2 2linear equations i

1,092 views
968 views

Published on

0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
1,092
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
Downloads
0
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

2 2linear equations i

  1. 1. Linear Equations I
  2. 2. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left.
  3. 3. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left.
  4. 4. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning?
  5. 5. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14.
  6. 6. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18.
  7. 7. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18. In symbols x – 4 =14 add 4 to both sides
  8. 8. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18. In symbols x – 4 =14 add 4 to both sides +4 +4
  9. 9. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18. In symbols x – 4 =14 add 4 to both sides +4 +4 so x = 18
  10. 10. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18. In symbols x – 4 =14 add 4 to both sides +4 +4 so x = 18 We bought 5 apples for $20. How much was each apple?
  11. 11. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18. In symbols x – 4 =14 add 4 to both sides +4 +4 so x = 18 We bought 5 apples for $20. How much was each apple? Again, let x be the price of an apple, then 5 apples cost $5x.
  12. 12. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18. In symbols x – 4 =14 add 4 to both sides +4 +4 so x = 18 We bought 5 apples for $20. How much was each apple? Again, let x be the price of an apple, then 5 apples cost $5x. Since the total cost was 20, by dividing 20/5 = 4 and we get that each apple was $4.
  13. 13. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18. In symbols x – 4 =14 add 4 to both sides +4 +4 so x = 18 We bought 5 apples for $20. How much was each apple? Again, let x be the price of an apple, then 5 apples cost $5x. Since the total cost was 20, by dividing 20/5 = 4 and we get that each apple was $4. In symbols, 5x = 20
  14. 14. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18. In symbols x – 4 =14 add 4 to both sides +4 +4 so x = 18 We bought 5 apples for $20. How much was each apple? Again, let x be the price of an apple, then 5 apples cost $5x. Since the total cost was 20, by dividing 20/5 = 4 and we get that each apple was $4. In symbols, 5x = 20 both sides divided by 5
  15. 15. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18. In symbols x – 4 =14 add 4 to both sides +4 +4 so x = 18 We bought 5 apples for $20. How much was each apple? Again, let x be the price of an apple, then 5 apples cost $5x. Since the total cost was 20, by dividing 20/5 = 4 and we get that each apple was $4. In symbols, 5x = 20 both sides divided by 5 5x = 20 so 5 5
  16. 16. Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14. Calculate backwards by adding 4 to14, we get 14 + 4 = 18. In symbols x – 4 =14 add 4 to both sides +4 +4 so x = 18 We bought 5 apples for $20. How much was each apple? Again, let x be the price of an apple, then 5 apples cost $5x. Since the total cost was 20, by dividing 20/5 = 4 and we get that each apple was $4. In symbols, 5x = 20 both sides divided by 5 5x = 20 so 5 5 so x=4
  17. 17. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome.
  18. 18. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable.
  19. 19. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable. An equation is two expressions set equal to each other.
  20. 20. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable. An equation is two expressions set equal to each other. Equations have the form: left expression = right expression
  21. 21. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable. An equation is two expressions set equal to each other. Equations have the form: left expression = right expression or LHS = RHS
  22. 22. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable. An equation is two expressions set equal to each other. Equations have the form: left expression = right expression or LHS = RHS In the two examples above x – 4 = 14 and 5x = 20, are equations.
  23. 23. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable. An equation is two expressions set equal to each other. Equations have the form: left expression = right expression or LHS = RHS In the two examples above x – 4 = 14 and 5x = 20, are equations. In particular these are linear equations.
  24. 24. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable. An equation is two expressions set equal to each other. Equations have the form: left expression = right expression or LHS = RHS In the two examples above x – 4 = 14 and 5x = 20, are equations. In particular these are linear equations. A linear equation is an equation where both sides are linear expressions.
  25. 25. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable. An equation is two expressions set equal to each other. Equations have the form: left expression = right expression or LHS = RHS In the two examples above x – 4 = 14 and 5x = 20, are equations. In particular these are linear equations. A linear equation is an equation where both sides are linear expressions. A linear equation does not contain any higher power of x such as x2, x3; it does not have x's in the denominator.
