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EXAMPLE 1

ANSWER. The final position is 3 . So , – 3 + 6 = 3. EXAMPLE 1 . Add two integers using a number line. Use a number line to find the sum . a . – 3 + 6. ANSWER. The final position is – 9 . So , – 4 + ( – 5 ) = – 9. EXAMPLE 1. Add two integers using a number line. b . – 4 + (– 5).

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EXAMPLE 1

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  1. ANSWER The final position is3. So, –3+6 = 3. EXAMPLE 1 Add two integers using a number line Use a number line to find the sum. a.– 3 + 6

  2. ANSWER The final position is–9. So, –4+ (–5) =–9. EXAMPLE 1 Add two integers using a number line b.– 4 + (– 5)

  3. ANSWER ANSWER 5 – 3 for Example 1 GUIDED PRACTICE Use a number line to find the sum. 1. 7 +(– 2) 2. 8 +(– 11)

  4. ANSWER ANSWER – 4 – 5 for Example 1 GUIDED PRACTICE Use a number line to find the sum. 3. –8 + 4 4.–1 + (– 4)

  5. = 19.3 – –12.2 EXAMPLE 2 Add real numbers Find the sum. a.– 5.3 + (– 4.9) = – ( – 5.3 + – 4.9 ) Rule of same signs = – (5.3 + 4.9) Take absolute values. = – 10.2 Add. b. 19.3 + (–12.2) Rule of different signs =19.3 – 12.2 Take absolute values. = 7.1 Subtract.

  6. Identify properties of addition EXAMPLE 3 Property illustrated Statement Associative property of addition a.(x + 9) + 2 = x + (9 + 2) b.8.3 + (– 8.3) = 0 Inverse property of addition c. – y + 0.7 = 0.7 + (– y) Commutative property of addition

  7. Solve a multi-step problem EXAMPLE 4 BUSINESS The table shows the annual profits of two piano manufacturers. Which manufacturer had the greater total profit for the three years?

  8. Solve a multi-step problem EXAMPLE 4 SOLUTION STEP1 Calculate the total profit for each manufacturer. Manufacturer A: Total profit = – 5.8 + 8.7 + 6.8 = – 5.8 + (8.7 + 6.8) = – 5.8 + 15.5 = 9.7

  9. ANSWER Manufacturer A had the greater total profit. Solve a multi-step problem EXAMPLE 4 Manufacturer B: = – 6.5 + 7.9 + 8.2 Total profit = – 6.5 + (7.9 + 8.2) = – 6.5 + 16.1 = 9.6 STEP2 Compare the total profits: 9.7 > 9.6.

  10. for Examples 2, 3 and 4 GUIDED PRACTICE Find the sum. 5. – 0.6 + (– 6.7) = – 7.3 6. 10.1 + (– 16.2) = – 6.1 7. – 13.1 + 8.7 = – 4.4 Identify the property being illustrated. Inverse property of addition 8.7 + (– 7) = 0 9.– 12 + 0 = – 12 Identity property of addition 10. 4+ 8 = 8 + 4 Commutative property of addition

  11. 11. WHAT IF?In Example 4, suppose that the profits for year 4 are – $1.7 million for manufacturer A and – $2.1 million for manufacturer B. Which manufacturer has the greater total profit for the four years? for Examples 2, 3 and 4 GUIDED PRACTICE

  12. for Examples 2, 3 and 4 GUIDED PRACTICE ANSWER Manufacturer A has the greater total profit.

  13. EXAMPLE 1 Subtract real numbers Find the difference. a.– 12–19 =– 12+ (–19) = – 31 b.18 – (–7) = 18 + 7 = 31

  14. 3. 1 1 – 1 – 3 2 6 = for Example 1 GUIDED PRACTICE Find the difference. 1.– 2–7 = – 9 2. 11.7– (–5) = 16.7

  15. EXAMPLE 2 Evaluate a variable expression Evaluate the expression y – x + 6.8 when x = – 2 and y = 7.2. y–x+ 6.8 =7.2– (–2)+ 6.8 Substitute –2 for xand 7.2 for y. =7.2 + 2 +6.8 Add the opposite –2. = 16 Add.

