1 / 59

Solve Equations and Functions Math Problems Easily | Helpful Step-by-Step Guide

Learn how to solve various equations and functions step-by-step with detailed explanations. Solve linear equations, find variable values, and understand functions easily.

alafave
Download Presentation

Solve Equations and Functions Math Problems Easily | Helpful Step-by-Step Guide

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 1. Solve the following equations: (i) 2x − 5 = 11 2x – 5 = 11 (Add 5 to both sides) 2x = 16 (Divide both sides by 2) x = 8 (ii) 4y − 3 = 21 4y – 3 = 21 (Add 3 to both sides) 4y = 24 (Divide both sides by 4) y = 6

  2. 1. Solve the following equations: (iii) 25 = 13 + 6p 25 = 13 + 6p (Subtract 13 from both sides) 12 = 6p (Divide both sides by 6) 2 = p (iv) 5a + 7 = 9a − 13 5a + 7 = 9a – 13 (Add 13 to both sides) 5a + 20 = 9a (Subtract 5a from both sides) 20 = 4a (Divide both sides by 4) 5 = a

  3. 1. Solve the following equations: (v) 6k − 1 = 3k + 14 6k – 1 = 3k + 14 (Add 1 to both sides) 6k = 3k + 15 (Subtract 3k from both sides) 3k = 15 (Divide both sides by 3) k = 5

  4. 1. Solve the following equations: (vi) 2t + 1 = 8t − 35 2t + 1 = 8t – 35 (Add 35 to both sides) 2t + 36 = 8t (Subtract 2t from both sides) 36 = 6t (Divide both sides by 6) 6 = t

  5. 1. Solve the following equations: (vii) 12p − 21 = 7p + 9 12p – 21 = 7p + 9 (Add 21 to both sides) 12p = 7p + 30 (Subtract 7p from both sides) 5p = 30 (Divide both sides by 5) p = 6

  6. 1. Solve the following equations: (viii) b + 17 = 32 − 9b b + 17 = 32 – 9b (Subtract 17 from both sides) b = – 9b + 15 (Add 9b to both sides) 10b = 15 (Divide both sides by 10) b = 1·5

  7. 1. Solve the following equations: (ix) 5x − 1 − 2x − 4 = x 5x – 1 – 2x – 4 = x (Collect terms) 3x – 5 = x (Add 5 to both sides) 3x = x + 5 (Subtract x from both sides) 2x = 5 (Divide both sides by 2) x = 2·5

  8. 2. f (x) = 2x + 7. Find: (i) the value of x when f (x) = 9 f (x) = 2x + 7 f (x) = 9 9 = 2x + 7 (Subtract 7 from both sides) 2 = 2x (Divide both sides by 2) 1 = x

  9. 2. f (x) = 2x + 7. Find: (ii) the value of x when f (x) = 1. f (x) = 1 1 = 2x + 7 (Subtract 7 from both sides) – 6 = 2x (Divide both sides by 2) – 3 = x

  10. 3. Solve the following, givenf (x) = 15x − 1 and g(x) = 9x − 19. (i) By letting f (x) = g(x), form an equation in terms of x. f (x) = 15x – 1 g(x) = 9x – 19 f (x) = g(x) 15x – 1 = 9x – 19 (Add 1 to both sides) 15x = 9x – 18 (Subtract 9x from both sides) 6x = – 18

  11. 3. Solve the following, givenf (x) = 15x − 1 and g(x) = 9x − 19. (ii) Hence, solve this equation to find the value of x, for which f (x) = g(x). 6x = – 18 (Divide both sides by 6) x = – 3

  12. 3. Solve the following, givenf (x) = 15x − 1 and g(x) = 9x − 19. (iii) If f (x) and g(x) were graphed using the same axes and scale, what would be the relationship between their graphs and the answer to part (ii)? Both graphs would be straight lines and they would intersect at x = – 3.

  13. 4. Given h(x) = 16 − 2xand k(x) = 4x + 4, find: (i) the value of x, for which h(x) = k(x) h(x) = 16 – 2xk(x) = 4x + 4 h(x)= k(x) 16 – 2x = 4x + 4 (Subtract 4 from both sides) 12 – 2x = 4x (Add 2x to both sides) 12 = 6x (Divide both sides by 6) 2 = x

  14. 4. Given h(x) = 16 − 2xand k(x) = 4x + 4, find: (ii) the value of x, for which h(x) + k(x) = 8. h(x) + k(x) = 8 (16 – 2x) + (4x + 4) = 8 (collect terms) 16 – 2x + 4x + 4 = 8 (remove brackets) 20 + 2x = 8 (Subtract 20 from both sides) 2x = – 12 (Divide both sides by 2) x = – 6

  15. 5. Your friend thinks of a number, multiplies it by five, then add 12. The result is 52. What is the number? Let the number be x 5x + 12 = 52 (Subtract 12 from both sides) 5x = 40 (Divide both sides by 5) x = 8