  26. 26. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable. An equation is two expressions set equal to each other. Equations have the form: left expression = right expression or LHS = RHS In the two examples above x – 4 = 14 and 5x = 20, are equations. In particular these are linear equations. A linear equation is an equation where both sides are linear expressions. A linear equation does not contain any higher power of x such as x2, x3; it does not have x's in the denominator. x2 – 3x = 2x – 3 is not a linear equation because of the x2.
  27. 27. Linear Equations I In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable. An equation is two expressions set equal to each other. Equations have the form: left expression = right expression or LHS = RHS In the two examples above x – 4 = 14 and 5x = 20, are equations. In particular these are linear equations. A linear equation is an equation where both sides are linear expressions. A linear equation does not contain any higher power of x such as x2, x3; it does not have x's in the denominator. x2 – 3x = 2x – 3 is not a linear equation because of the x2. Linear equations are the simpliest type of equations.
  28. 28. Linear Equations I We want to solve equations.
  29. 29. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal.
  30. 30. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation.
  31. 31. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 14 = 2x + 8, x = (3) is a solution
  32. 32. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 14 = 2x + 8, x = (3) is a solution because 14 = 2(3) + 8 ?
  33. 33. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 14 = 2x + 8, x = (3) is a solution because 14 = 2(3) + 8 ? 14 = 6 + 8 ? 14 = 14 yes
  34. 34. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 14 = 2x + 8, x = (3) is a solution because 14 = 2(3) + 8 ? 14 = 6 + 8 ? 14 = 14 yes Linear equations are the easiest equations to solve
  35. 35. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 14 = 2x + 8, x = (3) is a solution because 14 = 2(3) + 8 ? 14 = 6 + 8 ? 14 = 14 yes Linear equations are the easiest equations to solve and the simplest of these are the ones where the a single operation is applied to x.
  36. 36. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 14 = 2x + 8, x = (3) is a solution because 14 = 2(3) + 8 ? 14 = 6 + 8 ? 14 = 14 yes Linear equations are the easiest equations to solve and the simplest of these are the ones where the a single operation is applied to x. These are the one-step equations.
  37. 37. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 14 = 2x + 8, x = (3) is a solution because 14 = 2(3) + 8 ? 14 = 6 + 8 ? 14 = 14 yes Linear equations are the easiest equations to solve and the simplest of these are the ones where the a single operation is applied to x. These are the one-step equations. For example, x – 3 = 12,
  38. 38. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 14 = 2x + 8, x = (3) is a solution because 14 = 2(3) + 8 ? 14 = 6 + 8 ? 14 = 14 yes Linear equations are the easiest equations to solve and the simplest of these are the ones where the a single operation is applied to x. These are the one-step equations. For example, x – 3 = 12, 12 = x + 3,
  39. 39. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 14 = 2x + 8, x = (3) is a solution because 14 = 2(3) + 8 ? 14 = 6 + 8 ? 14 = 14 yes Linear equations are the easiest equations to solve and the simplest of these are the ones where the a single operation is applied to x. These are the one-step equations. For example, x – 3 = 12, 12 = x + 3, 3*x = 12,
  40. 40. Linear Equations I We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 14 = 2x + 8, x = (3) is a solution because 14 = 2(3) + 8 ? 14 = 6 + 8 ? 14 = 14 yes Linear equations are the easiest equations to solve and the simplest of these are the ones where the a single operation is applied to x. These are the one-step equations. For example, x – 3 = 12, 12 = x + 3, 3*x = 12, 12 = x are one-step equations. 3
  41. 41. Linear Equations I Basic principle for solving one- step-equations:
  42. 42. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation.
