, we have
25x – 24x = 30 – 25
or 

x = 5

∴ 5x = 5 * 5 = 25 

and 4x = 4 * 5 = 20

Therefore, Present age of Shaloo = 25 years
Present age of Preeti = 20 years.

Share with a friend

Complete Syllabus of Class 8

Dynamic Test

Content Category

Related Searches

pdf

,

Important questions

,

ppt

,

Sample Paper

,

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

,

practice quizzes

,

Exam

,

Extra Questions

,

shortcuts and tricks

,

Viva Questions

,

Summary

,

Previous Year Questions with Solutions

,

Objective type Questions

,

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

,

past year papers

,

study material

,

Free

,

Semester Notes

,

mock tests for examination

,

video lectures

,

MCQs

;