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Warm-Up Equations Practice Presentation

Prepare for California Standards - Solve equations, find common denominators, identify operations, and solve multistep problems. Understand equivalent equations and operations.

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Warm-Up Equations Practice Presentation

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  1. Preview Warm Up California Standards Lesson Presentation

  2. Warm Up Solve each equation. 1. 3 + x = 11 2. x – 7 = 19 3. 6x = 15 4. Find a common denominator. 5. 6. 26 8 12 Possible answer: 6 Possible answer: 20

  3. California Standards Preparation for 5.0 Students solve multistep problems, including word problems, involvinglinear equations and linear inequalities in one variable and provide justification for each step.

  4. Vocabulary equivalent equations

  5. Operations in the equation To solve   Subtract 5 from both First x is multiplied by 2. sides of the equation.  Then 5 is added.  Then divide both sides by 2. Many equations contain more than one operation, such as 2x + 5 = 11. This equation contains multiplication and addition. Equations that contain two operations require two steps to solve. Identify the operations in the equation and the order in which they are applied to the variable. Then use inverse operations to undo them in reverse over one at a time. 2x + 5 = 11

  6. –5 –5 2x = 6 2x + 5 =11 Subtract 5 from both sides of the equation. Divide both sides of the equation by 2. x = 3 The solution set is {3}. Each time you perform an inverse operation, you create an equation that is equivalent to the original equation. Equivalent equations have the same solutions, or the same solution set. In the example above, 2x + 5 = 11, 2x = 6, and x = 3 are all equivalent equations.

  7. –10 –10 Additional Example 1A: Solving Two-Step Equations Solve 18 = 4a + 10. First a is multiplied by 4. Then 10 is added. 18 = 4a + 10 Subtract 10 from both sides. 8 = 4a 8 = 4a is equivalent to 18 = 4a + 10. Since a is multiplied by 4, divide both sides by 4 to undo the multiplication. 2 = a The solution set is {2}.

  8. +2 +2 Additional Example 1B: Solving Two-Step Equations Solve 5t – 2 = –32. First t is multiplied by 5. Then 2 is subtracted. 5t – 2 = –32 Add 2 to both sides. 5t = –30 5t = –30 is equivalent to 5t – 2 = –32. Since t is multiplied by 5, divide both sides by 5 to undo the multiplication. t = –6 The solution set is {–6}.

  9. + 4 +4 Check It Out! Example 1a Solve the equation. Check your answer. –4+ 7x = 3 First x is multiplied by 7. Then –4 is added. –4+ 7x = 3 Add 4 to both sides. 7x = 7 7x = 7 is equivalent to –4 + 7x = 3. Since x is multiplied by 7, divide both sides by 7 to undo the multiplication. x = 1 The solution set is {1}.

  10. Check It Out! Example 1a Continued Solve the equation. Check your answer. Check –4+ 7x = 3 To check your solution, substitute 1 for x in the original equation. –4 + 7(1) 3 –4 + 7 3 33 

  11. + 5.7 +5.7 Check It Out! Example 1b Solve the equation. Check your answer. 1.5 = 1.2y – 5.7 First y is multiplied by 1.2. Then 5.7 is subtracted. 1.5 = 1.2y – 5.7 Add 5.7 to both sides. 7.2 = 1.2y 7.2 = 1.2y is equivalent to 1.5 = 1.2y – 5.7. Since y is multiplied by 1.2, divide both sides by 1.2 to undo the multiplication. 6= y The solution set is {6}.

  12. Check It Out! Example 1b Continued Solve the equation. Check your answer. Check 1.5 = 1.2y – 5.7 To check your solution, substitute 6 for y in the original equation. 1.5 1.2(6) – 5.7 1.5 7.2 – 5.7 1.51.5 

  13. –2 –2 = 0 is equivalent to + 2 = 2. = 0 Check It Out! Example 1c Solve the equation. Check your answer. First n is divided by 7. Then 2 is added. Subtract 2 from each side. Since n is divided by 7, multiply both sides by 7 to undo the division. n = 0 The solution set is {0}.

