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Page 1 Exercise 8.2 Page No: 8.12 Solve each of the following equations and check your answers: 1. x – 3 = 5 Solution: Given x – 3 = 5 Adding 3 to both sides we get, x – 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS, we get LHS = x – 3 and RHS = 5 LHS = 8 – 3 = 5 and RHS = 5 LHS = RHS Hence, verified. 2. x + 9 = 13 Solution: Given x + 9 = 13 Subtracting 9 from both sides i.e. LHS and RHS, we get x + 9 – 9 = 13 – 9 x = 4 Verification: Substituting x = 4 on LHS, we get LHS = 4 + 9 = 13 = RHS LHS = RHS Hence, verified. 3. x – (3/5) = (7/5) Solution: Given x – (3/5) = (7/5) Add (3/5) to both sides, we get x – (3/5) + (3/5) = (7/5) + (3/5) x = (7/5) + (3/5) Page 2 Exercise 8.2 Page No: 8.12 Solve each of the following equations and check your answers: 1. x – 3 = 5 Solution: Given x – 3 = 5 Adding 3 to both sides we get, x – 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS, we get LHS = x – 3 and RHS = 5 LHS = 8 – 3 = 5 and RHS = 5 LHS = RHS Hence, verified. 2. x + 9 = 13 Solution: Given x + 9 = 13 Subtracting 9 from both sides i.e. LHS and RHS, we get x + 9 – 9 = 13 – 9 x = 4 Verification: Substituting x = 4 on LHS, we get LHS = 4 + 9 = 13 = RHS LHS = RHS Hence, verified. 3. x – (3/5) = (7/5) Solution: Given x – (3/5) = (7/5) Add (3/5) to both sides, we get x – (3/5) + (3/5) = (7/5) + (3/5) x = (7/5) + (3/5) x = (10/5) x = 2 Verification: Substitute x = 2 in LHS of given equation, then we get 2 – (3/5) = (7/5) (10 – 3)/5 = (7/5) (7/5) = (7/5) LHS = RHS Hence, verified 4. 3x = 0 Solution: Given 3x = 0 On dividing both sides by 3 we get, (3x/3) = (0/3) x = 0 Verification: Substituting x = 0 in LHS we get 3 (0) = 0 And RHS = 0 Therefore LHS = RHS Hence, verified. 5. (x/2) = 0 Solution: Given x/2 = 0 Multiplying both sides by 2, we get (x/2) × 2 = 0 × 2 x = 0 Verification: Substituting x = 0 in LHS, we get LHS = 0/2 = 0 and RHS = 0 LHS = 0 and RHS = 0 Therefore LHS = RHS Hence, verified. Page 3 Exercise 8.2 Page No: 8.12 Solve each of the following equations and check your answers: 1. x – 3 = 5 Solution: Given x – 3 = 5 Adding 3 to both sides we get, x – 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS, we get LHS = x – 3 and RHS = 5 LHS = 8 – 3 = 5 and RHS = 5 LHS = RHS Hence, verified. 2. x + 9 = 13 Solution: Given x + 9 = 13 Subtracting 9 from both sides i.e. LHS and RHS, we get x + 9 – 9 = 13 – 9 x = 4 Verification: Substituting x = 4 on LHS, we get LHS = 4 + 9 = 13 = RHS LHS = RHS Hence, verified. 3. x – (3/5) = (7/5) Solution: Given x – (3/5) = (7/5) Add (3/5) to both sides, we get x – (3/5) + (3/5) = (7/5) + (3/5) x = (7/5) + (3/5) x = (10/5) x = 2 Verification: Substitute x = 2 in LHS of given equation, then we get 2 – (3/5) = (7/5) (10 – 3)/5 = (7/5) (7/5) = (7/5) LHS = RHS Hence, verified 4. 3x = 0 Solution: Given 3x = 0 On dividing both sides by 3 we get, (3x/3) = (0/3) x = 0 Verification: Substituting x = 0 in LHS we get 3 (0) = 0 And RHS = 0 Therefore LHS = RHS Hence, verified. 5. (x/2) = 0 Solution: Given x/2 = 0 Multiplying both sides by 2, we get (x/2) × 2 = 0 × 2 x = 0 Verification: Substituting x = 0 in LHS, we get LHS = 0/2 = 0 and RHS = 0 LHS = 0 and RHS = 0 Therefore LHS = RHS Hence, verified. 