Solutions to NCERT Class 8 Mathematics Chapter 2 Linear Equations in One Variable:

Welcome students! Our Chapter 2 solutions are an excellent place to continue your study on equations. The problem-solving techniques and pretty much all that you need to know are explained clearly in the solutions and extra materials. We only advise that you solve the problems yourselves alongside looking at the solutions. This will help you learn the material faster and retain it better.

Solutions to Exercise 2.1 (Page No 17) of NCERT Class 8 Math Chapter 2 Linear Equations in One Variable –

Solve the following equations and check your results.

1. 3x = 2x + 18
Answer:
3x = 2x + 18
or, 3x – 2x = 18 (Transposing 2x to LHS)
or, x = 18 (Answer)
Verification:
Putting x = 18 in the equation we get,
3 × 18 = 2 × 18 + 18
or, 54 = 36 + 18
or, 54 = 54 (LHS = RHS, hence verified)
Summary: The solution of the equation 3x = 2x + 18 is x = 18.

2. 5t – 3 = 3t – 5
Answer:
5t – 3 = 3t – 5
or, 5t – 3t = 3 – 5 (Transposing 3t to LHS and -3 to RHS)
or, 2t = – 2
or, t = (-2)/2 (Dividing both sides by 2)
or, t = -1 (Answer)
Verification:
Putting t =-1 in the equation we get,
5 × (-1) – 3 = – 3 × (-1) – 5
or, -5 – 3 = -3 – 5
or, -8 = -8 (LHS = RHS, hence verified)
Summary: The solution of the equation 5t – 3 = 3t – 5 is t = -1.

3. 5x + 9 = 5 + 3x
Answer:
5x + 9 = 5 + 3x
or, 5x – 3x = 5 – 9 (Transposing 3x to LHS and 9 to RHS)
or, 2x = -4
or, x = (-4)/2 (Dividing both sides by 2)
or, x = -2 (Answer)
Verification:
Putting x = -2 in the equation we get,
5 × (-2) + 9 = 5 + 3 × (-2)
or, -10 + 9 = 5 – 6
or, – 1 = – 1 (LHS = RHS, hence verified)
Summary: The solution of the equation 5x + 9 = 5 + 3x is x = -2.

4. 4z + 3 = 6 + 2z
Answer:
4z + 3 = 6 + 2z
or, 4z – 2z = 6 – 3 (Transposing 2z to LHS and 3 to RHS)
or, 2z = 3
or, z = 3/2 (Dividing both sides by 2) (Answer)
Verification:
Putting z = 3/2 in the equation we get,
4 × 3/2 + 3 = 6 + 2 × 3/2
or, 6 + 3 = 6 + 3
or, 9 = 9 (LHS = RHS, hence verified)
Summary: The solution of the equation 4z + 3 = 6 + 2z is z = 3/2.

5. 2x – 1 = 14 – x
Answer:
2x – 1 = 14 – x
or, 2x + x = 14 + 1 (Transposing -x to LHS and -1 to RHS)
or, 3x = 15
or, x = 15/3 (Dividing both sides by 3)
or, x = 5 (Answer)
Verification:
Putting x = 5 in the equation we get,
2 × 5 – 1 = 14 – 5
or, 10 – 1 = 14 – 5
or, 9 = 9 (LHS = RHS, hence verified)
Summary: The solution of the equation 2x – 1 = 14 – x is x = 5.

6. 8x + 4 = 3 (x – 1) + 7
Answer:
8x + 4 = 3 (x – 1) + 7
or, 8x + 4 = 3x – 3 + 7
or, 8x – 3x = -3 + 7 – 4 (Transposing 3x to LHS and 4 to RHS)
or, 5x = 0
or, x = 0/5 (Dividing both sides by 5)
or, x = 0 (Answer)
Verification:
Putting x = 0 in the equation we get,
8 × (0) + 4 = 3 × (0 – 1) + 7
or, 4 = -3 + 7
or, 4 = 4 (LHS = RHS, hence verified)
Summary: The solution of the equation 8x + 4 = 3 (x – 1) + 7 is x = 0.

7. x = 4/5 (x + 10)
Answer:
x = 4/5 (x + 10)
or, x = 4/5x + 4/5 × 10
or, x = 4/5x + 8
or, x – 4/5x = 8 (Transposing 4/5x to LHS)
or, (5x – 4x)/5 = 8
or, x/5 = 8
or, x = 8 × 5
or, x = 40 (Answer)
Verification:
Putting x = 40 in the equation we get,
40 = 4/5 × (40 + 10)
or, 40 = 4/5 × 50
or, 40 = (4 × 50)/5
or, 40 = 40 (LHS = RHS, hence verified)
Summary: The solution of the equation x = 4/5 (x + 10) is x = 40.

8. 2x/3 + 1 = 7x/15 + 3
Answer:
2x/3 + 1 = 7x/15 + 3
or, 2x/3 – 7x/15 = 3 – 1 (Transposing 7x/15 to LHS and 1 to RHS)
or, 10x/15 – 7x/15 = 3 – 1
or, (10x – 7x)/15 = 2
or, 3x/15 = 2
or, x = (15 × 2)/3 (Multiplying both sides by 15 and then dividing both sides by 3)
or, x = 10 (Answer)
Verification:
Putting x = 10 in the equation we get,
(2 × 10)/3 + 1 = (7 × 10)/15 + 3
or, 20/3 + 1 = 70/15 + 3
or, 20/3 + 1 = 14/3 + 3
or, 23/3 = 23/3 (LHS = RHS, hence verified)
Summary: The solution of the equation 2x/3 + 1 = 7x/15 + 3 is x = 23/3.

9. 2y + 5/3 = 26/3 – y
Answer:
2y + 5/3 = 26/3 – y
or, 2y + y = 26/3 – 5/3 (Transposing -y to LHS and 5/3 to RHS)
or, 3y = (26 – 5)/3
or, 3y = 21/3
or, y = 21/(3 × 3) (Dividing both sides by 3)
or, y = 7/3 (Answer)
Verification:
Putting y = 7/3 in the equation we get,
2 × 7/3 + 5/3 = 26/3 – 7/3
or, 14/3 + 5/3 = (26 – 7)/3
or, (14 + 5)/3 = 19/3
or, 19/3 = 19/3 (LHS = RHS, hence verified)
Summary: The solution of the equation 2y + 5/3 = 26/3 – y is y = 7/3.

10. 3m = 5m – 8/5
Answer:
3m = 5m – 8/5 can be written as:
5m – 8/5 = 3m
or, 5m – 3m = 8/5 (Transposing 3m to LHS and – 8/5 to RHS)
or, 2m = 8/5
or, m = 8/(5 × 2) (Dividing both sides by 2)
or, m = 4/5 (Answer)
Verification:
Putting m = 4/5 in the equation we get,
3 × 4/5 = 5 × 4/5 – 8/5
or, 12/5 = 4 – 8/5
or, 12/5 = (4 × 5 – 8)/5
or, 12/5 = (20 – 8)/5
or, 12/5 = 12/5 (LHS = RHS, hence verified)
Summary: The solution of the equation 3m = 5m – 8/5 is m = 4/5.