  43. 43. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 b. x + 3 = –12 c. 3x = 15
  44. 44. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 b. x + 3 = –12 c. 3x = 15
  45. 45. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 x = 15 b. x + 3 = –12 c. 3x = 15
  46. 46. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 ? x = 15 check: 15 – 3 = 12 b. x + 3 = –12 c. 3x = 15
  47. 47. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 ? x = 15 check: 15 – 3 = 12 12 = 12 (yes) b. x + 3 = –12 c. 3x = 15
  48. 48. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 ? x = 15 check: 15 – 3 = 12 12 = 12 (yes) b. x + 3 = –12 Subtract 3 from both sides –3 –3 c. 3x = 15
  49. 49. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 ? x = 15 check: 15 – 3 = 12 12 = 12 (yes) b. x + 3 = –12 Subtract 3 from both sides –3 –3 x = –15 c. 3x = 15
  50. 50. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 ? x = 15 check: 15 – 3 = 12 12 = 12 (yes) b. x + 3 = –12 Subtract 3 from both sides –3 –3 ? x = –15 check: –15 + 3 = –12 c. 3x = 15
  51. 51. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 ? x = 15 check: 15 – 3 = 12 12 = 12 (yes) b. x + 3 = –12 Subtract 3 from both sides –3 –3 ? x = –15 check: –15 + 3 = –12 –12 = –12 (yes) c. 3x = 15
  52. 52. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 ? x = 15 check: 15 – 3 = 12 12 = 12 (yes) b. x + 3 = –12 Subtract 3 from both sides –3 –3 ? x = –15 check: –15 + 3 = –12 –12 = –12 (yes) c. 3x = 15 Both sides divided by 3 3x = 15 3 3
  53. 53. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 ? x = 15 check: 15 – 3 = 12 12 = 12 (yes) b. x + 3 = –12 Subtract 3 from both sides –3 –3 ? x = –15 check: –15 + 3 = –12 –12 = –12 (yes) c. 3x = 15 Both sides divided by 3 3x = 15 3 3 x=5
  54. 54. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 ? x = 15 check: 15 – 3 = 12 12 = 12 (yes) b. x + 3 = –12 Subtract 3 from both sides –3 –3 ? x = –15 check: –15 + 3 = –12 –12 = –12 (yes) c. 3x = 15 Both sides divided by 3 3x = 15 3 3 ? x=5 check: 3(5) = 15
  55. 55. Linear Equations I Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x – 3 = 12 Add 3 to both sides +3 +3 ? x = 15 check: 15 – 3 = 12 12 = 12 (yes) b. x + 3 = –12 Subtract 3 from both sides –3 –3 ? x = –15 check: –15 + 3 = –12 –12 = –12 (yes) c. 3x = 15 Both sides divided by 3 3x = 15 3 3 ? x=5 check: 3(5) = 15 15 = 15 (yes)
  56. 56. Linear Equations I d. x = –12 3
  57. 57. Linear Equations I x = –12 Multiply both sides by 3 3 ( x = –12) (3) 3 d.
  58. 58. Linear Equations I x = –12 Multiply both sides by 3 3 ( x = –12) (3) 3 x = –36 d.
  59. 59. Linear Equations I x = –12 Multiply both sides by 3 3 ( x = –12) (3) 3 x = –36 Check: – 36 = –12 3 d.
  60. 60. Linear Equations I x = –12 Multiply both sides by 3 3 ( x = –12) (3) 3 x = –36 Check: – 36 = –12 3 d. Fact: Given a linear equation if we +, –, * , /, to both sides by the same quantity, the new equation will have the same solution.
  61. 61. Linear Equations I x = –12 Multiply both sides by 3 3 ( x = –12) (3) 3 x = –36 Check: – 36 = –12 3 d. Fact: Given a linear equation if we +, –, * , /, to both sides by the same quantity, the new equation will have the same solution. Next we tackle equations that require two steps.
  62. 62. Linear Equations I x = –12 Multiply both sides by 3 3 ( x = –12) (3) 3 x = –36 Check: – 36 = –12 3 d. Fact: Given a linear equation if we +, –, * , /, to both sides by the same quantity, the new equation will have the same solution. Next we tackle equations that require two steps. These are the ones that we have to collect the x-terms or the constant terms first.
