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

Class 8 Mathematics by Full Circle

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

The document NCERT Solutions(Part- 3)- 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

SOLVING EQUATIONS HAVING THE VARIABLE ON BOTH SIDES

Example:

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

Ans: We have NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev

Transposing   NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev  to RHS, we have

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

Transpiring 5x to LHS, we have

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

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

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

Dividing both sides byNCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRevwe have

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

= -35/7 = -5

∴ x = -5

EXERCISE 2.3 

Ques: Solve the following equations and check your results.
1. 3x = 2x + 18
2. 5t – 3 = 3t – 5
3. 5x + 9 = 5 + 3x
4. 4z + 3 = 6 + 2z
5. 2x – 1 = 14 – x
6. 8x + 4 = 3(x – 1) + 7
7. NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
8.NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
9.NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
10. 3m = 5m 8/5

Ans: 1. 3x = 2x + 18

Transposing 2x from RHS to LHS, we have
3x – 2x = 18
or  
x = 18

2. 5t – 3 = 3t – 5 

Transposing (–3) to RHS, we have
5t = 3t – 5 + 3
or
5t = 3t – 2 

Transposing 3t to LHS, we have

5t – 3t = –2
or
2t = –2

Diving both sides by 2, we have
NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
or
t = –1

3. 5x + 9 = 5 + 3x

Transposing 9 to RHS, we have

5x = 5 + 3x – 9
or

5x = –4 + 3x
Transposing 3x to LHS, we have
5x – 3x = –4
or
2x = –4

Dividing both sides by 2, we have

x = -4/2 = -2

or
x = –2

4. 4z + 3 = 6 + 2z 

Transposing 3 to RHS, we have 

4z = 6 – 3 + 2z
or
4z = 3 + 2z 

Transposing 2z to LHS, we have
4z – 2z = 3
or
2z = 3 

Dividing both sides by 2, we have
z = 3/2

5. 2x – 1 = 14 – x 

Transposing (–1) to RHS, we have 

2x = 14 – x + 1
or
2x = 15 – x 

Transposing (–x) to LHS, we have
2x + x = 15
or
3x = 15 

Dividing both sides by 3, we have

x = 15/3 = 5

∴ x = 5

6.  8x + 4 = 3(x – 1) + 7
or
8x + 4 = 3x – 3 + 7
or
8x + 4 = 3x + 4 

Transposing 4 to RHS, we have 

8x = 3x + 4 – 4 

Transposing 3x to LHS, we have

8x – 3x = 0
or
5x = 0
∴  x = 0

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

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

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

Transposing NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev x to LHS, we have 

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

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

or     x/5 = 8

Multiplying both sides by 5, we have
x = 8 * 5 = 40
∴ x = 40

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

Transposing 1 to RHS, we have

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

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

Transposing 7x/15 to LHS, we have

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

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

or  3x/15 = 2

Multiplying both sides by 15, we have

3x = 2 * 15 = 30

Dividing both sides by 3, we have

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

∴ x = 10

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

Transposing 5/3  to RHS, we have

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

Transposing (–y) to LHS, we have 

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

or  3y = 7

Dividing both sides by 3, we have

3y = 3 = 7/3

∴ y = 7/3

10. 3m = 5m NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev 

Transposing 5m to LHS, we have

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

or   –2m = NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev

Dividing both sides by (–2), we have

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

∴ m = 4/5

Some More Application

Example: The digits of a two digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, then we get 99. Find the original number.

Ans: Let the digit at unit place be ‘x’.

∴ The digit at the tens place = (x + 5)
∴ Original number = 10(x + 5) + x

With interchange of digits, the new number = 10x + (x + 5)
Now, According to the condition, we have

[original number] + [New number] = 99
or  [10(x + 5) + x] + [10x + (x + 5)] = 99
or  [10x + 50 + x] + [10x + x + 5] = 99
or  11x + 50 + 11x + 5 = 99
or  22x + 55 = 99

Transposing 55 to RHS, we have
22x = 99 – 55 = 44
Dividing both sides by 22, we have 

x = 44 ÷ 22 = 2
∴ x = 2 

i.e. Unit place digit = 2
∴Tens place digit = 2 + 5 = 7

Thus the original number = 72.

Note: The statement of the above example is valid for both 72 and 27 and both are correct answers.

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