  16. EXAMPLE 3 Evaluate Change Temperature One of the most extreme temperaturechanges in United States history occurred in Fairfield, Montana, on December 24, 1924. At noon, the temperature was 63°F.By midnight, the temperature fell to – 21°F. What was the change in temperature? SOLUTION The change Cin temperature is the difference of the temperature mat midnight and the temperature nat noon.

  17. C = – m n m – n C = EXAMPLE 3 Evaluate Change STEP1 Write a verbal model. Then write an equation. STEP2 Find the change in temperature. Write equation. = – 21 – 63 Substitute values. = – 21 + (– 63) Add the opposite of 63. = – 84 Add –21and –63.

  18. ANSWER The change in temperature was – 84°F. EXAMPLE 3 Evaluate Change

  19. 7. Car values x – y + 8 y – (x –2) (y – 4)– x 4. 5. 6. ANSWER – $2700 for Example 2 and 3 GUIDED PRACTICE Evaluate the expression when x = – 3 and y = 5.2. =–0.2 =10.2 =4.2 A new car is valued at $15,000. One year later, the car is valued at $12,300. What is the change in value of car?

  20. (–10) (–4) = 1 2 1 c. – (–4) (–3) Multiply – and –4 2 EXAMPLE 1 Multiply real numbers Find the product. a. – 3 (6) = – 18 Different signs; product is negative. b. 2 (–5) (–4) Multiply2and –5. Same signs; product is positive. = 40 = 2 (– 3) = – 6 Different signs; product is negative.

  21. 4 3 3. (–3) (7) for Example 1 GUIDED PRACTICE Find the product. 1. – 2 (– 7) = 14 2. – 0.5 (– 4) (– 9) = – 18 = – 28

  22. a.(x 7) 0.5 = x (7 0.5) b.8 0 = 0 y (– 6) c.– 6 y = d.9 (– 1) = – 9 e. 1 v = v EXAMPLE 2 Identify properties of multiplication Statement Property illustrated Associative property of multiplication Multiplicative property of zero Commutative property ofmultiplication Multiplicative property of – 1 Identity property of multiplication

  23. 4. –1 8 = – 8 5. x 12 12 x = (y 4) 9 = y (4 9) 6. 7. 0 (– 41) = 0 – 6 – 5 (– 6) = 8. (– 5) –13 (– 1) – 8 9. = for Example 2 GUIDED PRACTICE Identify the property illustrated. Multiplicative property of – 1 Commutative property ofmultiplication Associative property of multiplication Multiplicative property of zero Commutative property ofmultiplication Multiplicative property of – 1

  24. Find the product (–4x)0.25. Justify your steps. (–4x)0.25 = 0.25 (–4x) = (0.25 (–4))x =–1 x EXAMPLE 3 Identify properties of multiplication Commutative property of multiplication Associative property of multiplication Product of 0.25 and –4 is –1. Multiplicative property of –1. =–x

  25. EXAMPLE 4 Solve a multi-step problem Lakes In 1900 the elevation of Mono Lake in California was about 6416 feet. From 1900 to 1950, the average rate of change in elevation was about – 0.12 foot per year.From 1950 to 2000, the average rate of change was about – 0.526 foot per year. Approximate the elevation in 2000.

  26. New elevation (feet) Original elevation (feet) Time passed (years) Average rate of change (feet/year) + • = EXAMPLE 4 Solve a multi-step problem SOLUTION STEP 1 Write a verbal model. STEP 2 Calculate the elevation in 1950. Use the elevation in 1900 as the original elevation. The time span 1950 – 1900 = 50years. New elevation = 6416 +(–0.12)(50) Substitute values. = 6416 + (–6) Multiply –0.12 and 50. =6410 Add 6416 and –6.