  63. 63. Linear Equations I x = –12 Multiply both sides by 3 3 ( x = –12) (3) 3 x = –36 Check: – 36 = –12 3 d. Fact: Given a linear equation if we +, –, * , /, to both sides by the same quantity, the new equation will have the same solution. Next we tackle equations that require two steps. These are the ones that we have to collect the x-terms or the constant terms first. We do this by addition or subtraction.
  64. 64. Linear Equations I x = –12 Multiply both sides by 3 3 ( x = –12) (3) 3 x = –36 Check: – 36 = –12 3 d. Fact: Given a linear equation if we +, –, * , /, to both sides by the same quantity, the new equation will have the same solution. Next we tackle equations that require two steps. These are the ones that we have to collect the x-terms or the constant terms first. We do this by addition or subtraction. Example C. Solve for x a. x – 6 = 3x
  65. 65. Linear Equations I x = –12 Multiply both sides by 3 3 ( x = –12) (3) 3 x = –36 Check: – 36 = –12 3 d. Fact: Given a linear equation if we +, –, * , /, to both sides by the same quantity, the new equation will have the same solution. Next we tackle equations that require two steps. These are the ones that we have to collect the x-terms or the constant terms first. We do this by addition or subtraction. Example C. Solve for x a. x – 6 = 3x –x –x –6 = 2x Collect the x's by subtracting x from both sides Divide by 2
  66. 66. Linear Equations I x = –12 Multiply both sides by 3 3 ( x = –12) (3) 3 x = –36 Check: – 36 = –12 3 d. Fact: Given a linear equation if we +, –, * , /, to both sides by the same quantity, the new equation will have the same solution. Next we tackle equations that require two steps. These are the ones that we have to collect the x-terms or the constant terms first. We do this by addition or subtraction. Example C. Solve for x a. x – 6 = 3x –x –x –6 = 2x 2 2 –3 = x Collect the x's by subtracting x from both sides Divide by 2
  67. 67. Linear Equations I b. 3x – 6 = 3
  68. 68. Linear Equations I b. 3x – 6 = 3 +6 +6 collect the numbers by adding 6
  69. 69. Linear Equations I b. 3x – 6 = 3 +6 +6 3x = 9 collect the numbers by adding 6
  70. 70. Linear Equations I b. 3x – 6 = 3 collect the numbers by adding 6 +6 +6 3x = 9 div. by 3 3x/3 = 9/3
  71. 71. Linear Equations I b. 3x – 6 = 3 collect the numbers by adding 6 +6 +6 3x = 9 div. by 3 3x/3 = 9/3 x=3
  72. 72. Linear Equations I b. 3x – 6 = 3 collect the numbers by adding 6 +6 +6 3x = 9 div. by 3 3x/3 = 9/3 x=3 The more general linear equations have the form #x ± # = #x ± #, where # can be any number.
  73. 73. Linear Equations I b. 3x – 6 = 3 collect the numbers by adding 6 +6 +6 3x = 9 div. by 3 3x/3 = 9/3 x=3 The more general linear equations have the form #x ± # = #x ± #, where # can be any number. We solve it by following steps:
  74. 74. Linear Equations I b. 3x – 6 = 3 collect the numbers by adding 6 +6 +6 3x = 9 div. by 3 3x/3 = 9/3 x=3 The more general linear equations have the form #x ± # = #x ± #, where # can be any number. We solve it by following steps: 1. Add or subtract to move the x-term to one side of the equation and get: #x ± # = # or # = #x ± #
  75. 75. Linear Equations I b. 3x – 6 = 3 collect the numbers by adding 6 +6 +6 3x = 9 div. by 3 3x/3 = 9/3 x=3 The more general linear equations have the form #x ± # = #x ± #, where # can be any number. We solve it by following steps: 1. Add or subtract to move the x-term to one side of the equation and get: #x ± # = # or # = #x ± # 2. Add or subtract the # to separate the number-term from the x-term to get: #x = # or # = #x.