Solutions to Exercise 2.2 (Page No 19) of NCERT Class 8 Math Chapter 2 Linear Equations in One Variable –

Solve the following linear equations.

1. x/2 – 1/5 = x/3 + 1/4
Answer:
x/2 – 1/5 = x/3 + 1/4
or, x/2 – x/3 = 1/5 + 1/4 (Transposing x/3 to LHS and -1/5 to RHS)
or, 3x/6 – 2x/6 = 4/20 + 5/20
or, (3x – 2x)/6 = (4 + 5)/20
or, x/6 = 9/20
or, x = (9 × 6)/20 (Multiplying both sides by 6)
or, x = 27/10 (Answer)
Check:
LHS: 1/2 × 27/10 – 1/5
= (1 × 27)/(2 × 10) – 1/5 = 27/20 – 4/20 = (27 – 4)/20 = 23/20
RHS: 27/10 × 1/3 + 1/4 = 9/10 + 1/4 = 18/20 + 5/20 = 23/20
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation x/2 – 1/5 = x/3 + 1/4 is x = 27/10.

2. n/2 – 3n/4 + 5n/6 = 21
Answer:
n/2 – 3n/4 + 5n/6 = 21
or, 12 × n/2 – 12 × 3n/4 + 12 × 5n/6 = 21 × 12 (Multiplying both sides by 12 which is the LCM of the denominators 2, 4, 6)
or, 6n – 9n + 10n = 252
or, 7n = 252
or, n = 252/7 (Dividing both sides by 7)
or, n = 36 (Answer)
Check:
LHS: n/2 – 3n/4 + 5n/6
= 36/2 – (3 × 36)/4 + (5 × 36)/6 = 18 – 27 + 30 = 21
RHS = 21
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation n/2 – 3n/4 + 5n/6 = 21 is n = 36.

3. x + 7 – 8x/3 = 17/6 – 5x/2
Answer:
x + 7 – 8x/3 = 17/6 – 5x/2
or, 6x + 7 × 6 – 8x/3 × 6 = 17/6 × 6 – 5x/2 × 6 (Multiplying both sides by 6 which is the LCM of the denominators 2, 4, 6)
or, 6x + 42 – 16x = 17 – 15x
or, 6x – 16x + 15x = 17 – 42 (Transposing – 15x to LHS and 42 to RHS)
or, 5x = -25
or, x = (-25)/5 (Dividing both sides by 5)
or, x = -5 (Answer)
Check:
LHS: x + 7 – 8x/3 = -5 + 7 – (8 ×(-5))/3 = -5 + 7 + 40/3 = 2 + 40/3 = (6 + 40)/3 = 46/3
RHS: 17/6 – 5x/2 = 17/6 – (5 × (-5))/2 = 17/6 + 25/2 = (17 + 25 × 3 )/6 = (17 + 75 )/6 = (92 )/6 = (46 )/3
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation x + 7 – 8x/3 = 17/6 – 5x/2 is x = -5.

4. (x – 5)/3 = (x – 3)/5
Answer:
(x – 5)/3 = (x – 3)/5
or, ((x – 5))/3 × 15 = ((x – 3))/5 × 15 (Multiplying both sides by 15 which is the LCM of the denominators 3, 5)
or, 5(x – 5) = 3(x – 3)
or, 5x – 25 = 3x – 9
or, 5x – 3x = 25 – 9 (Transposing 3x to LHS and -25 to RHS)
or, 2x = 16
or, x = 16/2 (Dividing both sides by 2)
or, x = 8 (Answer)
Check:
LHS: (x – 5)/3 = (8 – 5)/3 = 3/3 = 1
RHS: (x – 3)/5 = (8 – 3)/5 = 5/5 = 1
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation (x – 5)/3 = (x – 3)/5 is x = 8

5. (3t – 2)/4 – (2t + 3)/3 = 2/3 – t
Answer:
(3t – 2)/4 – (2t + 3)/3 = 2/3 – t
or, ((3t – 2)/4) × 12 – ((2t + 3)/3) × 12 = 2/3 × 12 – t × 12 (Multiplying both sides by 12 which is the LCM of the denominators 4, 3, 3)
or, (3t – 2) × 3 – (2t + 3) × 4 = (2 × 12)/3 – t × 12
or, 9t – 6 – 8t – 12 = 8 – 12t (Transposing 12t to LHS and -6 and -12 to RHS)
or, 9t – 8t + 12t = 6 + 12 + 8
or, 13t = 26
or, t = 26/13 (Dividing both sides by 13)
or, t = 2 (Answer)
Check:
LHS: (3t – 2)/4 – (2t + 3)/3 = (3 × 2 – 2)/4 – (2 × 2 + 3)/3 = 1 – 7/3 = ( 3 – 7)/3 = (-4)/3
RHS: 2/3 – t = 2/3 – 2 = (2 – 6)/3 = (-4)/3
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation (3t – 2)/4 – (2t + 3)/3 = 2/3 – t is t = 2.

6. m – (m – 1)/2 = 1 – (m – 2)/3
Answer:
m – (m – 1)/2 = 1 – (m – 2)/3
or, 6m – ((m – 1)/2) × 6 = 6 – ((m – 2)/3) × 6 (Multiplying both sides by 6 which is the LCM of the denominators 2, 3)
or, 6m – 3(m – 1) = 6 – 2(m – 2)
or, 6m – 3m + 3 = 6 – 2m + 4
or, 6m – 3m + 2m = 6 + 4 – 3 (Transposing -2m to LHS and 3 to RHS)
or, 5m = 7
or, m = 7/5 (Dividing both sides by 5)
or, m = 7/5 (Answer)
Check:
LHS: m – (m – 1)/2 = 7/5 – (7/5 – 1)/2 = 7/5 – (2/5)/2 = 7/5 – 1/5 = 6/5
RHS: 1 – (m – 2)/3 = 1 – (7/5 – 2)/3 = 1 – ((-3)/5)/3 = 1 + 1/5 = 6/5
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation m – (m – 1)/2 = 1 – (m – 2)/3 is m = 7/5 .

Simplify and solve the following linear equations.

7. 3(t – 3) = 5(2t + 1)
Answer:
3(t – 3) = 5(2t + 1)
or, 3t – 9 = 10t + 5
or, 10t + 5 = 3t – 9
or, 10t – 3t = – 9 – 5 (Transposing 3t to LHS and 5 to RHS)
or, 7t = -14
or, t = (-14)/7 (Dividing both sides by 7)
or, t = -2 (Ans)
Check:
LHS: 3(t – 3) = 3(-2 – 3) = 3 × (-5) = -15
RHS: 5(2t + 1) = 10t + 5 = 10 × (-2) + 5 = -20 + 5 = -15
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation 3(t – 3) = 5(2t + 1) is t = -2.