  16. 6. A number is trebled and added to 11. The result is 50. What is the number? Let the number be x 3x + 11 = 50 (Subtract 11 from both sides) 3x = 39 (Divide both sides by 3) x = 13

  17. 7. The three angles of a triangle are θ, θ– 25 and θ+ 85. Find θ. The three angles of a triangle sum to 180˚  + ( – 25) + ( + 85) = 180 (collect terms) 3 + 60 = 180 (Subtract 60 from both sides) 3 = 120 (Divide both sides by 3)  = 40

  18. 8. Solve the following equations: (i) 3(x − 1) = 2x 3(x – 1) = 2x (expand) 3x – 3 = 2x (Add 3 to both sides) 3x = 2x + 3 (Subtract 2x from both sides) x = 3

  19. 8. Solve the following equations: (ii) 4(a + 1) = 16 − 2a 4(a + 1) = 16 – 2a (expand) 4a + 4 = 16 – 2a (Subtract 4 from both sides) 4a = 12 – 2a (Add 2a to both sides) 6a = 12 (Divide both sides by 6) a = 2

  20. 8. Solve the following equations: (iii) 3(4p − 7) = 7p + 9 3(4p – 7) = 7p + 9 (expand) 12p – 21 = 7p + 9 (Add 21to both sides) 12p = 7p + 30 (Subtract 7p from both sides) 5p = 30 (Divide both sides by 5) p = 6

  21. 8. Solve the following equations: (iv) 4(x − 1) = 2x + 3 4(x – 1) = 2x + 3 (expand) 4x – 4 = 2x + 3 (Add 4to both sides) 4x = 2x + 7 (Subtract 2x from both sides) 2x = 7 (Divide both sides by 2)

  22. 8. Solve the following equations: (v) 2(k + 7) − 5(k − 1) = 13 2(k + 7) – 5(k – 1) = 13 (expand) 2k + 14 – 5k + 5 = 13 (collect like terms) –3k + 19 = 13 (Add 3k to both sides) 19 = 13 + 3k (Subtract 13 from both sides) 6 = 3k (Divide both sides by 3) 2 = k

  23. 8. Solve the following equations: (vi) 2(3t + 7) = 7(4t − 1) − 1 2(3t + 7) = 7(4t – 1) – 1 (expand) 6t + 14 = 28t – 7 – 1 (collect like terms) 6t + 14 = 28t – 8 (Add 8to both sides) 6t + 22 = 28t (Subtract 6t from both sides) 22 = 22t (Divide both sides by 22) 1 = t

  24. 8. Solve the following equations: (vii) 5(7x + 3) = 2(3 − x) − 3 + x 5(7x + 3) = 2(3 – x) – 3 + x (expand) 35x + 15 = 6 – 2x – 3 + x (collect like terms) 35x + 15 = 3 – x (Subtract 15 from both sides) 35x = – 12 – x (Add x to both sides) 36x = – 12 (Divide both sides by 36)

  25. 8. Solve the following equations: (viii) 11 = 7(y + 1) − 2(3 − 8y) − 3y 11 = 7(y + 1) – 2(3 – 8y) – 3y (expand) 11 = 7y + 7 – 6 + 16y – 3y (collect like terms) 11 = 20y + 1 (Subtract 1 from both sides) 10 = 20y (Divide both sides by 20)

  26. 9. Find the value of x, which satisfies the equation 3(4x + 2) − x = −2x − 7 and verify your solution. 3(4x + 2) – x = – 2x – 7 (expand) 12x + 6 – x = – 2x – 7 (collect) 11x + 6 = – 2x – 7 (Subtract 6 from both sides) 11x = – 2x – 13 (Add 2x to both sides) 13x = – 13 (Divide both sides by 13) x = – 1

  27. 9. Find the value of x, which satisfies the equation 3(4x + 2) − x = −2x − 7 and verify your solution. Verify the solution of x = – 1: 3(4x + 2) – x = – 2x – 7 3(4(–1) + 2) – (–1) = – 2(–1) – 7 3(– 4 + 2) + 1 = 2 – 7 3(–2) + 1 = –5 – 6 + 1 = –5 –5 = –5

  28. 10. Find the value of y, which satisfies the equation 7 + 4y = 3(y − 5) and verify your solution. 7 + 4y = 3(y – 5) (expand) 7 + 4y = 3y – 15 (Subtract 7 from both sides) 4y = 3y – 22 (Subtract 3y from both sides) y = – 22 Verify the solution of y = – 22: 7 + 4y = 3(y – 5) 7 + 4(– 22) = 3(– 22 – 5) 7 – 88 = 3(– 27) – 81 = – 81

  29. 11. Find the value of k, which satisfies the equation 9(3 − k) + 5 + 4k = − k and verify your solution. 9(3 – k) + 5 + 4k = – k (expand) 27 – 9k + 5 + 4k = – k (collect) 32 – 5k = – k (Add 5k toboth sides) 32 = 4k (Divide both sides by 4) 8 = k