  14. Check It Out! Example 1c Continued Solve the equation. Check your answer. Check To check your solution, substitute 0 for n in the original equation. 

  15. Since is subtracted from , add to both sides to undo the subtraction. Additional Example 2A: Solving Two-Step Equations That Contain Fractions Solve the equation. Method 1 Use fraction operations.

  16. Additional Example 2A Continued Since y is divided by 8 multiply both sides by 8. Simplify. y = 16 The solution set is {16}.

  17. +6 +6 Additional Example 2A Continued Method 2 Multiply by the least common denominator (LCD) to clear fractions. Multiply both sides by 8, the LCD of the fractions. Distribute 8 on the left side. Simplify. Since 6 is subtracted from y, add 6 to both sides to undo the subtraction. y – 6 = 10 y = 16 The solution set is {16}.

  18. Check Additional Example 2A Continued Check your answer. To check your solution, substitute 16 for y in the original equation. 

  19. Since is added to , subtract from both sides to undo the addition. Additional Example 2B: Solving Two-Step Equations That Contain Fractions Solve the equation. Method 1 Use fraction operations.

  20. Since r is multiplied by multiply both sides by , the reciprocal. The solution set is . Additional Example 2B Continued Simplify.

  21. Additional Example 2B Continued Method 2 Multiply by the least common denominator (LCD) to clear the fractions. Multiply both sides by 12, the LCD of the fractions. Distribute 12 on the left side.

  22. – 9 –9 8r =–2 The solution set is . Additional Example 2B Continued 8r + 9 = 7 Simplify. Since 9 is added 8r, subtract 9 from both sides to undo the addition. Since r is multiplied by 8, divide both sides 8 to undo the multiplication.

  23. To check your solution, substitute for r in the original equation. Additional Example 2B Continued Check your answer. Check 

  24. Helpful Hint You can multiply both sides of the equation by any common denominator of the fractions. Using the LCD is the most efficient.

  25. Since is subtracted from , add to both sides to undo the subtraction. Check It Out! Example 2a Solve the equation. Check your answer. Method 1 Use fraction operations.

  26. Since x is multiplied by multiply both sides by , the reciprocal. The solution set is . Check It Out! Example 2a Continued Simplify.

  27. Check It Out! Example 2a Continued Method 2 Multiply by the least common denominator (LCD) to clear the fractions. Multiply both sides by 10, the LCD of the fractions. Distribute 10 on the left side. 4x – 5 = 50

  28. + 5 +5 4x = 55 The solution set is . Check It Out! Example 2a Continued Solve the equation. Check your answer. 4x – 5 = 50 Simplify. Since 5 is subtracted from 4x add 5 to both sides to undo the subtraction. Simplify. Since x is multiplied by 4, divide both sides 4 to undo the multiplication.

  29. To check your solution, substitute for x in the original equation. Check It Out! Example 2a Continued Check  5 5

  30. Since is added to , subtract from both sides to undo the addition. Check It Out! Example 2b Solve the equation. Check your answer. Method 1 Use fraction operations.

  31. Since u is multiplied by multiply both sides by the reciprocal, . The solution set is . Check It Out! Example 2b Continued Simplify.

  32. Check It Out! Example 2b Continued Method 2 Multiply by the least common denominator (LCD) to clear fractions. Multiply both sides by 8, the LCD of the fractions. Distribute 8 on the left side. 6u + 4 = 7

  33. – 4 – 4 6u = 3 The solution set is . Check It Out! Example 2b Continued Solve the equation. Check your answer. 6u + 4 = 7 Simplify. Since 4 is added to 6u subtract 4 from both sides to undo the addition. Simplify. Since u is multiplied by 6, divide both sides 6 to undo the multiplication.

  34. To check your solution, substitute for u in the original equation. Check It Out! Example 2b Continued Check 

  35. Since is subtracted from , add to both sides to undo the subtraction. Check It Out! Example 2c Solve the equation. Check your answer. Method 1 Use fraction operations.