8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0
Answer:
15(y – 4) –2(y – 9) + 5(y + 6) = 0
or, 15y – 60 – 2y + 18 + 5y + 30 = 0
or, 18y – 12 = 0
or, 18y = 12 (Transposing -12 to RHS)
or, y = 12/18 (Dividing both sides by 18)
or, y = 2/3 (Answer)
Check:
LHS: 15(y – 4) – 2(y – 9) + 5(y + 6)
= 15(2/3 – 4) – 2(2/3 – 9) + 5(2/3 + 6)
= 15 ((2 – 12)/3) – 2((2 – 27)/3) + 5((2 + 18)/3)
= 15 × ((-10)/3) – 2 × ((-25)/3) + 5 × (20/3)
= -50 + 50/3 + 100/3 = -50 + (50 + 100)/3 = -50 + 150/3 = -50 + 50 = 0
RHS: 0
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation 15(y – 4) –2(y – 9) + 5(y + 6) = 0 is y = 2/3.

9. 3 (5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Answer:
3 (5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
or, 15z – 21 – 18z + 22 = 32z – 52 – 17
or, – 3z + 1 = 32z – 69
or, 32z – 69 = – 3z + 1
or, 32z + 3z = 69 + 1 (Transposing – 3z to LHS and -69 to RHS)
or, 35z = 70
or, z = 70/35 (Dividing both sides by 35)
or, z = 2 (Answer)
Check:
LHS: 3 (5z – 7) – 2(9z – 11) = 3(5 × 2 – 7) – 2(9 × 2 – 11) = 3(10 – 7) – 2(18 – 11) = 9 – 14 = -5
RHS: 4(8z – 13) – 17 = 4(8 × 2 – 13) – 17 = 4(16 – 13) – 17 = 4 × 3 – 17 = 12 – 17 = -5
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation 3 (5z – 7) – 2(9z – 11) = 4(8z – 13) – 17 is z = 2.

10. 0.25(4f – 3) = 0.05(10f – 9)
Answer:
0.25(4f – 3) = 0.05(10f – 9)
or, 1/4 (4f – 3) = 5/100 (10f – 9)
or, 1/4 × 4f – 1/4 × 3 = 5/100 × 10f – 5/100 × 9
or, f – 3/4 = (5 × 10f)/100 – (5 × 9)/100
or, f – 3/4 = f/2 – 9/20
or, f – f/2 = 3/4 – 9/20 (Transposing f/2 to LHS and – 3/4 to RHS)
or, f/2 = 15/20 – 9/20
or, f/2 = (15 – 9)/20
or, f/2 = 6/20
or, f = (6 × 2)/20
or, f = 3/5 (Answer)
Check:
LHS: 0.25(4f – 3) = 1/4 (4 × 3/5 – 3) = 1/4 ( (4 × 3)/5 – 3) = 1/4 × 12/5 – 3/4 = 3/5 – 3/4 = 12/20 – 15/20 = (12 – 15)/20 = (-3)/20
RHS: 0.05(10f – 9) = 5/100 (10f – 9) = 1/20 (10 × 3/5 – 9) = 1/20 × 10 × 3/5 – 1/20 × 9 = 3/10 – 9/20 = (6 – 9)/20 = (-3)/20
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation 0.25(4f – 3) = 0.05(10f – 9) is f = 3/5.

Important Questions from Previous NCERT Textbook:

Exercise 2.1 Page No 23 (Old Textbook):

Solve the following equations.

1. x – 2 = 7
Answer:
x – 2 = 7
or, x – 2 + 2 = 7 + 2 (Adding 2 to both sides)
or, x = 9 (Answer)
Summary: The solution of the equation x – 2 = 7 is x = 9.

2. y + 3 = 10
Answer:
y + 3 = 10
or, y = 10 – 3 (Transposing 3 to RHS)
or, y = 7 (Answer)
Summary: The solution of the equation y + 3 = 10 is y = 7.

3. 6 = z + 2
Answer:
6 = z + 2
The above equation can be written as:
z + 2 = 6
or, z = 6 – 2 (Transposing 2 to RHS)
or, z = 4 (Answer)
Summary: The solution of the equation 6 = z + 2 is z = 4.

4. 3/7 + x = 17/7
Answer:
3/7 + x = 17/7
or, x = 17/7 – 3/7 (Transposing 3/7 to RHS)
or, x = (17 – 3)/7
or, x = 14/7
or, x = 2 (Answer)
Summary: The solution of the equation 3/7 + x = 17/7 is x = 2.

5. 6x = 12
Answer:
6x = 12
or, x = 12/6 (Dividing both sides by 6)
or, x = 2 (Answer)
Summary: The solution of the equation 6x = 12 is x = 2.

6. t/5 = 10
Answer:
t/5 = 10
or, t = 10 × 5 (Multiplying both sides by 5)
or, t = 50 (Answer)
Summary: The solution of the equation t/5 = 10 is t = 50.

7. 2x/3 = 18
Answer:
2x/3 = 18
or, 2x = 18 × 3 (Multiplying both sides by 3)
or, x = (18 × 3)/2 (Dividing both sides by 2)
or, x = 27 (Answer)
Summary: The solution of the equation 2x/3 = 18 is x = 27.

8. 1.6 = y/(1.5)
Answer:
1.6 = y/1.5 can be written as:
y/1.5 = 1.6
or, y = 1.6 × 1.5 (Multiplying both sides by 1.5)
or, y = 2.4 (Answer)
Summary: The solution of the equation 1.6 = y/1.5 is y = 2.4.

9. 7x – 9 = 16
Answer:
7x – 9 = 16
or, 7x – 9 + 9 = 16 + 9 (Adding 9 to both sides)
or, 7x = 25
or, x = 25/7 (Dividing both sides by 7) (Answer)
Summary: The solution of the equation 7x – 9 = 16 is x = 25/7.

10. 14y – 8 = 13
Answer:
14y – 8 = 13
or, 14y – 8 + 8 = 13 + 8 (Adding 8 to both sides)
or, 14y = 21
or, y = 21/14 (Dividing both sides by 14)
or, y = 3/2 (Answer)
Summary: The solution of the equation 14y – 8 = 13 is y = 3/2.

11. 17 + 6p = 9
Answer:
17 + 6p = 9
or, 6p = 9 – 17 (Transposing 17 to RHS)
or, 6p = –8
or, p = (–8)/6 (Dividing both sides by 6)
or, p = (–4)/3 (Answer)
Summary: The solution of the equation 17 + 6p = 9 is p = (–4)/3.

