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# 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.

2.

3.

4.

5.

6.

Ans:
1.

Transposing to RHS and x/3  to LHS, we have

or

or

or

(Multiplying both sides by 6)

Check:

âˆ´ LHS = RHS

2.

âˆµ  LCM of 2, 4 and 6 = 12

âˆ´  Multiplying both sides by 12, we have

or

6n â€“ 9n + 10n = 252

or

7n = 252

or

n = 252/7 = 36

âˆ´  n = 36

Check:

âˆ´ LHS = RHS

3.

âˆµ LCM of 3, 6 and 2 is 6.

âˆ´ Multiplying both sides by 6, we have

or 6x + 42 â€“ 16x = 17 â€“ 15x

or (6 â€“ 16)x + 42 = 17 â€“ 15x

or â€“10x + 42 = 17 â€“ 15x

Transposing 42 to RHS and â€“15x to LHS, we have

â€“10x + 15x = 17 â€“ 42 or 5x = â€“25

or

5x = â€“25

or

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

âˆ´ x  = â€“5

Check:

âˆ´ LHS = RHS

4.

âˆµ LCM of 3 and 5 is 15.

âˆ´ Multiplying both sides by 15, we have

or

5(x â€“ 5) = 3(x â€“ 3)

or

5x â€“ 25 = 3x â€“ 9

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

or

2x = 16

or

x = 16/2   (Dividing both sides by 2)

âˆ´ x = 8

Check:

= 3/3 = 1

âˆ´ LHS = RHS

5.

âˆµ  LCM of 4 and 3 is 12.

âˆ´  Multiplying both sides by 12, we have

or

3(3t â€“ 2) â€“ 4(2t + 3) = (4 x2) â€“ 12t
or

9t â€“ 6 â€“ 8t â€“ 12 = 8 â€“ 12t
or

(9 â€“ 8)t â€“ (6 + 12) = 8 â€“ 12t

or

t â€“ 18 = 8 â€“ 12t

Transposing â€“18 to RHS and â€“12t to LHS, we have
t + 12t = 8 + 18

or

13t = 26

or

t = 26/13

âˆ´ t = 2

Check:

âˆ´ LHS = RHS

6.

Since, LCM of 2 and 3 is 6.

âˆ´   Multiplying both sides by 6, we have

or

6m â€“ 3(m â€“ 1) = 6 â€“ 2(m â€“ 2)
or

6m â€“ 3m + 3 = 6 â€“ 2m + 4
or

(6 â€“ 3)m + 3 = (6 + 4) â€“ 2m
or

3m + 3 = 10 â€“ 2m
Transposing 3 to RHS and â€“2m to LHS, we have
3m + 2m = 10 â€“ 3

or  5m = 7

or

m = 7/5    (Dividing both sides by 5)

Check:

âˆ´ LHS = RHS

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

Ans: 7. 3(t â€“ 3) = 5(2t + 1)
or

3t â€“ 9 = 10t + 5

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

â€“7t = 14

or
t = 14/-7 = -2               (Dividing both sides by â€“7)

âˆ´  t = â€“2

8. 15(y â€“ 4) â€“ 2(y â€“ 9) + 5(y + 6) = 0

Opening the brackets, we have

15y â€“ 60 â€“ 2y + 18 + 5y + 30 = 0

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

18y + (â€“12) = 0

Transposing (â€“12) to RHS, we have

18y = 12

or

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

âˆ´  y = 2/3

9. 3(5z â€“ 7) â€“ 2(9z â€“ 11) = 4(8z â€“ 13) â€“ 17

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

or

â€“3z + 1 = 32z â€“ 69
Transposing 1 to RHS and 32z to LHS, we have
â€“3z â€“ 32z = â€“69 â€“ 1

or

â€“35z = â€“70

or   z = -70/-35 = 2   (Dividing both sides by â€“35)

âˆ´  z = 2

10. 0.25(4f â€“ 3) = 0.05(10f â€“ 9)

Opening the brackets, we have

0.25 * 4f â€“ 3 * 0.25 = 0.05 *10f â€“ 0.05 * 9
or

or

Transposing   to RHS and    to LHS, we have

or

or

(Dividing both sides by 1/2)

or

## Mathematics (Maths) Class 8

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