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Class 8 NCERT Solutions – Chapter 2 Linear Equations in One Variable – Exercise 2.5

  • Last Updated : 26 Nov, 2020

Solve the following linear equations.

Question 1.  x/2 – 1/5 = x/3 + 1/4

Solution:

(5x – 2)/10 = (4x + 3)/12   …(Taking LCM on both the sides)

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12(5x – 2) = 10 (4x + 3)  …(Cross multiplying)



60x – 24 = 40x + 30   …(Solving the brackets)

60x – 40x = 30 + 24  …(Transposing terms of x to LHS and others to RHS)

20x = 54

x = 54/20 or 27/10 … (Solution)

Verification:

Putting value of “x” in the equation to check if our answer is correct

27/20 – 1/5 = 27/30 + 1/4

(27 – 4)/20 = (108 + 30)/120

23/20 = 138/120

23/20 = 23/20

LHS = RHS (Hence Proved that solution is correct) 

Question 2.  n/2 – 3n/4 + 5n/6 = 21

Solution:

(6n – 9n + 10n)/12 = 21   …(Taking LCM and solving LHS)

7n/12 = 21 (Solving LHS)

7n = 21 × 12

n = 36  …(Solution)

Verification:

Putting value of “n” in the equation to check if our answer is correct



36/2 – 108/4 + 180/6 = 21

18 – 27 + 30 = 21

21 = 21

LHS = RHS (Hence Proved that solution is correct)

Question 3.  x + 7 – 8x/3 = 17/6 – 5x/2

Solution:

x – 8x/3 + 5x/2 = 17/6 – 7  …(Transposing terms of x to LHS and others to RHS)

(6x – 16x + 15x)/6 = (17 – 42)/6  …(Taking LCM and solving)

5x/6 = -25/6

x = -5  …(Solution)

Verification –

Putting value of “x” in the equation to check if our answer is correct

-5 + 7 – (-40)/3 = 17/6 – (-25)/2

2 + 40/3 = 17/6 + 25/2

46/3 = (17 + 75)/6

46/3 = 92/6

46/3 = 46/3

LHS = RHS  (Hence Proved that solution is correct)

Question 4.  (x – 5)/3 = (x – 3)/5

Solution:

5(x – 5) = 3(x – 3)  …(Cross multiply)

5x – 25 = 3x – 9  



2x = 16

x = 8  …(Solution)

Verification – 

Putting value of “x” in the equation to check if our answer is correct

(8 – 5)/3 = (8 – 3)/5

3/3 = 5/5

1 = 1

LHS = RHS (Hence Proved that solution is correct)

Question 5.  (3t – 2)/4 – (2t + 3)/3 = 2/3 – t

Solution:

3t/4 – 1/2 – 2t/3 -1 = 2/3 – t  …(Solving brackets)

3t/4 – 2t/3 + t = 2/3 + 1 + 1/2  …(Transposing terms of x to LHS and others to RHS)

(9t – 8t + 12t)/12 = (4 + 6 + 3)/6  …(Taking LCM both sides)

13t/12 = 13/6

t = 2   …(Solution)

Verification –

Putting value of “t” in the equation to check if our answer is correct

(3 × 2 – 2)/4 – (2 × 2 + 3)/3 = 2/3 – 2

4/4 – 7/3 = 2/3 – 2

(12 – 28)/12 = (2 – 6)/3

-16/12 = -4/3

-4/3 = -4/3

LHS = RHS (Hence Proved that solution is correct)

Question 6.  m – (m – 1)/2 = 1 – (m – 2)/3

Solution:

(2m – m + 1)/2 = (3 – m + 2)/3  …(Taking LCM both sides)

(m + 1)/2 = (5 – m)/3

3(m + 1) = 2(5 – m)  …(Cross multiplying)

3m + 3 = 10 – 2m

5m = 7

m = 7/5  …(Solution)

Verification –



Putting value of “m” in the equation to check if our answer is correct

7/5 – (7/5 – 1)/2 = 1 – (7/5 – 2)/3

7/5 – 1/5 = 1 – (-3)/15

6/5 = 1 + 1/5

6/5 = 6/5

LHS = RHS (Hence Proved that solution is correct)

Question 7. 3(t – 3) = 5(2t + 1)

Solution:

3t – 9 = 10t + 5  …(Opening brackets)

3t – 10t = 9 + 5

-7t = 14

t = -2  …(Solution)

Verification –

Putting value of “t” in the equation to check if our answer is correct 

3(-2 – 3) = 5(2(-2) + 1)

3(-5) = 5(-4 +1)

-15 = -15

LHS = RHS (Hence Proved that solution is correct)

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

Solution:

15y – 60 – 2y + 18 + 5y + 30 = 0

18y – 12 = 0

y = 12/18 or 2/3  …(Solution)

Verification –

Putting value of “y” in the equation to check if our answer is correct 

15(2/3 – 4) – 2(2/3 – 9) + 5(2/3 + 6) = 0

10 – 60 – 4/3 +18 + 10/3 + 30 = 0

-50 -4/3 + 48 + 10/3 = 0

-2 + 6/3 = 0

-2 + 2 = 0

0 = 0

LHS = RHS (Hence Proved that solution is correct)

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

Solution:

15z – 21 – 18z + 22 = 32z – 52 – 17  …(Solving the brackets)

-3z + 1 = 32z – 69

-35z = -70

z = 2  …(Solution)

Verification –

Putting value of “z” in the equation to check if our answer is correct 

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

3(3) – 2(7) = 4(3) – 17

9 – 14 = 12 – 17



-5 = -5

LHS = RHS (Hence Proved that solution is correct)

Question 10. 0.25(4f – 3) = 0.05(10f – 9)

Solution:

f – 0.25(3) = 0.5f – 0.05(9)

f – 0.75 = 0.5f – 0.45

0.5f = 0.75 – 0.45

f = 3/5 or 0.6 (Solution)

Verification – 

Putting value of “f” in the equation to check if our answer is correct

0.25(4(0.6) – 3) = 0.05(10(0.6) – 9)

0.25(2.4 – 3) = 0.05(6 – 9)

0.25 × (-0.6) = 0.05 × (-3)

-0.15 = -0.15

LHS = RHS (Hence Proved that solution is correct)




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