12. x/3 + 1 = 7/15
Answer:
x/3 + 1 = 7/15
or, x/3 = 7/15 – 1 (Transposing 1 to RHS)
or, x/3 = (7 – 15)/15
or, x/3 = (-8)/15
or, x = (-8)/15 × 3 (Multiplying both sides by 3)
or, x = (-8)/5 (Answer)
Summary: The solution of the equation x/3 + 1 = 7/15 is x = (-8)/5.

Exercise 2.2 Page No: 28 (Old Textbook):

1. If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. What is the number?
Answer:
Let the required number be x.
If 1/2 is subtracted from a number we get the number (x – 1/2).
Therefore, from the problem statement we get:
1/2(x – 1/2) = 1/8
or, (x – 1/2) = 1/4 (Multiplying both sides by 2)
or, x = 1/4 + 1/2
or, x = 1/4 + 2/4
or, x = 3/4
Summary: If you subtract 1/2 from a number and multiply the result by 1/2 to get 1/8, the required number = 3/4.

2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and breadth of the pool?
Answer:
Let the breadth of the rectangle be x m.
So, the length of the rectangle is (2x + 2) m.
Perimeter of the rectangle = 2(length + breadth) = 154 m.
Therefore,
2(2x + 2 + x) = 154
or, 2(3x + 2) = 154
or, (3x + 2) = 154/2 (Dividing both sides by 2)
or, 3x + 2 = 77
or, 3x = 77 – 2 (Transposing 2 to RHS)
or, 3x = 75
or, x = 75/3 (Dividing both sides by 3)
or, x = 25
Therefore, breadth = 25 m and the length = (25 × 2 + 2) = 52 m.
Summary: If the perimeter of a swimming pool is 154 m and length is 2 m more than twice its breadth, the breadth = 25 m and the length = 52 m.

3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 42/15 cm. What is the length of either of the remaining equal sides?
Answer:
Let each equal side be x cm. The base = 4/3 cm.
Therefore, x + x + 4/3 = 42/15
or, 2x + 4/3 = 62/15
or, 2x = 62/15 – 4/3 (Transposing 4/3 to RHS)
or, 2x = 62/15 – 20/15
or, 2x = (62 – 20)/15
or, 2x = 42/15
or, x = 42/15 × 1/2 (Dividing both sides by 2)
or, x = 7/5
Summary: If the base of an isosceles triangle is 4/3 cm and perimeter is 42/15 cm. the length of each remaining equal side = 7/5 cm.

4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Answer:
Let the smaller number be x. Then the bigger number is (x + 15).
Therefore,
x + (x + 15) = 95
or, 2x + 15 = 95
or, 2x = 95 – 15 (Transposing 15 to RHS)
or, 2x = 80
or, x = 80/2 (Dividing both sides by 2)
or, x = 40
Therefore, the smaller number = 40 and the bigger number = (40 + 15) = 55.
Summary: If the sum of two numbers is 95 and if one exceeds the other by 15, the smaller number = 40 and the bigger number = 55.

5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Answer:
Let the two numbers be 5x and 3x.
Therefore,
5x – 3x = 18
or, 2x = 18
or, x = 18/2 (Dividing both sides by 2)
or, x = 9
Therefore, the numbers are 5x = (5 × 9) = 45 and 3x = (3 × 3) = 27.
Summary: If two numbers are in the ratio 5:3 and differ by 18, the greater number = 45 and the smaller number = 27.

6. Three consecutive integers add up to 51. What are these integers?
Answer:
Let the three consecutive integers be x, (x + 1), (x + 2).
Therefore,
x + (x + 1) + (x + 2) = 51
or, 3x + 3 = 51
or, 3x = 51 – 3 (Transposing 3 to RHS)
or, 3x = 48
or, x = 48/3 (Dividing both sides by 3)
or, x = 16
Therefore, the three consecutive integers are x = 16, (x + 1) = 17, (x + 2) = 18.
Summary: If three consecutive integers add up to 51, the integers are 16, 17 and 18.

7. The sum of three consecutive multiples of 8 is 888. Find the multiples.
Answer:
Let the smallest multiple of 8 be x. Then the next two multiples = (x + 8) and (x + 16).
Therefore,
x + (x + 8) + (x + 16) = 888
or, 3x + 24 = 888
or, 3x = 888 – 24 (Transposing 24 to RHS)
or, 3x = 864
or, x = 864/3 (Dividing both sides by 3)
or, x = 288
Therefore, the first multiple is 288, the second multiple = (288 + 8) = 296 and the third multiple = (288 + 16) = 304.
Summary: If the sum of three consecutive multiples of 8 is 888, the integers are 288, 296 and 304.

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Answer:
Let the three consecutive integers be x, (x + 1), (x + 2).
From the given problem statement we get:
2x + 3(x + 1) + 4(x + 2) = 74
or, 2x + 3x + 3 + 4x + 8 = 74
or, 9x + 11 = 74
or, 9x = 74 – 11 (Transposing 11 to RHS)
or, 9x = 63
or, x = 63/9 (Dividing both sides by 9)
or, x = 7
Therefore, (x + 1) = 7 + 1 = 8 and (x + 2) = 7 + 2 = 9.
Summary: If three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively they add up to 74, the numbers are 7, 8 and 9.

9. The ages of Rahul and Haroon are in the ratio 5:7. Four years later, the sum of their ages will be 56 years. What are their present ages?
Answer:
Let the ages of Rahul and Haroon be 5x years and 7x years respectively.
Therefore, 4 years later Rahul’s age = (5x + 4) years and Haroon’s age = (7x +4) years.
From the given problem statement,
(5x + 4) + (7x + 4) = 56
or, 12x + 8 = 56
or, 12x = 56 – 8 (Transposing 8 to RHS)
or, 12x = 48
or, 12x = 48/12 (Dividing both sides by 12)
or, x = 4
Therefore, 5x = 5 × 4 = 20 and 7x = 7 × 4 = 28.
Summary: If the ages of Rahul and Haroon are in the ratio 5:7 and four years later the sum of their ages is 56 years, then Rahul’s age = 20 years and Haroon’s age = 28 years.

10. The number of boys and girls in a class is in the ratio of 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?
Answer: Let the number of boys in the class = 7x and the number of girls = 5x.
From the problem statement we get:
7x – 5x = 8
or, 2x = 8
or, x = 8/2 (Dividing both sides by 2)
or, x = 4
Therefore, number of boys = 7x = 7 × 4 = 28 and number of girls = 5x = 5 × 4 = 20.
Therefore, total class strength = (28 + 20) = 48.
Summary: If the number of boys and girls in a class is in the ratio of 7:5 and the number of boys is 8 more than the number of girls, the total class strength = 48.

11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Answer:
Let the age of Baichung’s father be x years.
Therefore, the age of Baichung’s grandfather = (x + 26) years and the age of Baichung = (x – 29) years.
From the problem statement we get:
x + (x + 26) + (x – 29) = 135
or, 3x – 3 = 135
or, 3x = 135 + 3 (Transposing – 3 to RHS)
or, 3x = 138
or, x = 138/3 (Dividing both sides by 3)
or, x = 46
Therefore, (x + 26) = 46 + 26 = 72 and (x – 29) = (46 – 29) = 17.
Summary: Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung and the sum of their ages is 135 years. Then Baichung’s father’s age = 46 years, Baichung’s grandfather’s age = 72 years and Baichung’s = 17 years.

