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

Class 8 Mathematics by Full Circle

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

The document NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by Full Circle.
All you need of Class 8 at this link: Class 8

Ques: Solve the following equations.
1. x – 2 = 17                
2. y + 3 = 10              
3. 6 = z + 2
4. 
NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev         
5. 6x = 12

NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev               
9. 7x – 9 = 16
10. 14y – 8 = 13        
11. 17 + 6p = 9

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

Ans: 1. x – 2 = 7 
Transposing (–2) to RHS, we have
x = 7 + 2

∴ x = 9

2. y + 3 = 10
Transposing 3 to RHS, we have
y = 10 – 3
∴ y = 7

3. 6 = z + 2
Transposing 2 to LHS, we have
6 – 2 = z or
4 = z
∴ z = 4

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

Transposing 3/7 to RHS, we have

NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev
= 14/7 = 2

= 2

5. 6x = 12 Dividing both sides by 6, we have:

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

∴ x = 2

6. t/5 = 10
Multiplying both sides by 5, we have 

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

∴ t = 50

7. 2x/3 = 18

Multiplying both sides by 3, we have

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

or 2x = 54

Dividing both sides by 2, we have

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

x = 54/2 = 27

∴ x = 27 

Dividing both sides by 2, we hav
8. 1.6 =  y/1.5

Multiplying both sides by 1.5, we have

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

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

or   

2.4 = y

∴ y = 2.4

9. 7x – 9 = 16
Transposing (–9) to RHS, we have
7x = 16 + 9 = 25

Dividing both sides by 7, we have

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

∴ x = 25/7

10. 14y – 8 = 13
Transposing –8 to RHS, we have
14y = 13 + 8 = 21
Dividing both sides by 14, we have

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

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

∴ y = 3/2

11. 17 + 6p = 9
Transposing 17 to RHS, we have
6p = 9 – 17 = –8
Dividing both sides by 6, we have

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

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

NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev
Transposing 1 to RHS, we have

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

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

Multiplying both sides by 3, we have
NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev

∴ x = -8/5

SOME APPLICATIONS

Example 1: What should be subtracted from twice the rational number -5/3 to get 2/5 ?

Ans: Twice of  NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev
Let the required number to be subtracted be x.
∴  NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev

Transposing -10/3 to RHS, we have

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

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

Thus, the required number is   NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev


Example 2: The perimeter of a rectangle is 17 cm. If its width is NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRevcm, then find its length.

Ans:

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

Let the length of the rectangle = l cm
∴ Since, width of the rectangle (b) = 
NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev  cm
and perimeter = 17 cm
We know that perimeter of a rectangle = 2(l + b) cm

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

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

Dividing both sides by 2, we have 

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

Length of the rectangle =  NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev


Example 3: The sum of three consecutive multiples of 7 is 63. Find these multiples.
Ans: Let x be a multiple of 7.
∴ The next two consecutive multiples of 7 are x + 7 and x + 7 + 7 (i.e. x + 14).
∴ Three consecutive multiples of 7 are: x, x + 7 and x + 14
According to the condition,

x + (x + 7) + (x + 14) = 63
∴ 3x + 21 = 63

Transposing 21 to RHS, we have

3x = 63 – 21 = 42

Dividing both sides by 3, we have

3x/3 = 42/14

x = 14

∴  x + 7 = 14 + 7 = 21
and x + 14 = 14 + 14 = 28

Thus, the three consecutive multiples of 7 are: 14, 21 and 28.

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