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Warm Up

Warm Up. Problem of the Day. Lesson Presentation. Lesson Quizzes. 4. __. c. Warm Up Write an algebraic expression for each word phrase. 1. A number x decreased by 9 2. 5 times the sum of p and 6 3. 2 plus the product of 8 and n 4. the quotient of 4 and a number c. x – 9.

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Warm Up

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  1. Warm Up Problem of the Day Lesson Presentation Lesson Quizzes

  2. 4 __ c Warm Up Write an algebraic expression for each word phrase. 1. A number x decreased by 9 2. 5 times the sum of p and 6 3. 2 plus the product of 8 and n 4. the quotient of 4 and a number c x – 9 5(p + 6)‏ 2 +8n

  3. Problem of the Day How many pieces do you have if you cut a log into six pieces and then cut each piece into 4 pieces? 24

  4. Learn to solve equations using multiplication and division.

  5. You can solve a multiplication equation using the Division Property of Equality.

  6. ? 6(8) = 48 ? 48 = 48 Additional Example 1A: Solving Equations Using Division Solve 6x = 48. 6x = 48 Use the Division Property of Equality: Divide both sides by 6. 6x = 48 6 6 1 • x = x 1x = 8 x = 8 Check 6x = 48 Substitute 8 for x. 

  7. ? –9(–5) = 45 ? 45 = 45 Additional Example 1B: Solving Equations Using Division Solve –9y = 45. –9y = 45 –9y = 45 Use the Division Property of Equality: Divide both sides by –9. –9 –9 1 • y = y 1y = –5 y = –5 Check –9y = 45 Substitute –5 for y. 

  8. ? 9(4) = 36 ? 36 = 36 Check It Out: Example 1A Solve 9x = 36. 9x = 36 Use the Division Property of Equality: Divide both sides by 9. 9x = 36 9 9 1 • x = x 1x = 4 x = 4 Check 9x = 36 Substitute 4 for x. 

  9. ? –3(–12) = 36 ? 36 = 36 Check It Out: Example 1B Solve –3y = 36. –3y = 36 Use the Division Property of Equality: Divide both sides by –3. –3y = 36 –3 –3 1 • y = y 1y = –12 y = –12 Check –3y = 36 Substitute –12 for y. 

  10. You can solve a division equation using the Multiplication Property of Equality.

  11. b –4 b = 5 –4 b = 5 –4 –20 ? = 5 –4 ? 5 = 5 Additional Example 2: Solving Equations Using Multiplication Solve = 5. Use the Multiplication Property of Equality: Multiply both sides by –4. –4 • –4 • b = –20 Check Substitute –20 for b. 

  12. c –3 c = 5 –3 c = 5 –3 –15 ? = 5 –3 ? 5 = 5 Check It Out: Example 2 Solve = 5. Use the Multiplication Property of Equality: Multiply both sides by –3. –3 • –3 • c = –15 Check Substitute –15 for c. 

  13. 1 4 1 x = 670 4 1 x = 670 4 Additional Example 3: Money Application To go on a school trip, Helene has raised $670, which is one-fourth of the amount she needs. What is the total amount needed? fraction of amount raised so far amount raised so far total amount needed  = • = x 670 Write the equation. Multiply both sides by 4. 4• 4• Helene needs $2680 total. x = 2680 Check: The amount raised so far is about $700. She needs about 4 times this amount, or $2800. The answer is reasonable.

  14. fraction of total amount raised so far 1 8 1 x = 750 8 1 x = 750 8 Check It Out: Example 3 The school library needs money to complete a new collection. So far, the library has raised $750, which is only one-eighth of what they need. What is the total amount needed? amount raised so far total amount needed =  • = x 750 Write the equation. 8• Multiply both sides by 8. 8• The library needs $6000 total. x = 6000 Check: The amount raised so far is about $800. She needs about 8 times this amount, or $6400. The answer is reasonable.

  15. Sometimes it is necessary to solve equations by using two inverse operations. For instance, the equation 6x 2 = 10 has multiplication and subtraction. Variable term 6x 2 = 10 Multiplication Subtraction To solve this equation, add to isolate the term with the variable in it. Then divide to solve.

  16. Additional Example 4: Solving a Two-Step Equation Solve 3x + 2 = 14. Subtract 2 to both sides to isolate the term with x in it. Step 1: 3x + 2 = 14 – 2 – 2 3x = 12 Step 2: 3x = 12 Divide both sides by 3. 3 3 x = 4

  17. Check It Out: Example 4 Solve 4y + 5 = 29. Subtract 5 from both sides to isolate the term with y in it. Step 1: 4y + 5 = 29 – 5 – 5 4y = 24 Step 2: 4y = 24 Divide both sides by 4. 4 4 y = 6

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

  19. x –4 Lesson Quiz Solve. 1.3t = 9 2. –15 = 3b 3. = –7 4.z ÷ 4 = 22 5. A roller coaster descends a hill at a rate of 80 feet per second. The bottom of the hill is 400 feet from the top. How long will it take the coaster rides to reach the bottom? t = 3 b = –5 x = 28 z = 88 5 seconds

  20. Lesson Quiz for Student Response Systems 1. Solve the equation. 7e = 28 A.e = 1 B.e = 2 C.e = 3 D.e = 4

  21. Lesson Quiz for Student Response Systems 2. Solve the equation. –72 = 4g A.g = –18 B.g = –17 C.g = –16 D.g = –15

  22. Lesson Quiz for Student Response Systems 3. Solve the equation. = –12 A.i = –72 B.i = –18 C.i = 72 D.i = 80 i–6

  23. Lesson Quiz for Student Response Systems 4. Solve the equation. k ÷ 9 = 13 A.k = 104 B.k = 117 C.k = 118 D.k = 123

  24. Lesson Quiz for Student Response Systems 5. A surf boat descends a stream running downhill at a rate of 15 ft/minute. The bottom of the hill is at 90 ft from the top. How long will it take the surf boat to reach the bottom? A. 3 min B. 4 min C. 5 min D. 6 min

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