  30. 11. Find the value of k, which satisfies the equation 9(3 − k) + 5 + 4k = − k and verify your solution. Verify the solution of k = 8: 9(3 – k) + 5 + 4k = – k 9(3 – 8) + 5 + 4(8) = – 8 9(– 5) + 5 + 32 = – 8 – 45 + 5 + 32 = – 8 – 8 = – 8

  31. 12. When 7 times a certain number is subtracted from 15, the result is − 6. Find the number. Let the number be x: 15 – 7x = – 6 (Subtract 15 from both sides) – 7x = – 21 (Divide both sides by –7) x = 3

  32. 13. Solve the following equations: (i)

  33. 13. Solve the following equations: (ii)

  34. 13. Solve the following equations: (iii)

  35. 13. Solve the following equations: (iv)

  36. 13. Solve the following equations: (v)

  37. 13. Solve the following equations: (vi)

  38. 14. Find the value of x,which satisfies the equation and verify your solution.

  39. 14. Find the value of x,which satisfies the equation and verify your solution. Verify:

  40. 15. Find the value of p,which satisfies the equation and verify your solution. 2(3p – 2) – 1(3) = 5(p – 1) (Expand) 6p – 4 – 3 = 5p – 5 6p – 7 = 5p – 5 (Add 7 to both sides) 6p = 5p + 2 (Subtract 5p from both sides) p = 2

  41. 15. Find the value of p,which satisfies the equation and verify your solution. Verify: Therefore p = 2 Satisfies the equation.

  42. 16. The perimeter of the rectangle shown is 18 cm.Write an equation for the perimeter of the rectangle and solve it to find the length of all the sides. Perimeter = 2(length) + 2(width) 18 = 2(x + 3) + 2(2x) (expand) 18 = 2x + 6 + 4x (collect like terms) 18 = 6x + 6 (Subtract 6 from both sides) 12 = 6x (Divide both sides by 6) 2 cm = x Length = x + 3 Width = 2x = 2 + 3 = 2(2) = 5 cm = 4 cm

  43. 17. A wedding band charges €250 plus €180 per hour that they play. The cost of the band can be represented by the equation C = 180h + 250, where Cis the total cost and his the time played, in hours. Find how many hours the band played, if the total cost is €1,060. C = 180h + 250 1060 = 180h + 250 (Subtract 250 from both sides) 810 = 180h (Divide both sides by 180) 4·5 hours = 4 hours 30 minutes

  44. 18. An airline allows each person to carry 15 kg of baggage. Extra baggage is charged at €7·50 per kilo. The airline uses the following formula to calculate the cost, C, for extra luggage: C = 7·5(w − 15). (i) Write down what you think w stands for. w = weight of the baggage in total.

  45. 18. An airline allows each person to carry 15 kg of baggage. Extra baggage is charged at €7·50 per kilo. The airline uses the following formula to calculate the cost, C, for extra luggage: C = 7·5(w − 15). (ii) Josh has a suitcase, which weighs 17·5 kg. How much will Josh have to pay for his excess baggage? C = 75(w – 15) = 75(175 – 15) = 75(25) = €1875

  46. 18. An airline allows each person to carry 15 kg of baggage. Extra baggage is charged at €7·50 per kilo. The airline uses the following formula to calculate the cost, C, for extra luggage: C = 7·5(w − 15). (iii) Leo is charged €82·50 for extra baggage. How much extra baggage did he have? C = 75(w – 15) 82·50 = 75(w – 15) (Divide both sides by 75) 11 = w – 15 (Add 15 to both sides) 26 kg = w Total weight of baggage = 26 kg Extra baggage weight = 11 kg

  47. 18. An airline allows each person to carry 15 kg of baggage. Extra baggage is charged at €7·50 per kilo. The airline uses the following formula to calculate the cost, C, for extra luggage: C = 7·5(w − 15). (iv) Donna is charged €30 for extra baggage. Find the total weight of Donna’s luggage. C = 75(w – 15) 30 = 75(w – 15) (Divide both sides by 75) 4 = w – 15 (Add 15 to both sides) 19 kg = w Total weight of baggage = 19 kg.

  48. 19. Ida is an estate agent. She charges a flat fee of €800 plus 2% of the sales price of the home. The total commission that Ida earns on the sale of a home can be found by using the following function: C = 800 + 0·02P. Where Cis the commission earned, in euro, Pis the sale price of the home, in euro. (i) Find the commission that Ida would receive on a home that sold for €325,000. C = 800 + 0∙02P = 800 + 002(325000) = 800 + 6500 = €7300

  49. 19. Ida is an estate agent. She charges a flat fee of €800 plus 2% of the sales price of the home. The total commission that Ida earns on the sale of a home can be found by using the following function: C = 800 + 0·02P. Where Cis the commission earned, in euro, Pis the sale price of the home, in euro. (ii) If the total commission paid was €4,100, what was the sale price of the home? C = 800 + 002P 4100 = 800 + 002P (Subtract 800 from both sides) 3300 = 002P (Divide both sides by 0·02) €165,000 = P

More Related