6. x – (1/3) = (2/3) Solution: Given x – (1/3) = (2/3) Adding (1/3) to both sides, we get x – (1/3) + (1/3) = (2/3) + (1/3) x = (2 + 1)/3 x = (3/3) x =1 Verification: Substituting x = 1 in LHS, we get 1 – (1/3) = (2/3) (3 – 1)/3 = (2/3) (2/3) = (2/3) Therefore LHS = RHS Hence, verified. 7. x + (1/2) = (7/2) Solution: Given x + (1/2) = (7/2) Subtracting (1/2) from both sides, we get x + (1/2) – (1/2) = (7/2) – (1/2) x = (7 – 1)/2 x = (6/2) x = 3 Verification: Substituting x = 3 in LHS we get 3 + (1/2) = (7/2) (6 + 1)/2 = (7/2) (7/2) = (7/2) Therefore LHS = RHS Hence, verified. 8. 10 – y = 6 Solution: Page 4 Exercise 8.2 Page No: 8.12 Solve each of the following equations and check your answers: 1. x – 3 = 5 Solution: Given x – 3 = 5 Adding 3 to both sides we get, x – 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS, we get LHS = x – 3 and RHS = 5 LHS = 8 – 3 = 5 and RHS = 5 LHS = RHS Hence, verified. 2. x + 9 = 13 Solution: Given x + 9 = 13 Subtracting 9 from both sides i.e. LHS and RHS, we get x + 9 – 9 = 13 – 9 x = 4 Verification: Substituting x = 4 on LHS, we get LHS = 4 + 9 = 13 = RHS LHS = RHS Hence, verified. 3. x – (3/5) = (7/5) Solution: Given x – (3/5) = (7/5) Add (3/5) to both sides, we get x – (3/5) + (3/5) = (7/5) + (3/5) x = (7/5) + (3/5) x = (10/5) x = 2 Verification: Substitute x = 2 in LHS of given equation, then we get 2 – (3/5) = (7/5) (10 – 3)/5 = (7/5) (7/5) = (7/5) LHS = RHS Hence, verified 4. 3x = 0 Solution: Given 3x = 0 On dividing both sides by 3 we get, (3x/3) = (0/3) x = 0 Verification: Substituting x = 0 in LHS we get 3 (0) = 0 And RHS = 0 Therefore LHS = RHS Hence, verified. 5. (x/2) = 0 Solution: Given x/2 = 0 Multiplying both sides by 2, we get (x/2) × 2 = 0 × 2 x = 0 Verification: Substituting x = 0 in LHS, we get LHS = 0/2 = 0 and RHS = 0 LHS = 0 and RHS = 0 Therefore LHS = RHS Hence, verified. 6. x – (1/3) = (2/3) Solution: Given x – (1/3) = (2/3) Adding (1/3) to both sides, we get x – (1/3) + (1/3) = (2/3) + (1/3) x = (2 + 1)/3 x = (3/3) x =1 Verification: Substituting x = 1 in LHS, we get 1 – (1/3) = (2/3) (3 – 1)/3 = (2/3) (2/3) = (2/3) Therefore LHS = RHS Hence, verified. 7. x + (1/2) = (7/2) Solution: Given x + (1/2) = (7/2) Subtracting (1/2) from both sides, we get x + (1/2) – (1/2) = (7/2) – (1/2) x = (7 – 1)/2 x = (6/2) x = 3 Verification: Substituting x = 3 in LHS we get 3 + (1/2) = (7/2) (6 + 1)/2 = (7/2) (7/2) = (7/2) Therefore LHS = RHS Hence, verified. 8. 10 – y = 6 Solution: Given 10 – y = 6 Subtracting 10 from both sides, we get 10 – y – 10 = 6 – 10 -y = -4 Multiplying both sides by -1, we get -y × -1 = - 4 × - 1 y = 4 Verification: Substituting y = 4 in LHS, we get 10 – y = 10 – 4 = 6 and RHS = 6 Therefore LHS = RHS Hence, verified. 9. 7 + 4y = -5 Solution: Given 7 + 4y = -5 Subtracting 7 from both sides, we get 7 + 4y – 7 = -5 -7 4y = -12 Dividing both sides by 4, we get y = -12/ 4 y = -3 Verification: Substituting y = -3 in LHS, we get 7 + 4y = 7 + 4(-3) = 7 – 12 = -5, and RHS = -5 Therefore LHS = RHS Hence, verified. 10. (4/5) – x = (3/5) Solution: Given (4/5) – x = (3/5) Subtracting (4/5) from both sides, we get (4/5) – x – (4/5) = (3/5) – (4/5) - x = (3 -4)/5 - x = (-1/5) Page 5 Exercise 8.2 Page No: 8.12 Solve each of the following equations and check your answers: 1. x – 3 = 5 Solution: Given x – 3 = 5 Adding 3 to both sides we get, x – 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS, we get LHS = x – 3 and RHS = 5 LHS = 8 – 3 = 5 and RHS = 5 LHS = RHS Hence, verified. 