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Warm Up Evaluate. 1.  + 4 2. 0.51 + (0.29) Give the opposite of each number. 3. 8 4.  Evaluate each expre

3. 1. 2. 2. 2. 3. 3. 3. 3. 2. 3. Warm Up Evaluate. 1.  + 4 2. 0.51 + (0.29) Give the opposite of each number. 3. 8 4.  Evaluate each expression for a = 3 and b =  2. 5. a + 5 6. 12  b. 0.8. –8. 14 . 8 . Objective.

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Warm Up Evaluate. 1.  + 4 2. 0.51 + (0.29) Give the opposite of each number. 3. 8 4.  Evaluate each expre

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  1. 3 1 2 2 2 3 3 3 3 2 3 Warm Up Evaluate. 1. + 4 2. 0.51 + (0.29) Give the opposite of each number. 3. 8 4.  Evaluate each expression for a = 3 and b = 2. 5. a + 5 6. 12  b 0.8 –8 14 8

  2. Objective Solve one-step equations in one variable by using addition or subtraction.

  3. An equation is a mathematical statement that two expressions are equal. A solution of an equation is a value of the variable that makes the equation true. To find solutions, isolate the variable. A variable is isolated when it appears by itself on one side of an equation, and not at all on the other side.

  4. Isolate a variable by using inverse operations which "undo" operations on the variable. An equation is like a balanced scale. To keep the balance, perform the same operation on both sides. Addition Subtraction Subtraction Addition

  5. Example 1A: Solving Equations by Using Addition Solve the equation. Check your answer. y – 8 = 24 + 8+ 8 Since 8 is subtracted from y, add 8 to both sides to undo the subtraction. y = 32

  6. 5 7 7 7 = z– Since is subtracted from z, add to both sides to undo the subtraction. 16 16 16 16 3 + + 4 7 7 = z 16 16 Example 1B: Solving Equations by Using Addition Solve the equation. Check your answer.

  7. Example 2B: Solving Equations by Using Subtraction Solve the equation. Check your answer. 4.2 = t +1.8 –1.8– 1.8 Since 1.8 is added to t, subtract 1.8 from both sides to undo the addition. 2.4= t

  8. 3 5 3 3 3 5 + z = – 4 4 4 4 4 4 Since – is added to z, add to both sides. + + 3 3 4 4 Check It Out! Example 3b Solve – + z = .Check your answer. z = 2

  9. decrease in population original population current population minus is Example 4: Application Over 20 years, the population of a town decreased by 275 people to a population of 850. Write and solve an equation to find the original population. p– d = c Write an equation to represent the relationship. p – d = c p – 275 = 850 Since 275 is subtracted from p, add 275 to both sides to undo the subtraction. + 275+ 275 p =1125 The original population was 1125 people.

  10. Check It Out! Example 4 A person's maximum heart rate is the highest rate, in beats per minute, that the person's heart should reach. One method to estimate maximum heart rate states that your age added to your maximum heart rate is 220. Using this method, write and solve an equation to find a person's age if the person's maximum heart rate is 185 beats per minute.

  11. added to maximum heart rate 220 age is Check It Out! Example 4 Continued a+ r = 220 a + r = 220 Write an equation to represent the relationship. a + 185 = 220 Substitute 185 for r. Since 185 is added to a, subtract 185 from both sides to undo the addition. – 185– 185 a = 35 A person whose maximum heart rate is 185 beats per minute would be 35 years old.

  12. Properties of Equality

  13. Properties of Equality

  14. Warm-Up Solve each equation. 1.r – 4 = –8 2. 3. This year a high school had 578 sophomores enrolled. This is 89 less than the number enrolled last year. Write and solve an equation to find the number of sophomores enrolled last year.

  15. Objective Solve one-step equations in one variable by using multiplication or division.

  16. Solving an equation that contains multiplication or division is similar to solving an equation that contains addition or subtraction. Use inverse operations to undo the operations on the variable. Multiplication Division Division Multiplication

  17. j –8 = 3 Example 1A: Solving Equations by Using Multiplication Solve the equation. Since j is divided by 3, multiply both sides by 3 to undo the division. –24 = j

  18. Check It Out! Example 2b Solve the equation. Check your answer. 0.5y = –10 Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication. y = –20

  19. Remember that dividing is the same as multiplying by the reciprocal. When solving equations, you will sometimes find it easier to multiply by a reciprocal instead of dividing. This is often true when an equation contains fractions.

  20. 5 6 5 6 6 5 6 5 The reciprocal of is . Since w is multiplied by , multiply both sides by . Example 3A: Solving Equations That Contain Fractions Solve the equation. 5 w= 20 6 w = 24

  21. 1 5 4 4 1 5 1 1 5 5 The reciprocal of is 5. Since b is multiplied by , multiply both sides by 5. – = b Check It Out! Example 3a Solve the equation. Check your answer. –= b

  22. 1 Ciro puts of the money he earns from mowing lawns into a college education fund. This year Ciro added $285 to his college education fund. Write and solve an equation to find how much money Ciro earned mowing lawns this year. 4 Example 4: Application one-fourth times earnings equals college fund Write an equation to represent the relationship. Substitute 285 for c. Since m is divided by 4, multiply both sides by 4 to undo the division. Ciro earned $1140 mowing lawns. m = $1140

  23. Check it Out! Example 4 The distance in miles from the airport that a plane should begin descending, divided by 3, equals the plane's height above the ground in thousands of feet. A plane began descending 45 miles from the airport. Use the equation to find how high the plane was flying when the descent began. Distance divided by 3 equals height in thousands of feet Write an equation to represent the relationship. Substitute 45 for d. 15 = h The plane was flying at 15,000 ft when the descent began.

  24. Properties of Equality

  25. 8 8 = a a 4 4 = c c WORDS Division Property of Equality You can divide both sides of an equation by the same nonzero number, and the statement will still be true. NUMBERS 8 = 8 2 = 2 ALGEBRA a = b (c ≠ 0) Properties of Equality

  26. Lesson Quiz: Part 1 Solve each equation. 1. 2. 3. 8y = 4 4. 126 = 9q 5. 6. 21 2.8 –14 40

  27. 7. A person's weight on Venus is about his or her weight on Earth. Write and solve an equation to find how much a person weighs on Earth if he or she weighs 108 pounds on Venus. 9 10 Lesson Quiz: Part 2

  28. Warm Up Evaluate each expression. 1. 9 –3(–2) 2. 3(–5 + 7) 3. 4. 26 – 4(7 – 5) Simplify each expression. 5. 10c + c 6. 8.2b + 3.8b – 12b 7. 5m + 2(2m – 7) 8. 6x – (2x + 5) 15 6 –4 18 11c 0 9m – 14 4x – 5

  29. Objective Solve equations in one variable that contain more than one operation.

  30. Cost per CD Total cost Cost of discount card Alex belongs to a music club. In this club, students can buy a student discount card for $19.95. This card allows them to buy CDs for $3.95 each. After one year, Alex has spent $63.40. To find the number of CDs c that Alex bought, you can solve an equation. Notice that this equation contains multiplication and addition. Equations that contain more than one operation require more than one step to solve. Identify the operations in the equation and the order in which they are applied to the variable. Then use inverse operations and work backward to undo them one at a time. 3.95c + 19.95 = 63.40

  31. –10 – 10 8 = 4a 4 4 Example 1A: Solving Two-Step Equations Solve 18 = 4a + 10. 18 = 4a + 10 8 = 4a 2 = a

  32. + 4 + 4 7x = 7 7 7 Check it Out! Example 1a Solve –4 + 7x = 3. –4 + 7x = 3 7x = 7 x = 1

  33. + 5.7 + 5.7 7.2 = 1.2y 1.2 1.2 Check it Out! Example 1b Solve 1.5 = 1.2y – 5.7. 1.5 = 1.2y – 5.7 7.2 = 1.2y 6 = y

  34. Example 2A: Solving Two-Step Equations That Contain Fractions Solve . Method 1 Use fraction operations.

  35. Example 2A Continued Simplify.

  36. Example 2B: Solving Two-Step Equations That Contain Fractions Solve . Method 1 Use fraction operations.

  37. Example 2B Continued

  38. Equations that are more complicated may have to be simplified before they can be solved. You may have to use the Distributive Property or combine like terms before you begin using inverse operations.

  39. + 21 +21 Example 3A: Simplifying Before Solving Equations Solve 8x – 21 - 5x = –15. 8x – 21 – 5x = –15 8x – 5x – 21 = –15 Use the Commutative Property of Addition. 3x– 21 = –15 Combine like terms. Since 21 is subtracted from 3x, add 21 to both sides to undo the subtraction. 3x = 6 Since x is multiplied by 3, divide both sides by 3 to undo the multiplication. x = 2

  40. + 8 + 8 6y = –12 6 6 Example 3B: Simplifying Before Solving Equations Solve 10y – (4y + 8) = –20 Write subtraction as addition of the opposite. 10y + (–1)(4y + 8) = –20 10y + (–1)(4y) + (–1)( 8) = –20 Distribute –1 on the left side. 10y – 4y – 8 = –20 Simplify. 6y – 8 = –20 Combine like terms. Since 8 is subtracted from 6y, add 8 to both sides to undo the subtraction. 6y = –12 Since y is multiplied by 6, divide both sides by 6 to undo the multiplication. y = –2

  41. Check It Out! Example 4 Sara paid $15.95 to become a member at a gym. She then paid a monthly membership fee. Her total cost for 12 months was $735.95. How much was the monthly fee?

  42. monthly fee total cost initial membership + = Check It Out! Example 4Continued Make a Plan Let m represent the monthly membership fee that Sara must pay. That means that Sara must pay 12m. However, Sara must also add the amount she spent to become a gym member. 735.95 = 12m + 15.95

  43. 3 Solve – 15.95 – 15.95 720 = 12m 12 12 Check It Out! Example 4Continued Since 15.95 is added to 12m, subtract 15.95 from both sides to undo the addition. 735.95 = 12m + 15.95 720 = 12m Since m is multiplied by 12, divide both sides by 12 to undo the multiplication. 60 = m

  44. +9 +9 Example 5B: Solving Equations to Find an Indicated Value If 3d – (9 – 2d) = 51, find the value of 3d. Step 1 Find the value of d. 3d – (9 – 2d) = 51 3d – 9 + 2d = 51 5d – 9 = 51 Since 9 is subtracted from 5d, add 9 to both sides to undo the subtraction. 5d = 60 Since d is multiplied by 5, divide both sides by 5 to undo the multiplication. d = 12

  45. Example 5B Continued If 3d – (9 – 2d) = 51, find the value of 3d. Step 2 Find the value of 3d. d = 12 3(12) To find the value of 3d, substitute 12 for d. 36 Simplify.

  46. Warm Up Simplify. 1. –7(x – 3) 2. 3. 15 – (x – 2) Solve. 4. 3x + 2 = 8 5.

  47. Objective Solve equations in one variable that contain variable terms on both sides.

  48. Helpful Hint Equations are often easier to solve when the variable has a positive coefficient. Keep this in mind when deciding on which side to "collect" variable terms. To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation.

  49. –5n –5n + 2 + 2 Example 1: Solving Equations with Variables on Both Sides Solve 7n – 2 = 5n + 6. 7n – 2 = 5n + 6 2n – 2 = 6 2n = 8 n = 4

  50. –0.3y –0.3y +0.3 + 0.3 Check It Out! Example 1b Solve 0.5 + 0.3y = 0.7y – 0.3. 0.5 + 0.3y = 0.7y – 0.3 0.5 = 0.4y – 0.3 0.8 = 0.4y 2 = y

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