  76. 76. Linear Equations I b. 3x – 6 = 3 collect the numbers by adding 6 +6 +6 3x = 9 div. by 3 3x/3 = 9/3 x=3 The more general linear equations have the form #x ± # = #x ± #, where # can be any number. We solve it by following steps: 1. Add or subtract to move the x-term to one side of the equation and get: #x ± # = # or # = #x ± # 2. Add or subtract the # to separate the number-term from the x-term to get: #x = # or # = #x. 3. Divide or multiply to get x: x = solution or solution = x
  77. 77. Linear Equations I b. 3x – 6 = 3 collect the numbers by adding 6 +6 +6 3x = 9 div. by 3 3x/3 = 9/3 x=3 The more general linear equations have the form #x ± # = #x ± #, where # can be any number. We solve it by following steps: 1. Add or subtract to move the x-term to one side of the equation and get: #x ± # = # or # = #x ± # 2. Add or subtract the # to separate the number-term from the x-term to get: #x = # or # = #x. 3. Divide or multiply to get x: x = solution or solution = x The idea is to use + or –, to separate the x-term from the number term,
  78. 78. Linear Equations I b. 3x – 6 = 3 collect the numbers by adding 6 +6 +6 3x = 9 div. by 3 3x/3 = 9/3 x=3 The more general linear equations have the form #x ± # = #x ± #, where # can be any number. We solve it by following steps: 1. Add or subtract to move the x-term to one side of the equation and get: #x ± # = # or # = #x ± # 2. Add or subtract the # to separate the number-term from the x-term to get: #x = # or # = #x. 3. Divide or multiply to get x: x = solution or solution = x The idea is to use + or –, to separate the x-term from the number term, then use * or / to find the value of x.
  79. 79. Linear Equations I Example D. Solve 3x – 4 = 5x + 2
  80. 80. Linear Equations I Example D. Solve 3x – 4 = 5x + 2 –3x –3x subtract 3x to remove the x from one side.
  81. 81. Linear Equations I Example D. Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2 subtract 3x to remove the x from one side.
  82. 82. Linear Equations I Example D. Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2 –2 –2 subtract 3x to remove the x from one side. subtract 2 to move the 2 to the other side.
  83. 83. Linear Equations I Example D. Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2 –2 –2 – 6 = 2x subtract 3x to remove the x from one side. subtract 2 to move the 2 to the other side.
  84. 84. Linear Equations I Example D. Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2 –2 –2 – 6 = 2x –6 2x = 2 2 subtract 3x to remove the x from one side. subtract 2 to move the 2 to the other side. divide by 2 get x.
  85. 85. Linear Equations I Example D. Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2 –2 –2 – 6 = 2x –6 2x = 2 2 –3 = x subtract 3x to remove the x from one side. subtract 2 to move the 2 to the other side. divide by 2 get x.
  86. 86. Linear Equations I Example D. subtract 3x to remove Solve 3x – 4 = 5x + 2 the x from one side. –3x –3x subtract 2 to move the 2 to – 4 = 2x + 2 the other side. –2 –2 – 6 = 2x –6 2x divide by 2 get x. = 2 2 –3 = x Finally, all linear equations can be reduced to the format #x ± # = #x ± # by simplifying each side by itself first.
  87. 87. Linear Equations I Example D. subtract 3x to remove Solve 3x – 4 = 5x + 2 the x from one side. –3x –3x subtract 2 to move the 2 to – 4 = 2x + 2 the other side. –2 –2 – 6 = 2x –6 2x divide by 2 get x. = 2 2 –3 = x Finally, all linear equations can be reduced to the format #x ± # = #x ± # by simplifying each side by itself first. Example E. Solve the equation. 2x – 3(1 – 3x) = 3(x – 6) – 1
  88. 88. Linear Equations I Example D. subtract 3x to remove Solve 3x – 4 = 5x + 2 the x from one side. –3x –3x subtract 2 to move the 2 to – 4 = 2x + 2 the other side. –2 –2 – 6 = 2x –6 2x divide by 2 get x. = 2 2 –3 = x Finally, all linear equations can be reduced to the format #x ± # = #x ± # by simplifying each side by itself first. Example E. Solve the equation. 2x – 3(1 – 3x) = 3(x – 6) – 1 Simplify each side by combining like-term first.