12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Answer:
Let Ravi’s present age be x.
From the problem statement we get:
x + 15 = 4x
or, 4x = x + 15
or, 4x – x = x + 15 – x (Subtracting x from both sides)
or, 3x = 15
or, x = 15/3 (Dividing both sides by 3)
or, x = 5
Summary: If fifteen years from now Ravi’s age will be four times his present age, then Ravi’s present age = 5 years.

13. A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get – 7/12. What is the number?
Answer:
Let the rational number be x.
From the problem statement we get:
5/2x + 2/3 = – 7/12
or, 5/2x = – 7/12 – 2/3
or, 5/2x = (–7)/12 – 8/12
or, 5/2x = (–7-8)/12
or, 5/2x = (-15)/12
or, x = (-15)/12 × 2/5 (Dividing both sides by 5/2)
or, x = (-15 × 2)/(12 × 5)
or, x = (-1)/2
Summary: If a rational number is such that when you multiply it by 5/2 and add 2/3 to the product you get – 7/12, the required rational number = (-1)/2.

14. Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, ₹50 and ₹10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹4,00,000. How many notes of each denomination does she have?
Answer:
Let the number of ₹100 notes be 2x, the number of ₹50 notes be 3x and the number of ₹10 notes be 5x.
From the problem statement we get:
100 × 2x + 50 × 3x + 10 × 5x = 400000
or, 200x + 150x + 50x = 400000
or, 400x = 400000
or, x = 400000/400 (Dividing both sides by 400)
or, x = 1000
Therefore, 2x = 2 × 1000 = 2000, 3x = 3 × 1000 = 3000 and 5x = 5 × 1000 = 5000.
Summary: Lakshmi has currency notes of denominations ₹100, ₹50 and ₹10 respectively, the ratio of the number of these notes is 2:3:5 and the total cash with Lakshmi is ₹4,00,000. Then the number of ₹100 notes = 2000, the number of ₹50 notes = 3000 and the number of ₹10 notes = 5000.

15. I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of ₹2 coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Answer:
Let the number of ₹5 coins is x. Therefore, the number of ₹2 rupee coins is 3x.
Since the total number of coins is 160, the number of ₹1 rupee coins = (160 – 3x – x) = (160 – 4x).
The value of all the ₹5 coins = 5x, the value of all 2 rupee coins = 3x × 2 = 6x and the value of ₹1 rupee coins = (160 – 4x).
From the problem statement we get:
5x + 6x + (160 – 4x) = 300
or, 5x + 6x – 4x + 160 = 300
or, 7x + 160 = 300
or, 7x = 300 – 160 (Transposing 160 to RHS)
or, 7x = 140
or, x = 140/7 (Dividing both sides by 7)
or, x = 20
Therefore, 3x = 3 × 20 = 60 and (160 – 4x) = (160 – 4 × 20) = 160 – 80 = 80.
Summary: The number of coins of denomination ₹1 = 80, the number of coins of denomination ₹2 = 60 and the number of coins of denomination ₹5 = 20.

16. The organisers of an essay competition decide that a winner in the competition gets a prize of ₹100 and a participant who does not win gets a prize of ₹25. The total prize money distributed is ₹3,000. Find the number of winners, if the total number of participants is 63.
Answer:
Let number of winners be x. Therefore, the number of participant who do not win (63 – x).
Total money given to winners = 100x and total money given to participants who do not win = 25(63 – x).
From the problem statement:
100x + 25(63 – x) = 3000
or, 100x + 1575 – 25x = 3000
or, 75x + 1575 = 3000
or, 75x = 3000 – 1575 (Transposing 1575 to RHS)
or, 75x = 1425
or, x = 1425/75 (Dividing both sides by 75)
or, x = 19
Summary: The winner gets a prize of ₹100, a participant who does not win gets a prize of ₹25 and the total prize money distributed is ₹3,000. Therefore, the number of winners = 19.

Exercise 2.4 Page: 31 (Old Textbook):

1. Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Answer:
Let the number be x.
From the problem statement we get:
(x – 5/2) × 8 = 3x
or, 8x – (5 × 8)/2 = 3x
or, 8x – 20 = 3x
or, 8x – 3x = 20 (Transposing 3x to LHS and -20 to RHS)
or, 5x = 20
or, x = 20/5 (Dividing both sides by 4)
or, x = 4 (Answer)
Summary: Amina thinks of a number, subtracts 5/2 from it and multiplies the result by 8. The result now obtained is 3 times the same number she thought of. The required number = 4.

2. A positive number is 5 times another number. If 21 is added to both numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Answer:
Let the smaller number be x and hence the greater number = 5x.
If 21 is added to both numbers, the smaller number = (x + 21) and the greater number = (5x + 21)
From the problem statement we get:
5x + 21 = 2(x + 21)
or, 5x + 21 = 2x + 42
or, 5x – 2x = 42 – 21 (Transposing 2x to LHS and 21 to RHS)
or, 3x = 21
or, x = 21/3 (Dividing both sides by 3)
or, x = 7 (Answer)
Therefore, 5x = 35
Summary: A positive number is 5 times another number. If 21 is added to both numbers, then one of the new numbers becomes twice the other new number. The two numbers in question are 7 and 35.

3. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Answer:
Let us take the two-digit number such that the digit in the ten’s place is b.
Since the sum of the digits of a two-digit number is 9, the digit in the unit’s place is (9 – b).
Therefore, the original number is 10b + (9 – b).
After interchanging the digits the new number is 10(9 – b) + b.
The resulting new number is greater than the original number by 27.
From the problems statement we get:
10(9 – b) + b = 10b + (9 – b) + 27
or, 90 – 10b + b = 10b + 9 – b + 27
or, – 10b + b – 10b + b = 9 + 27 – 90 (Transposing the terms containing b to LHS and the numerical terms to RHS)
or, – 18b = 54
or, b = 54/(-18) (Dividing both sides by -18)
or, b = 3
Therefore, the two-digit number = 10b + (9 – b) = 10 × 3 + (9 – 3) = 30 + 6 = 36.
Summary: Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. The required original two-digit number = 36.

4. One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
Answer:
Let us take the two-digit number such that the digit in the ten’s place is b.
Therefore the digit in the unit’s place = 3b.
Therefore, the original number is 10b + 3b.
After interchanging the digits the new number is 30b + b.
From the given problem statement:
(10b + 3b) + (30b + b) = 88
or, 44b = 88
or, b = 88/44 (Dividing both sides by 44)
or, b = 2
Therefore, 10b + 3b = 10 × 2 + 3 × 2 = 26.
Summary: One of the two digits of a two digit number is three times the other digit. If we interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. The required original number = 26.