  36. Since n is multiplied by multiply both sides by the reciprocal, . Check It Out! Example 2c Continued Simplify. The solution set is {15}. n = 15

  37. Check It Out! Example 2c Continued Check To check your solution, substitute 15 for n in the original equation. 

  38. 1 Understand the Problem Additional Example 3: Problem-Solving Application Jan joined the dining club at the local café for a fee of $29.95. Being a member entitles her to save $2.50 every time she buys lunch. Jan calculates that she has saved a total of $12.55 so far by joining the club. Write and solve an equation to find how many times Jan has eaten lunch at the café. The answer will be the number of meals Jan had eaten.

  39. Make a Plan 2 Additional Example 3 Continued List the important information: • Jan paid a $29.95 dining club fee. • Jan saved $2.50 on each meal. • Her total savings was $12.55. Let m represent the number of meals that Jan had purchased. That means that Jan saved $2.50m. She must also subtract the cost the dining club fee from the amount she saved for the meals. Write an equation to represent this situation.

  40. 3 Solve = 12.55 2.50(m) – 29.95 2.50(m) – 29.95 = 12.55 + 29.95 +29.95 2.50m = 42.50 Additional Example 3 Continued Savings on meals is initial fee total savings minus Since 29.95 is subtracted from 2.50m, add 29.95 to both sides to undo the subtraction. Since m is multiplied by 2.50, divide both sides by 2.50 to undo the multiplication. m = 17 meals

  41. 4 Look Back Additional Example 3 Continued Check that the answer is reasonable. The savings per meal is $2.50, so if Jan had 17 meals minus the initial cost of joining the dining club which is $29.95, it is equal to a savings of $12.55.

  42. 1 Understand the Problem Check It Out! Example 3a 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? The answer will be the monthly fee that Sara had paid during the year.

  43. Make a Plan 2 Check It Out! Example 3a Continued List the important information: • Sara paid $15.95 for the membership. • Sara paid a monthly fee for 12 months. • Her total cost for 12 months was $735.95. Let m represent the monthly fee that Sara paid. Sara paid that fee for 12 months. She must also add the cost of the membership. Write an equation to represent this situation.

  44. 3 Solve + 15.95 12m = 735.95 12m + 15.95 = 735.95 –15.95 –15.95 12m = 720.00 Check It Out! Example 3a Continued initial fee Monthly fee is total cost. plus Since 15.95 is added to 12m, subtract 15.95 from both sides to undo the addition. Since m is multiplied by 12, divide both sides by 12 to undo the multiplication. m = $60 month

  45. 4 Look Back Check It Out! Example 3a Continued Check that the answer is reasonable. The cost per month is $60.00, so if Sara paid for 12 months plus the initial membership fee which is $15.95, it is equal to the total cost of $735.95.

  46. 1 Understand the Problem Check It Out! Example 3b Lynda has 12 records in her collection. She adds the same number of new records to her collection each month. After 7 months Lynda has 26 records. How many records does Lynda add each month? The answer will be the number of records that Lynda adds each month.

  47. Make a Plan 2 Check It Out! Example 3b Continued List the important information: • Lynda started with 12 records. • Lynda adds records each month. • At the end of 7 months she had 26 records. Let r represent the number of records added monthly. Lynda added new records for 7 months to her starting number of records 12. Write an equation to represent this situation.

  48. 3 Solve 7r + 12 = 26 7r + 12 = 26 –12 –12 7r = 14 Check It Out! Example 3b Continued Months initial records plus is total records Since 12 is added to 7r, subtract 12 from both sides to undo the addition. Since r is multiplied by 7, divide both sides by 7 to undo the multiplication. r = 2 records a month

  49. 4 Look Back Check It Out! Example 3b Continued Check that the answer is reasonable. The number of records added per month is 2, so if Lynda started with 12 records and added the 2 per month for 7 months, the total number is equal to 26.

  50. Lesson Quiz Solve each equation. 1. 4y + 8 = 2 2. 3 + 2x = 11 3.4. 4 –8 5. Nancy bought 5 rolls of color film and 6 rolls of black-and-white film. The 5 rolls of color film cost $15, and Nancy’s total was $39. Write and solve an equation to find the cost of one roll of black-and-white film. 6b + 15 = 39; $4

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