1. A LGEBRA I(A) CH3 N OTES Winter, 2010-2011 Ms. Ellmer 1

2. C H 3-1 S OLVING TWO -S TEP E QUATIONS Background: By writing an algebraic equation, you can use it for any values of the variable. So it has endless uses! Vocab: Equation: Expression with an = sign “Solve for x:” x is by itself on one side of = sign Term: a number, a variable, or the product of a number and any variable(s). “Combine Like Terms:” x’s combine with x’s, y’s combine with y’s, x 2 combines with x 2, numbers combine with numbers ONLY Opposite Functions: called inverse operations, operations that undo each other. Opposite Functions: +-+- * ∙ / x 2 √x 2 2

3. C H 3-1 S OLVING TWO -S TEP E QUATIONS IN ORDER TO SOLVE EQUATIONS, YOU MUST BE ABLE TO COMBINE LIKE TERMS PROPERLY!!!! LET’S PRACTICE! 3

4. C H 3-1 S OLVING TWO -S TEP E QUATIONS Ex.1 Combine all like terms in the following expressions. a. 2x + 36 2x + 36 b. (42+4x) + 83 4x +125 c. (42+4x) + 83 + 3(y-3) 4x +3y +116 d. 4x+3y+2x 2 +16+9y 2x 2 +4x +12y +16 4

5. C H 3-1 S OLVING TWO -S TEP E QUATIONS IN ORDER TO SOLVE EQUATIONS, YOU MUST BE ABLE TO IDENTIFY AND USE OPPOSITE FUNCTIONS CORRECTLY!!!! LET’S PRACTICE! You must use the OPPOSITE FUNCTION to move stuff from one side of the = to the other. 5

6. C H 3-1 S OLVING TWO -S TEP E QUATIONS Ex.2: Identify the opposite function needed to get the variable isolated to one side. a. x – 6 = 13 +6 = +6 ADDING b. 2x = 52 ---- = ---- 2 2 DIVIDING c. x + 8 = 56 - 8 = -8 SUBTRACTING d. x 2 = 36 √ x 2 = √ 36 TAKE THE SQUARE ROOT Opposite Functions +- *∙ / x 2 √x 2 6

7. C H 3-1 S OLVING TWO -S TEP E QUATIONS Now that you know what “Combine Like Terms” means and what opposite functions do, you can follow a simple recipe to solve equations. 7

8. C H 3-1 S OLVING TWO -S TEP E QUATIONS 8 Recipe to Solve Equations Step1: Get x term(s) alone on one side. Step2: Combine Like Terms. Step3: Isolate x using opposite functions. Step4: Plug x value back in to original question and check answer.

9. C H 3-1 S OLVING TWO -S TEP E QUATIONS 9 Ex.3 Solve the equation. 36.9 = 3.7b – 14.9 +14.9 = +14.9Step 1 & 2 51.8 = 3.7b ----- = -----Step 3 3.7 3.7 14 = b 36.9 = 3.7(14) – 14.9Step 4 36.9 = 36.9

10. CH3-1 S OLVING T WO -S TEP E QUATIONS 10 ALWAYS WORK DOWN YOUR PAGE

11. CH3-1 S OLVING T WO -S TEP E QUATIONS Now you try EVENS 2-34 11

12. CH3-2 S OLVING M ULTI -S TEP E QUATIONS If you follow the recipe, it doesn’t matter how big the equation gets….you can always know what to do. 12

13. CH3-2 S OLVING M ULTI -S TEP E QUATIONS Ex. 1 Solve each equation. Check your answer. 2w - 5w + 6.3 = -14.4 -6.3 = - 6.3 2w – 5w = -20.7 -3w = -20.7 ----- ------ -3 w = 6.9 2(6.9) -5(6.9) + 6.3 = -14.4 13.8-34.5+6.3 = -14.4 -14.4 = -14.4 13 Recipe to Solve Equations Step1: Get x term(s) alone on one side. Step2: Combine Like Terms. Step3: Isolate x using opposite functions. Step4: Plug x value back in to original question and check answer.