5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?
Answer:
Let Shobo’s present age be x. Hence, Shobo’s mother’s present age is 6x.
Shobo’s age five years from now = (x + 5).
From the problem statement we get:
(x + 5) = 1/3 × 6x
or, x + 5 = 2x
or, 2x = x + 5
or, 2x – x = 5 (Transposing x to LHS)
or, x = 5
Therefore, 6x = 30
Summary: The present age of Shobo is 5 years and the present age of his mother is 30 years.

6. There is a narrow rectangular plot reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate ₹100 per metre it will cost the village panchayat ₹75000 to fence the plot. What are the dimensions of the plot?
Answer:
Let the length of the plot be 11x and the breath of the plot be 4x.
The perimeter of the rectangular plot = 2(length + breadth) = 2(11x + 4x) = 30x.
The rate of fencing is ₹100 per metre.
From the problem statement we get:
30x × 100 = 75000
or, 3000x = 75000
or, x = 75000/3000
or, x = 25
Therefore, 11x = 11 × 25 = 275 m and 4x = 4 × 25 = 100 m.
Summary: The length of the plot = 275 m and the breadth of the plot = 100 m.

7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹50 per metre and trouser material that costs him ₹90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹36,600. How much trouser material did he buy?
Answer:
Let the total length of shirt material Hasan buys = 3x and the total length of trouser material he buys = 2x.
The cost of the total length of shirt material he buys = 3x × 50 = ₹150x.
The cost of the total length of trouser material he buys = 2x × 90 = ₹180x.
He sells the materials at 12% and 10% profit respectively.
The selling price of the total length of shirt material = 150x + 12/100 × 150x = 150x + 18x = ₹168x.
The selling price of the total length of trouser material = 180x + 10/100 × 180x = 180x + 18x = ₹198x.
His total sale is ₹36,600.
From the problem statement we get:
168x + 198x = 36600
or, 366x = 36600
or, x = 36600/366 (Dividing both sides by 366)
or, x = 100
Therefore, 2x = 200.
Summary: Based on the information given in the problem statement, the amount of trouser material that Hasan bought = 200 m.

8. Half of a herd of deer is grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Answer:
The total number of deer in the herd be x.
The number of deer grazing in the field = x/2.
Three fourths of the remaining are playing nearby.
The remaining number of deer = (x – x/2) = x/2.
Therefore, the number of deer playing nearby = 3/4 × x/2 = 3x/8.
The rest 9 are drinking water from the pond.
From the problem statement we get:
x – x/2 – 3x/8 = 9
or, 8x/8 – 4x/8 – 3x/8 = 9
or, (8x – 4x – 3x)/8 = 9
or, x/8 = 9
or, x = 9 × 8
or, x = 72
Summary: Half of a herd of deer is grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. The total number of deer is found to be = 72.

9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Answer:
Let the granddaughter’s age be x years.
Therefore, the grandfather’s age is 10x years.
From the problem statement:
10x – x = 54
or, 9x = 54
or, x = 54/9 (Dividing both sides by 9)
or, x = 6
Therefore, 10x = 60.
Summary: The present age of the granddaughter is 6 years and the present age of the grandfather is 60 years.

10. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Answer:
Let the son’s age be x. Then the age of Aman is 3x.
Ten years ago the son’s age = (x – 10) and Aman’s age = (3x – 10).
Ten years ago he was five times his son’s age.
From the given statement:
3x – 10 = 5(x – 10)
or, 3x – 10 = 5x – 50
or, 5x – 50 = 3x – 10
or, 5x – 3x = 50 – 10 (Transposing 3x to LHS and -50 to RHS)
or, 2x = 40
or, x = 40/2 (Dividing both sides by 2)
or, x = 20
Then 3x = 60.
Summary: Aman’s and his son’s present ages are 60 years and 20 years respectively.

Exercise 2.6 Page No: 35 (Old Textbook):

Solve the following equations.

1. (8x – 3)/3x = 2
Answer:
(8x – 3)/3x = 2
We put the above equation in the form of a linear equation by multiplying both sides of the equation by 3x,
8x – 3 = 2 × 3x
or, 8x – 6x = 3 (Transposing 6x to LHS)
or, 2x = 3
or x = 3/2 (Ans)
Check:
Numerator of LHS = 8 × 3/2 – 3 = 12 – 3 = 9
Denominator of LHS = 3 × 3/2 = 9/2
LHS = Numerator ÷ Denominator = 9 ÷ 9/2 = 2
Hence, LHS = RHS (Solution verified)
Summary: The solution of the equation (8x – 3)/3x = 2 is x = 3/2 .

2. 9x/( 7 – 6x) = 15
Answer:
9x/( 7 – 6x) = 15
We put the above equation in the form of a linear equation by multiplying both sides of the equation by (7 – 6x):
9x = 15(7 – 6x)
or, 9x = 105 – 90x
or, 9x + 90x = 105 (Transposing -90x to LHS)
or, 99x = 105
or, x = 105/99 (Dividing both sides by 99)
or x = 35/33 (Ans)
Check:
Numerator of LHS = 9 × 35/33 = 105/11
Denominator of LHS = 7 – 6x = 7 – 6 × 35/33 = 7 – 70/11 = 7/11
LHS = Numerator ÷ Denominator = 105/11 ÷ 7/11 = 105/11 × 11/7 = 105/7 = 15
Hence, LHS = RHS (Solution verified)
The solution of the equation 9x/( 7 – 6x) = 15 is x = 35/33 .

3. z/(z + 15) = 4/9
Answer:
z/(z + 15) = 4/9
We put the above equation in the form of a linear equation by cross multiplying:
9z = 4 × (z + 15)
or, 9z = 4z + 60
or, 9z – 4z = 60
or, 5z = 60
or z = 60/5
or, z = 12 (Answer)
Check:
Numerator of LHS = z = 12
Denominator of LHS = z + 15 = 12 + 15 = 27
LHS = Numerator ÷ Denominator = 12 ÷ 27 = 12/27 = 4/9
Hence, LHS = RHS (Solution verified)
The solution of the equation z/(z + 15) = 4/9 is z = 12.

4. (3y + 4)/(2 – 6y) = (-2)/5
Answer:
(3y + 4)/(2 – 6y) = (-2)/5
We put the above equation in the form of a linear equation by cross multiplying:
5(3y + 4) = -2(2 – 6y)
or, 15y + 20 = – 4 + 12y
or, 15y – 12y = -20 – 4
or, 3y = -24
or, y = (-24)/3
or, y = – 8 (Answer)
Check:
Numerator of LHS = 3y + 4 = 3 × (-8) + 4 = -20
Denominator of LHS = 2 – 6y = 2 – 6 × (-8) = 2 + 48 = 50
LHS = Numerator ÷ Denominator = -20 ÷ 50 = (-20)/50 = (-2)/5
Hence, LHS = RHS (Solution verified)
The solution of the equation (3y + 4)/(2 – 6y) = (-2)/5 is y = – 8.