  27. EXAMPLE 4 Solve a multi-step problem STEP 3 Calculate the elevation in 2000. Use the elevation in 1950 as the original elevation. The time span 2000 – 1950 = 50years. 6410 +(– 0.526)(50) New elevation = Substitute values. = 6410 + (–26.3) Multiply –0.526 and 50. =6383.7 Add 6410 and –26.3.

  28. ANSWER The elevation in 2000 was about 6383.7 feet above sea level. EXAMPLE 4 Solve a multi-step problem

  29. 3 3 (5y) 10. ( 5)y = 10 10 3 3 3 Product of and 5 is . 10 2 2 y Multiplicative property of . y 3 3 2 2 = = for Example 3 and 4 GUIDED PRACTICE Find the product. Justify your steps. Associative property of multiplication

  30. 11. 0.8 (–1)(–x) 0.8 (–x)(–1) = for Example 3 and 4 GUIDED PRACTICE Commutative property of multiplication Multiplicative property of –1 = – 0.8 (–x) = 0.8x Multiply

  31. (–y)(–0.5)(–6) 12. for Example 3 and 4 GUIDED PRACTICE Product of –0.5 and –6 is 3. = (–y)(3) = –3y Commutative property of multiplication

  32. ANSWER 13. Using the data in Example 4, approximate the elevation of Mono Lake in 1925 and in 1965. about 6413 ft; about 6402.11 ft for Example 3 and 4 GUIDED PRACTICE

  33. – – – – – – – 5 is because a. The multiplicative inverse of =1. (– 5) 7 1 6 1 6 7 is because b. The multiplicative inverse of 7 5 5 6 7 6 =1. EXAMPLE 1 Find multiplicative inverses of numbers

  34. 5 3 1 4 3 5 –16 –16 4 a. = –4 = b. –20 – –20 – = EXAMPLE 2 Divide real numbers Find the quotient. = 12

  35. 4. – 3. – 1 7 1 4 – – 27 4 7 3 1 – 8 for Examples 1 and 2 GUIDED PRACTICE Find the multiplicative inverse of the number. 1. – 27 2. – 8 – 3

  36. 5. – 64 (– 4) 6. – 2 3 2 1 9 5 4 8 3 10 7. 18 – =–1 1 8. – 18 45 =– for Examples 1 and 2 GUIDED PRACTICE Find the quotient. = 16 =– 81

  37. EXAMPLE 3 Find the mean TEMPERATURE The table gives the daily minimum temperatures (in degrees Fahrenheit) in Barrow, Alaska, for the first 5 days of February 2004. Find the mean daily minimum temperature.

  38. –21 + (–29) + (–39) + (–39) + (–22) = 5 150 – = 5 – 30 = The mean daily minimum temperature was –30°F. ANSWER EXAMPLE 3 Find the mean SOLUTION To find the mean daily minimum temperature, find the sum of the minimum temperatures for the 5 days and then divide the sum by 5. Mean

  39. 36x 24 36x 24 – – . Simplify the expression 6 6 ( ) 36x 24 – 6 = ( ) 36x 24 – = 1 1 1 6 6 6 36x – 24 = 6x– 4 = EXAMPLE 4 Simplify an expression Rewrite fraction as division. Division rule Distributive property Simplify.

  40. ANSWER The mean is 0.575. for Examples 3 and 4 GUIDED PRACTICE 9. Find the mean of the numbers –3, 4, 2.8, and –1.5 . Temperature 10. Find the mean daily maximum temperature (in degrees Fahrenheit) in Barrow, Alaska, for the first 5 days of February 2004.

  41. ANSWER The mean daily minimum temperature was –16.8°F. for Examples 3 and 4 GUIDED PRACTICE

  42. –10z– 20 13. –5 2x– 8 –6y+18 11. 12. –4 3 1 2 – x = + 2 for Examples 3 and 4 GUIDED PRACTICE Simplify the expression = –2y + 6 = 2z + 4

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