2. x + 9 = 13 Solution: Given x + 9 = 13 Subtracting 9 from both sides i.e. LHS and RHS, we get x + 9 – 9 = 13 – 9 x = 4 Verification: Substituting x = 4 on LHS, we get LHS = 4 + 9 = 13 = RHS LHS = RHS Hence, verified. 3. x – (3/5) = (7/5) Solution: Given x – (3/5) = (7/5) Add (3/5) to both sides, we get x – (3/5) + (3/5) = (7/5) + (3/5) x = (7/5) + (3/5) x = (10/5) x = 2 Verification: Substitute x = 2 in LHS of given equation, then we get 2 – (3/5) = (7/5) (10 – 3)/5 = (7/5) (7/5) = (7/5) LHS = RHS Hence, verified 4. 3x = 0 Solution: Given 3x = 0 On dividing both sides by 3 we get, (3x/3) = (0/3) x = 0 Verification: Substituting x = 0 in LHS we get 3 (0) = 0 And RHS = 0 Therefore LHS = RHS Hence, verified. 5. (x/2) = 0 Solution: Given x/2 = 0 Multiplying both sides by 2, we get (x/2) × 2 = 0 × 2 x = 0 Verification: Substituting x = 0 in LHS, we get LHS = 0/2 = 0 and RHS = 0 LHS = 0 and RHS = 0 Therefore LHS = RHS Hence, verified. 6. x – (1/3) = (2/3) Solution: Given x – (1/3) = (2/3) Adding (1/3) to both sides, we get x – (1/3) + (1/3) = (2/3) + (1/3) x = (2 + 1)/3 x = (3/3) x =1 Verification: Substituting x = 1 in LHS, we get 1 – (1/3) = (2/3) (3 – 1)/3 = (2/3) (2/3) = (2/3) Therefore LHS = RHS Hence, verified. 7. x + (1/2) = (7/2) Solution: Given x + (1/2) = (7/2) Subtracting (1/2) from both sides, we get x + (1/2) – (1/2) = (7/2) – (1/2) x = (7 – 1)/2 x = (6/2) x = 3 Verification: Substituting x = 3 in LHS we get 3 + (1/2) = (7/2) (6 + 1)/2 = (7/2) (7/2) = (7/2) Therefore LHS = RHS Hence, verified. 8. 10 – y = 6 Solution: Given 10 – y = 6 Subtracting 10 from both sides, we get 10 – y – 10 = 6 – 10 -y = -4 Multiplying both sides by -1, we get -y × -1 = - 4 × - 1 y = 4 Verification: Substituting y = 4 in LHS, we get 10 – y = 10 – 4 = 6 and RHS = 6 Therefore LHS = RHS Hence, verified. 9. 7 + 4y = -5 Solution: Given 7 + 4y = -5 Subtracting 7 from both sides, we get 7 + 4y – 7 = -5 -7 4y = -12 Dividing both sides by 4, we get y = -12/ 4 y = -3 Verification: Substituting y = -3 in LHS, we get 7 + 4y = 7 + 4(-3) = 7 – 12 = -5, and RHS = -5 Therefore LHS = RHS Hence, verified. 10. (4/5) – x = (3/5) Solution: Given (4/5) – x = (3/5) Subtracting (4/5) from both sides, we get (4/5) – x – (4/5) = (3/5) – (4/5) - x = (3 -4)/5 - x = (-1/5) x = (1/5) Verification: Substituting x = (1/5) in LHS we get (4/5) – (1/5) = (3/5) (4 -1)/5 = (3/5) (3/5) = (3/5) Therefore LHS =RHS Hence, verified. 11. 2y – (1/2) = (-1/3) Solution: Given 2y – (1/2) = (-1/3) Adding (1/2) from both the sides, we get 2y – (1/2) + (1/2) = (-1/3) + (1/2) 2y = (-1/3) + (1/2) 2y = (-2 + 3)/6 [LCM of 3 and 2 is 6] 2y = (1/6) Now divide both the side by 2, we get y = (1/12) Verification: Substituting y = (1/12) in LHS we get 2 (1/12) – (1/2) = (-1/3) (1/6) – (1/2) = (-1/3) (2 – 6)/12 = (-1/3) [LCM of 6 and 2 is 12] (-4/12) = (-1/3) (-1/3) = (-1/3) Therefore LHS = RHS Hence, verified. 12. 14 = (7x/10) – 8 Solution: Given 14 = (7x/10) – 8 Adding 8 to both sides we get, 14 + 8 = (7x/10) – 8 + 8 22 = (7x/10)Read More
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