5. (7y + 4)/(y + 2) = (-4)/3
Answer:
(7y + 4)/(y + 2) = (-4)/3
We put the above equation in the form of a linear equation by cross multiplying:
3(7y + 4) = -4(y + 2)
or, 21 y + 12 = -4y – 8
or, 21y + 4y = -12 – 8
or, 25y = – 20
or, y = (-20)/25
or, y = (-4)/5 (Answer)
Check:
Numerator of LHS = 7y + 4 = 7 × ((-4)/5) + 4 = (-28)/5 + 4 = (-28 + 20)/5 = (-8)/5
Denominator of LHS = y + 2 = (-4)/5 + 2 = (-4 + 10)/5 = 6/5
LHS = Numerator ÷ Denominator = (-8)/5 ÷ 6/5 = (-8)/6 = (-4)/3
Hence, LHS = RHS (Solution verified)
The solution of the equation (7y + 4)/(y + 2) = (-4)/3 is y = (-4)/5 .

6. The ages of Hari and Harry are in the ratio of 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.
Answer:
Let the ages of Hari be 5x and that of Harry be 7x.
Four years from now the age of Hari will be (5x + 4) and the age of Harry will be (7x + 4).
From the problem statement:
(5x + 4)/(7x + 4) = 3/4
We put the above equation in the form of a linear equation by cross multiplying:
4(5x + 4) = 3(7x + 4)
or, 20x + 16 = 21x + 12
or, 21x + 12 = 20x + 16
or, 21x – 20x = 16 – 12
or, x = 4
Then 5x = 20 and 7x = 28
Summary: The ages of Hari and Harry are in the ratio of 5:7 and four years from now the ratio of their ages will be 3:4. The present ages of Hari and Harry are 20 years and 28 years respectively.

7. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.
Answer:
Let the numerator of the rational number be x.
Then the denominator of the rational number is (x + 8).
If the numerator is increased by 17 the new numerator = x + 17.
If the denominator is decreased by 1 the new denominator = x + 8 – 1 = x + 7.
From the problem statement we get:
(x + 17)/(x + 7) = 3/2
We put the above equation in the form of a linear equation by cross multiplying:
2(x + 17) = 3(x + 7)
or, 2x + 34 = 3x + 21
or, 3x + 21 = 2x + 34
or, 3x – 2x = 34 – 21
or, x = 13
Then x + 8 = 21.
Rational number = Numerator/Denominator = 13/21
Summary: The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. The required rational number = 13/21.

Extra Questions to Complement Solutions to NCERT Class 8 Mathematics Chapter 2 Linear Equations in One Variable:

Very Short Answer Type Questions:

1. Name the operation to be performed to solve the following equation: 2x + 4 = x – 5.
Answer:
The operation to be performed is transposition.

2. How do you indicate that two expressions are equal?
Answer:
If two expressions are equal they form an equation. So, you put any one of the expressions on the LHS and the other one on the RHS and put an ‘=’ sign between them.

3. Is there any restriction on the number of terms a linear equation can have?
Answer:
No, there is no restriction on the number of terms a linear equation can have as long as the power of variable is 1.

4. What is the first operation you should do to solve the following equation:
x/3 – 3x/4 + 7x/12 = 10
Answer:
The first step is the reduce the equation to simpler form by multiplying by the LCM of the denominators 3, 4, 12 = 12.

5. What is the solution of the equation: x = x + 1?
Answer:
The given equation is:
x = x + 1
or, x – x = 1 (Transposing x to the LHS)
or, 0 = 1, which is not possible.
So, the given equation has no solution.

6. Is the following equation linear: a + b = (a – 1) + 5
Answer:
The given equation is not linear because there are two variables a and b in the equation.

7. Two times (a – 1) and five times (a + 1) are equal. Give the equation for the statement.
Answer:
The given equation can be written as: 2(a – 1) = 5(a + 1).

8. Half of a fraction of a number and the whole number are equal. Is this possible?
Answer:
The whole number is x. Therefore, half of a fraction of a number = x/2.
From the given statement: x = x/2 . When we divide both sides by x, we get 1 = 1/2 which is not possible.
Therefore, the given statement is not possible.

9. Is the equation of the form A + B + Cx linear if x is the variable?
Answer:
Yes, it is linear because there is only one variable x and the power of the variable is 1.

10. Is the equation of the form x/A + B = xC linear?
Answer:
Yes, it is linear because there is only one variable x and the power of the variable is 1.

Multiple Choice Questions (MCQ):

1. Which of the following equations is not a linear equation in one variable?

(a) a – 1 = 2a + 5
(b) ab + a = 2a – 3
(c) a/3 + 2a = (2 – a)/3 + 5
(d) 2(a – 1) = 5(2a + 1)

Answer: Correct answer: (b) ab + a = 2a – 3. The equation in (b) has two variables.

2. k1 and k2 are constants. What is the value of (k1 – k2) in the equation x – (x – 2) = x/3 + (k1 – k2) for x = 1?

(a) 2/3
(b) 5/3
(c) (-5)/3
(d) 1/3

Answer: Correct answer: (b) 5/3.
Putting the value of x = 1 in x – (x – 2) = x/3 + (k1 – k2) we get:
1 – (1 – 2) = 1/3 + (k1 – k2)
or, 1 + 1 = 1/3 + (k1 – k2)
or, 1/3 + (k1 – k2) = 2
or, (k1 – k2) = 2 – 1/3 (Transposing 1/3 to RHS)
or, (k1 – k2) = (6 – 1)/3
or, (k1 – k2) = 5/3

3. What can you say about the following equations?
Ax + B = – C and A/Bx + 1 = (-C)/B

(a) Different
(b) Both are non-linear
(c) Linear and Equivalent
(d) Linear only

Answer: Correct answer: (c) Linear and Equivalent.
A/Bx + 1 = (-C)/B
or, Ax + B = (-C)/B × B (Multiplying throughout by B)
or, Ax + B = – C
Therefore, the two equations are linear (x is the only variable) and equivalent.

4. Solve the following equation: 2(x – 6) = x/3 + 2.

(a) 42/5
(b) 0
(c) 32/5
(d) 19/5

Answer: Correct answer: (a) 42/5.
2(x – 6) = x/3 + 2
or, 2x – 12 = x/3 + 2
or, 2x – x/3 = 12 + 2 (Transposing)
or, (6x – x)/3 = 14
or, 5x/3 = 14
or, x = (3 × 14)/5
or, x = 42/5 (Ans)

5. The same linear equation is expressed in different ways. Depending on the way it is expressed, the solution will:

(a) Change
(b) Not change
(c) Depends on the specific way it is expressed
(d) None

Answer: Correct answer: (b) Not change.
A linear equation has the same solution regardless of how it is expressed.