  89. 89. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1
  90. 90. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x =
  91. 91. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1
  92. 92. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1 11x – 3 =
  93. 93. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1 11x – 3 = 3x – 19
  94. 94. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1 gather the x’s, subtract 3x from 11x – 3 = 3x – 19 both sides –3x –3x
  95. 95. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1 gather the x’s, subtract 3x from 11x – 3 = 3x – 19 both sides –3x –3x 8x – 3 = –19
  96. 96. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1 11x – 3 = 3x – 19 –3x –3x 8x – 3 = –19 +3 +3 gather the x’s, subtract 3x from both sides gather the number-terms, +3 to both sides
  97. 97. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1 11x – 3 = 3x – 19 –3x –3x 8x – 3 = –19 +3 +3 8x = –16 gather the x’s, subtract 3x from both sides gather the number-terms, +3 to both sides
  98. 98. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1 11x – 3 = 3x – 19 –3x –3x 8x – 3 = –19 +3 +3 8x = –16 gather the x’s, subtract 3x from both sides gather the number-terms, +3 to both sides divide by 8 to get x.
  99. 99. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1 11x – 3 = 3x – 19 –3x –3x 8x – 3 = –19 +3 +3 8x = –16 8x -16 8 = 8 gather the x’s, subtract 3x from both sides gather the number-terms, +3 to both sides divide by 8 to get x.
  100. 100. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1 11x – 3 = 3x – 19 –3x –3x 8x – 3 = –19 +3 +3 8x = –16 8x -16 8 = 8 -2 gather the x’s, subtract 3x from both sides gather the number-terms, +3 to both sides divide by 8 to get x.
  101. 101. Linear Equations I 2x – 3(1 – 3x) = 3(x – 6) – 1 2x – 3 + 9x = 3x – 18 – 1 11x – 3 = 3x – 19 –3x –3x 8x – 3 = –19 +3 +3 8x = –16 8x -16 8 = 8 x = -2 -2 gather the x’s, subtract 3x from both sides gather the number-terms, +3 to both sides divide by 8 to get x.
  102. 102. Linear Equations I Exercise A. Solve in one step by addition or subtraction . 3. –3 = x –5 1. x + 2 = 3 2. x – 1 = –3 4. x + 8 = –15 5. x – 2 = –1/2 6. 2 = x – 1 3 2 B. Solve in one step by multiplication or division. 7. 2x = 3 8. –3x = –1 11. –4 = x 2 14. 7 = –x 9. –3 = –5x 12. 7 = –x 3 15. –x = –7 10. 8 x = –15 13. –x = –4 3 C. Solve by collecting the x’s to one side first. (Remember to keep the x’s positive.) 18. –x = x – 8 16. x + 2 = 5 – 2x 17. 2x – 1 = – x –7 19. –x = 3 – 2x 20. –5x = 6 – 3x 22. –3x – 1= 3 – 6x 23. –x + 7 = 3 – 3x 21. –x + 2 = 3 + 2x 24. –2x + 2 = 9 + x
  103. 103. Linear Equations I D. Solve for x by first simplifying the equations to the form of #x ± # = #x ± #. 25. 2(x + 2) = 5 – (x – 1) 26. 3(x – 1) + 2 = – 2x – 9 27. –2(x – 3) = 2(–x – 1) + 3x 28. –(x + 4) – 2 = 4(x – 1) 29. x + 2(x – 3) = 2(x – 1) – 2 30. –2(x – 3) + 3 = 2(x – 1) + 3x + 13 31. –(x + 4) – 2(x+ 1) = 4(x – 1) – 2 32. x + 1 + 2(x – 3) = 2(x – 1) – (2 – 2x) 33. 4 – 3(2 – 2x) = 2(4x + 1) – 14 34. 5(x – 2) – 3(3 –x) = – 3(x +2) + 2(4x + 1) 35. –3(2 – 2x) + 3(3 – x) = 5(x – 1) + 2(2 – 3x) 36. 6(2x – 5) – 4 (3x +2) = 2x + 6(–3x – 4) – 8

×