1. Order of Operations with rational numbers

2. objective  use the order of operation to simplify numerical expression containing rational numbers

3. Example 1. Simplify: (-0.4) 3 (-5/2) 2 (-0.4)(-0.4)(-0.4) Simplify each power before multiplying each factor. Multiply like terms. -0.064 (-5/2)(-.5/2) (-0.4) 3 (-5/2) 2 25/4 Multiply.

4. Multiply the factors. Simplify. -0.064 -1.6/4 -0.4 25/4

5. Example 2. Simplify: -6(2/3- 5/9) ÷ [(2.4 5)(-1) 5 ]  To simplify expressions that contain more than one grouping symbol, begin computing with the innermost set Begin computing within parentheses. -6(2/3-5/9) ÷ [(2.4 5)(-1) 5 ] -6(1/9) ÷ [(12)(-1) 5 ] -6(1/9) ÷ [(12)(-1)] Simplify the power. Multiply.

6. -6/9 ÷ (-12) Simplify. Divide; multiply by the reciprocal. -2/3 Simplify. 2/36 = 1/18 -2/3 ÷ (-12) -6(1/9) ÷ [(12)(-1)] -1/12

7. 52 52 – 7 2/10  The division bar is a grouping symbol. To work on an expression with a division bar, first simplify the numerator, then the denominator, and finally divide. Subtract and rewrite the answer in simplest form. Simplify the power. + 2323 (2 – 15) 5 2 – 7 2/10 (2 – 15)+ 2 3 25 – 7 2/10 (2 – 15)+ 2 3 Simplify : Example 3

8. Subtract and rewrite the answer in simplest form. Compute within parentheses. 25 – 7 2/10 (2 – 15) + 2 3 17 4/5 (2 – 15) + 2 3 17 4/5 + 2323 -13 Simplify the power. 17 4/5 -13 + 8 Add. 17 4/5 -5 Divide.

9. 17 4/5 ÷ (-5) Rewrite in horizontal form. 17 4/5 -5 Divide. 17 4/5 ÷ (-5) Rename as fractions. 89/5 ÷ -5/1 Multiply by the reciprocal to divide 89/5 x -1/5 -89/25 Rename as a mixed number -3 14/25

10. -12.5 + 0.5  The division bar is a grouping symbol. To work on an expression with a division bar, first simplify the numerator, then the denominator, and finally divide. Rename 0.5 as 1/2. Add 0.5 3/4 -12.5 + 0.5 3/4 0.5 -12 3/4 0.5 Simplify : Example 4

11. Rename 0.5 as 1/2. -12 3/4 0.5 -12 3/4 1/2 Multiply. -12 3/8 Simplify. Write in horizontal form. -12 ÷ 3/8 Write as multiplication. -12/1 x 8/3 Simplify. -4/1 x 8/1 = -32/1 = -32

12. Homework PB, p 147-148

13. x – 2 5/8 = 1 1/4 x – 2 5/8 = 1 1/4 Addition/Subtraction Equations With Fractions Example 3. Solve and check. + 2 5/8 x = 1 2/8 + 2 5/8 x = 3 7/8 Substitute 3 7/8 for x to check. x – 2 5/8 = 1 1/4 3 7/8 – 2 5/8 = 1 1/4 1 2/8 = 1 1/4 Simplify. 1 1/4 = 1 1/4; true 3 7/8 is a solution.

14. Homework PB, p 149-150

15. -5/8 – 1/8 + n = 1 Addition/Subtraction Equations With Fractions Example 2. Solve and check. Combine like terms. -6/8 + n = 1 Simplify. Add 6/8 to both sides. +6/8 n1 6/8 = Simplify the fraction n = 1 3/4 Check the solution. Replace n with 1 3/4 -5/8 – 1/8 + 1 3/4 = 1 -6/8 + 1 3/4 = 1

16. Addition/Subtraction Equations With Fractions Example 2. Solve and check. n = 1 3/4 Check the solution. -5/8 – 1/8 + 1 3/4 = 1 -6/8 + 1 3/4 = 1 Combine Simplify -3/4 + 1 3/4 = 1 True, so 1 3/4 is a solution. Replace n with 1 3/4. -5/8 – 1/8 + n = 1 Add. 1 = 1

17. Multiplication and division equations with fractions objective:  apply the Multiplication Property of Equality Text, pp 136-137

18. 1/4 w + 2/4 w = 15 Example 1. Solve and check. Combine like terms. Multiply both sides by 4/3. 3/4 w = 154/3 w = 60/3 Divide. w = 20 Check. Substitute 20 for w. 1/4 (20) + 2/4 (20) = 15 Simplify. 5 + 10 = 15 True. So 20 is a solution

19. Homework PB, p 151-152

20. Two-Step equations with fractions objective:  apply the properties of equality to simplify two- step equations with fractions Text, pp 138-139

21. 1/2 p –16 1/2 = 15 Example 1. Solve and check. Add 16 ½ to both sides. Multiply both sides by 2/1. Check. +16 1/2 1/2 p = 31 1/2 2/1 p = 31 1/2 2/1 Rename 31 1/2 as improper fraction p = 63/2 2/1=63/1= 63 1/2 p –16 1/2 = 15 Substitute 63 for p. 1/2 63 –16 1/2 = 15

22. Example 1. Solve and check. 1/2 p –16 1/2 = 15 Substitute 63 for p. 1/2 63 –16 1/2 = 15 Multiply. 31 1/2 –16 1/2 = 15 Subtract. 15 = 15 True, so 63 is a true solution

23. 59 Example 2. Solve and check. Rename 2 1/4 as a fraction 42 d = 2 1/4 (-17) Simplify the grouping symbols. 59 = d 2 1/4 +17 Subtract 17 from both sides. -17 = d 2 1/4 42 d 9/4 = Multiply both sides by 4/9 42 9/4= d Multiply

24. Example 2. Solve and check. 42 9/4= d Simplify. 21 2 9/2 21= d Multiply. 189/2= d Rename as mixed number. 94 1/2= d Check. Use 94 1/2 in place of d. 59 d 2 1/4 = (- 17)

25. Example 2. Solve and check. Use 94 1/2 in place of d. 59 d 2 1/4 = (- 17) 59 94 1/2 2 1/4 (- 17) Simplify the parentheses. 59= 94 1/2 2 1/4 + 17 Write the division in horizontal form. 59=94 1/2÷ 2 1/4+ 17 59 = 189/2 ÷ 9/4+ 17 Write the division in horizontal form. Rename as fractions. Write as multiplication.

26. Example 2. Solve and check. 59 = 189/24/9 + 17 Simplify. 21 1 59 = 21/24/1+ 17 Simplify. 1 2 59 = 212+ 17 Multiply 59 = 42+ 17 Add. 59 = True. So 94 1/2 is a solution.

27. Homework PB, p 153-154 Class work PB, p 153

28. Customary units of measure objective:  rename customary units measure to a larger or smaller units Text, pp 138-139

29.  Customary units of length 1 foot (ft) = 12 inches (in) 1 yard (yd) = 3 ft or 36 in 1 mile (mi) = 5280 ft or 1760 yd

30.  Customary units of capacity 1 cup (c) = 8 fluid ounces (fl oz) 1 pint (pt) = 2 c 1 quart (qt) = 2 pt 1 gallon (gal) = 4 qt

31.  Customary units of weight 1 pound (lb) = 16 ounces (oz) 1 ton (T) = 2000 lb

32.  Customary units of Measure TT o rename larger units as smaller units, multiply by the conversion unit TT o rename smaller units as larger units, divide by the conversion unit

33. Example. How many yards are there in 2 ½ miles? Think.  2 1/2 mi = _________ yd 11 mi = 1760 yd  m i larger than yard ll arger to smaller, multiply 2 1/2 mi 1760 ydRename as fraction. 5/2 mi 1760 ydSimplify. 8801 5 mi 880 yd Multiply. 4400 yd 2 1/2 mi.

34. Homework PB, p 155-156 Class work PB, p 155

35. Problem solving strategy: objective:  s s olve word problems using the strategy “Make A Drawing” Text, pp 138-139 Make a drawing

36. Sample Problem 1. The clock tower in Liberty Square, known for its accuracy, chimes its bell every hour on the hour at equal intervals. If the clock strikes 6 chimes in 6 seconds, how long would it take for the clock to strike 12 chimes at 12 o’clock? (To complete the problem, assume that the chime itself takes no time) Hint: 12 seconds is not the answer.

37. Read Read to understand what is being asked. (List the facts and restate the question.) Facts:  Chime occurs in equal intervals.  6 chimes strike in 6 seconds at 6 o’clock.  The answer is not twelve seconds. Question:  H ow long would it take for the clock to strike 12 chimes at 12 o’clock.

38. Plan Select a strategy.  Guess and test.  Organize data.  Find a pattern. Problem-Solving Strategies  Make a drawing.  Reason logically  Work backward  Solve a simpler problem.  Adopt a different point of view.  Account for all possibilities.  Consider extreme cases. Using the strategy “Make a Drawing” will help you understand the situation.

39. Solve Apply the strategy. First make a drawing that help you understand the situation. Use dots to show the chimes that occur at 6 o’clock. 654321 12345 The 6 chimes occur in 6 seconds. There are 5 intervals in those 6 chimes, therefore each interval must be 6/5 seconds. Think: 6/5 5 = 6. 6 sec

40. Solve Apply the strategy. Now make a drawing to show the situation at 12 o’clock. Use dots also to show the chimes. 98765 12345 There are 11 intervals between the 12 chimes at 12 o’clock. If an interval is 6/5 of a second, then 6/5 11 will give us what it will take for the 12 chimes the clock will make at twelve. 1234101112 67891011

41. Solve Apply the strategy. There are 11 intervals between the 12 chimes at 12 o’clock. If an interval is 6/5 of a second, then 6/5 11 will give us the it will take for the 12 chimes the clock will make at twelve. 6/5 11 =66/5 =13 1/5 seconds The clock takes13 1/5 seconds to strike 12 chimes Check Check to make sure your answer makes sense There are twice as many chimes, so it ought to take twice as long. It appears to be so.

42. Check Check to make sure your answer makes sense There are twice as many chimes, so it ought to take twice as long. It appears to be so. T he 6 chimes occur in 6 seconds. T he 12 chimes occur in 13 1/5 seconds. T here are 5 intervals between the 6 chimes. T here are 11 intervals between the 12 chimes. There are more than twice as many intervals, so it ought to take more than twice as long. It appears to be so.

43. Sample Problem 2. There are 240 seven graders at Kingston Middle School. Of these students, 1/6 walk to school. Of those who do not walk, 3/4 take the bus to school. Of those who do not walk or take the bus half ride their bikes. How many seventh graders ride their bikes to school?

44. Read Read to understand what is being asked. (List the facts and restate the question.) Facts:  There are 240 seventh graders in all  1/6 walk to school.  3/4 of those who do not walk take the bus  1/2 of those who do not walk or take the bus ride their bike. Question:  H ow many seventh graders ride their bike to school?.

45. Plan Select a strategy. This problem has a lot of information. To make this information easier to understand, you can use the strategy “Make a Drawing”.

46. Solve Apply the strategy.  Draw a rectangle to represent the entire seventh grade.  Divide the rectangle to show those who walk and those who do not. 240 40 1/6 walks do not walk 200 Think. 1/6 of 240 is 40.

47.  Divide the section representing those who do not walk into fourths. do not walk 20040 50  Divide the section representing those who do not walk into fourths. walks Think. 1/4 of 200 is 50

48. 40 50 walks Divide the remaining fourth into two. 25 do not walk or take the bus So 25 students ride their bikes to school. Think. 1/2 of 50 is 25.

49. Check Check to make sure your answer makes sense. Look back at the final drawing. Make sure the numbers that represent each section satisfy the condition in the problem.  The total is 40 + 50 + + + 25 + = 240.  40 students walk. This is 1/6 of 240 students.  150 students ride the bus. This is 3/4 of the 200 students who do not walk.  25 students ride their bikes. This is 1/2 of the 50 who do not walk or ride the bus.

50. Different Ways to find GCF objective:  u u se two other ways of finding the GCF of two numbers. Text, pp 144

51. Method 1. division Example. Find the GCF of 72 and 56. 72 and 56. Divide the higher number by the lower number. 72 ÷ 56 = 1r16 If the remainder is 0, the lower number is the GCF. If not divide the divisor by the remainder. Continue this process until the remainder is 0. The last divisor is the GCF. 56 ÷ 16 = 3r 8 16 ÷ 8 = 2r 0

52. Method 2. Subtraction Example. Find the GCF of 72 and 56. 72 – 56 = 16. SS ubtract the lower number from the higher number. 56 – 16 = 40 CC ompare the three numbers. SS ubtract the lowest from the next lowest. CC ontinue the process until the last two numbers in the sentence are the same. That number is the GCF. 40 – 16 = 24 24 – 16 = 8 16 – 8 = 8

53. Homework PB, p 159 Class work PB, p 144