Short and Long Answer Type Questions:

1. You have learnt to manipulate equations by multiplying throughout by a suitable number and also dividing throughout by a suitable number. Give a number which does not allow you to do any one of the above operations.
Answer:
0 is a number with which we cannot divide both sides of an equation. Any expression divided by 0 will give us an undefined result. Technically we can multiply by 0, although it would give us 0 on both sides of the equation. This is useless when it comes to solving an equation.

2. x = 0 is the solution of 2x + 3 = x + 3. True/False?
Answer:
The given equation is:
2x + 3 = x + 3
or, 2x – x = 3 – 3
or, x = 0
Hence, the statement is True.

3. Three times Tina’s age 2 years ago is equal to two times Tina’s age 1 year in the future. What is Tina’s current age?
Answer:
Let Tina’s current age be x.
Then Tina’s age 2 years ago is (x – 2) and Tina’s age 1 year in the future is (x + 1).
From the problem statement we get:
3(x – 2) = 2(x + 1)
or, 3x – 6 = 2x + 2
or, 3x – 2x = 6 + 2 (Transposing 2x to LHS and 6 to RHS)
or, x = 8
Therefore, Tina’s current age is 8 years.

4. If x = 1 in the following equation x/4 – 2 = x/8 + k where k is a constant, what is the value of k?
Answer: ?
Putting x = 1 in the equation we get x/4 – 2 = x/8 + k =
1/4 – 2 = 1/8 + k
or, 1/8 + k = 1/4 – 2
or, k = 1/4 – 1/8 – 2
or, k = 2/8 – 1/8 – 16/8
or, k = (2 – 1 – 16)/8
or, k = (-15)/8

5. Is x = 1/x a linear equation?
Answer:
Given equation: x = 1/x .
Cross- multiplying we get:
(x) × (x) = 1
or, x2 = 1
The power of x is 2, whereas a linear equation has to have power = 1. Therefore, the above equation is not a linear equation.

6. Solve the following equations:

(i) 6y – (3y – 8) = 8 – (-2 – y)

(ii) y – [y – (-y – 1)] = y + (1 – y)

Answers:

(i) 6y – (3y – 8) = 8 – (-2 – y)
6y – 3y + 8 = 8 + 2 + y (Opening brackets)
or, 3y + 8 = 10 + y
or, 3y – y = 10 – 8 (Transposing y to LHS and 8 to RHS)
or, 2y = 2
or, y = 1 (Answer)

(ii) y – [y – (-y – 1)] = y + (1 – y)
y – [y + y + 1] = y + 1 – y [Opening the inner bracket in LHS and the bracket in RHS]
or, y – y – y – 1 = 1 [Opening the last bracket]
or, -y = 1 + 1 [Transposing -1 to RHS]
or, -y = 2
or, y = -2 (Multiplying both sides by -1) (Answer)

7. Solve the following equations:

(i) (x – 1)/2 + x/3 = (x + 2)/6 + x/2

(ii) m + (m – 1)/2 = (m + 2) + m/3

Answers:

(i) (x – 1)/2 + x/3 = (x + 2)/6 + x/2
or, 3(x – 1) + 2x = x + 2 + 3x (Multiplying throughout by 6 which is the LCM of 2, 3, 6)
or, 3x – 3 + 2x = 4x + 2
or, 5x – 3 = 4x + 2
or, 5x – 4x = 3 + 2 (Transposing 4x to LHS and -3 to RHS)
or, x = 5 (Answer)

(ii) m + (m – 1)/2 = (m + 2) + m/3
or, 6m + 3(m – 1) = 6(m + 2) + 2m (Multiplying throughout by 6 which is the LCM of 2, 3)
or, 6m + 3m – 3 = 6m + 12 + 2m
or, 9m – 3 = 8m + 12
or, 9m – 8m = 3 + 12 (Transposing 8m to LHS and -3 to RHS)
or, m = 15 (Answer)

8. Let n be a positive integer n. What can you say about the following equation:
n + (n + 1) + (n + 2) = 3
Answer:
Given: n + (n + 1) + (n + 2) = 3
or, 3n + 3 = 3
or, 3n = 3 – 3
or, 3n = 0
or, n = 0/3
or, n = 0
We are given that n is a positive integer. So, n cannot be 0 and the above equation is not possible.

9.Solve the following equations:

(i) x – x/3 + x/6 – x/9 = 1

(ii) (2x – 1) + (1 – x/3) = 1

Answers:

(i) x – x/3 + x/6 – x/9 = 1
or, 18x – x/3 × 18 + x/6 × 18 – x/9 × 18 = 18 (Multiplying throughout by 18 which is the LCM of 3, 6, 918x )
or, 18x – 6x + 3x – 2x = 18
or, 13x = 18
or, x = 18/13 (Answer)

(ii) (2x – 1) + (1 – x/3) = 1
or, 2x – x/3 – 1 + 1 = 1
or, (6x – x)/3 = 1
or, 5x/3 = 1
or, x = 3/5 (Answer)

10. Solve the following equations:

(i) 0.3 (x – 5) = 0.9 (x + 2)

(ii) (x – 5)/(0.5) = (x + 5)/(0.9)

Answers:

(i) 0.3 (x – 5) = 0.9 (x + 2)
or, 3/10 (x – 5) = 9/10 (x + 2)
or, 3(x – 5) = 9(x + 2) (Multiplying throughout by 10)
or, 3x – 15 = 9x + 18
or, 9x + 18 = 3x – 15
or, 9x – 3x = -15 – 18
or, 6x = -33
or, x = (-33)/6
or, x = (-11)/2 (Answer)

(ii) (x – 5)/(0.5) = (x + 5)/(0.9)
or, (x – 5)/0.5 = (x + 5)/0.9
or, (10(x – 5))/5 = (10(x + 5))/9
or, 90(x – 5) = 50(x + 5) (Multiplying throughout by the LCM of 5, 9 = 45)
or, 90x – 450 = 50x + 250
or, 90x – 50x = 450 + 250 (Transposing)
or, 40x = 700
or, x = 700/40
or, x = 35/2 (Answer)

Fill in the Blanks:

(a) The highest power of an equation is 2. The equation is not _______.
(b) The equation: n – 1 = n + 2 is ______.
(c) If x = 5, the value of k in the equation: x/2 = 5/k is ______.
(d) The numerical terms in a linear equation are called _____.
(e) Solving linear equations of different types can involve _____ operations.

Answers:

(a) The highest power of an equation is 2. The equation is not linear.
(b) The equation: n – 1 = n + 2 is not possible.
(c) If x = 5, the value of k in the equation: x/2 = 5/k is 2.
(d) The numerical terms in a linear equation are called constants.
(e) Solving linear equations of different types can involve multiple operations.

++++++++++++++

Frequently Asked Questions (FAQs) on NCERT Solutions to Class 8 Mathematics Chapter 2 Linear Equations in One Variable: