1 / 35

Warm Up

Warm Up. Lesson Presentation. Lesson Quiz. 3. 1. 2. 2. 2. 3. 3. 3. 3. 2. 3. Warm Up Simplify. 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. 8. 14.

jans
Télécharger la présentation

Warm Up

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. Warm Up Lesson Presentation Lesson Quiz

  2. 3 1 2 2 2 3 3 3 3 2 3 Warm Up Simplify. 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 8 14

  3. Sunshine State Standards Prep for MA.912.A.3.2 Identify and apply the…properties of equality. AlsoPrep forMA.912.A.3.1.

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

  5. Vocabulary equation solution of an equation

  6. 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. Isolate a variable by using inverse operations, which “undo” operations on the variable.

  7. An equation is like a balanced scale. To keep the balance, perform the same operation on both sides.

  8. y – 8 = 24 Check Additional 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 To check your solution, substitute 32 for y in the original equation. 32 – 824 2424 

  9. 5 7 7 7 5 7 5 5 7 5 = z– Since is subtracted from z, add to both sides to undo the subtraction. 16 16 16 16 16 16 16 16 16 16 3 + + 4 7 7 3 = z 16 16 4 3 = z – 4 Check To check your solution, substitute for z in the original equation. – Additional Example 1B: Solving Equations by Using Addition Solve the equation. Check your answer. 

  10. n – 3.2 = 5.6 Check Check It Out! Example 1a Solve the equation. Check your answer. n – 3.2 = 5.6 + 3.2+ 3.2 Since 3.2 is subtracted from n, add 3.2 to both sides to undo the subtraction. n = 8.8 To check your solution, substitute 8.8 for n in the original equation. 8.8 – 3.25.6 5.65.6 

  11. –6 = k– 6 Check Check It Out! Example 1b Solve the equation. Check your answer. –6 = k –6 + 6+ 6 Since 6 is subtracted from k, add 6 to both sides to undo the subtraction. 0= k To check your solution, substitute 0 for k in the original equation. –6 0– 6 –6 –6 

  12. 16 = m– 9 Check Check It Out! Example 1c Solve the equation. Check your answer. 16 = m –9 + 9+ 9 Since 9 is subtracted from m, add 9 to both sides to undo the subtraction. 25= m To check your solution, substitute 25 for m in the original equation. 16 25– 9 16 16 

  13. m + 17 = 33 Check Additional Example 2A: Solving Equations by Using Subtraction Solve the equation. Check your answer. m +17 = 33 – 17–17 Since 17 is added to m, subtract 17 from both sides to undo the addition. m = 16 To check your solution, substitute 16 for m in the original equation. 16 + 1733 3333 

  14. 4.2 = t+ 1.8 Check Additional 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 To check your solution, substitute 2.4 for t in the original equation. 4.2 2.4+ 1.8 4.2 4.2 

  15. 1 1 1 1 1 1 1 2 2 2 2 2 2 2 Since is added to d, subtract from both sides to undo the addition. – – d = 1 1 1 2 2 2 d + = 1 + 1 Check To check your solution, substitute for d in the original equation. Check It Out! Example 2a Solve the equation. Check your answer. d + = 1 1 1 

  16. –5 = k+ 5 Check Check It Out! Example 2b Solve the equation. Check your answer. –5 = k +5 – 5– 5 Since 5 is added to k, subtract 5 from both sides to undo the subtraction. –10= k To check your solution, substitute –10 for k in the original equation. –5 –10+ 5 –5 –5 

  17. 6 + t = 14 Check Check It Out! Example 2c Solve the equation. Check your answer. 6+t = 14 – 6– 6 Since 6 is added to t, subtract 6 from both sides to undo the addition. t = 8 To check your solution, substitute 8 for t in the original equation. 6 + 814 1414 

  18. Remember that subtracting is the same as adding the opposite. When solving equations, you will sometimes find it easier to add an opposite to both sides instead of subtracting. For example, this method may be useful when the equation contains negative numbers.

  19. Solve – + p =– . 5 5 2 2 2 5 2 5 Since – is added to p, add to both sides. 11 11 11 11 11 11 11 11 + + 3 p = 11 5 5 3 11 11 11 – + p = – Check To check your solution, substitute for p in the original equation. 2 5 3 – – + 11 11 11 – – Additional Example 3: Solving Equations by Adding the Opposite 

  20. –2.3+ m = 7 Check Check It Out! Example 3a Solve –2.3+m = 7. Check your answer. –2.3+m = 7 +2.3+ 2.3 Since –2.3 is added to m, add 2.3 to both sides. m = 9.3 To check your solution, substitute 9.3 for m in the original equation. –2.3 + 9.37 77 

  21. 5 5 5 3 3 3 5 5 3 3 + z = – 4 4 4 4 4 4 4 4 4 4 Since – is added to z, add to both sides. + + 3 3 4 4 – + z = Check 3 5 – + 2 4 4 Check It Out! Example 3b Solve – + z = .Check your answer. z = 2 To check your solution, substitute 2 for z in the original equation. 

  22. –11+ x = 33 Check Check It Out! Example 3c Solve –11+x = 33. Check your answer. –11+x = 33 +11+11 Since –11 is added to x, add 11 to both sides. x = 44 To check your solution, substitute 44 for x in the original equation. –11 + 4433 3333 

  23. decrease in population original population current population minus is Additional 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.

  24. 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.

  25. 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.

  26. The properties of equality allow you to perform inverse operations, as in the previous examples. These properties say that you can perform the same operation on both sides of an equation.

  27. Lesson Quizzes Standard Lesson Quiz Lesson Quiz for Student Response Systems

  28. Lesson Quiz Solve each equation. 1.r – 4 = –8 2. 3.m + 13 = 58 4. 0.75 = n + 0.6 5. –5 + c = 22 6. 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. –4 45 0.15 27 s – 89 = 578; s = 667

  29. Lesson Quiz for Student Response Systems 1. Solve the equation. x − 2 = –10 A. x = –12 C. x= 8 D. x = 12 B. x = –8

  30. Lesson Quiz for Student Response Systems 2. Solve the equation. A. C. D. B.

  31. Lesson Quiz for Student Response Systems 3. Solve the equation. n + 12 = 40 A. n =–52 C. n = 28 B. n = –28 D. n = 52

  32. Lesson Quiz for Student Response Systems 4. Solve the equation. 0.25 = q + 0.10 A. q =–0.35 C. q = 0.15 B. q = –0.15 D. q = 0.35

  33. Lesson Quiz for Student Response Systems 5. Solve the equation. –2 + d = 20 C. d = 18 A. d =–22 B. d = –18 D. d = 22

  34. Lesson Quiz for Student Response Systems 6. This year 250 high school students signed up for music lessons. This is 42 less than the number of students who signed up last year. Write and solve an equation to fine the number of students who signed up for music lessons last year. A. h − 42= 250; 208 students B. h − 42= 250; 292 students C. h + 42= 250; 292 students D. h + 42= 250; 208 students

More Related