14. CH3-2 S OLVING M ULTI -S TEP E QUATIONS Ex.2 Solve each equation. Check your answer. 2(m+1) = 16 2m +2 = 16 -2 = -2 2m = 14 ---- 2 m = 7 2(7+1) = 16 2(8) = 16 16 = 16 14 Recipe to Solve Equations Step1: Get x term(s) alone on one side. Step2: Combine Like Terms. Step3: Isolate x using opposite functions. Step4: Plug x value back in to original question and check answer.

15. CH3-2 S OLVING M ULTI -S TEP E QUATIONS Now, you try ODDS 1- 49 15

16. CH 3-3 E QUATIONS W /V ARIABLES ON B OTH S IDES What do we do if we have variables on both sides of the = sign?????? THE SAME RECIPE! 16 Recipe to Solve Equations Step1: Get x term(s) alone on one side. Step2: Combine Like Terms. Step3: Isolate x using opposite functions. Step4: Plug x value back in to original question and check answer.

17. CH 3-3 E QUATIONS W /V ARIABLES ON B OTH S IDES Ex. 1 Solve each equation. Check your answer. Write identity and no solution if applicable. #1. 7 – 2n = n – 14 -n = -n 7 – 3n = -14 -7 = - 7 - 3n = -21 ----- = ------ -3 -3 n = 7 Plug it back in to check answer: -7 = -7 17 Recipe to Solve Equations Step1: Get x term(s) alone on one side. Step2: Combine Like Terms. Step3: Isolate x using opposite functions. Step4: Plug x value back in to original question and check answer.

18. CH 3-3 E QUATIONS W /V ARIABLES ON B OTH S IDES Ex. 1 Solve each equation. Check your answer. Write identity and no solution if applicable. #7. 3(n-1) = 5n +3 -2n 3n -3 = 3n + 3 -3n = -3n -3 = +3 Where did n go? Canceled out, so NO SOLUTION 18 Recipe to Solve Equations Step1: Get x term(s) alone on one side. Step2: Combine Like Terms. Step3: Isolate x using opposite functions. Step4: Plug x value back in to original question and check answer.

19. CH 3-3 E QUATIONS W /V ARIABLES ON B OTH S IDES Ex. 1 Solve each equation. Check your answer. Write identity and no solution if applicable. #5. 8z – 7 = 3z – 7 + 5z 8z – 7 = 8z – 7 Is this the same exact thing On the left side as right Side? If yes, then IDENTITY 19 Recipe to Solve Equations Step1: Get x term(s) alone on one side. Step2: Combine Like Terms. Step3: Isolate x using opposite functions. Step4: Plug x value back in to original question and check answer.

20. C H 3-3 E QUATIONS W /V ARIABLES ON B OTH S IDES Now, you try EVENS 2-38 20

21. C H 3-4 R ATIO AND P ROPORTION Background: Ratios and proportions have many uses in many industries. They can be used to read a map, mix chemicals in painting and landscaping, mix cleaners in home improvement projects, scaled drawings, and finding unit prices while grocery shopping. Vocabulary: Ratio: A comparison of two numbers. Written in 3 ways: 1. a to b 2. a:b 3. a b Unit Rate: Any number over 1 with units “something per something else” 21

22. C H 3-4 R ATIO AND P ROPORTION How to Use It: In Science, unit rates allow you to “cancel your units,” or use dimensional analysis to get the units you want. Ex.1 40.56(km) ∙ (1 mi) = (hr) 1.6 (km) 25.35 mi/hr …..on a 10 speed bike!!!!! 22

23. C H 3-4 R ATIO AND P ROPORTION Ex.2 Page 143 Lance Armstrong! In 2004, Lance Armstrong won the Tour de France completing the 3391 km course in about 83.6 hours. Find Lance’s average speed using v=d/t. d=3391 km t = 83.6 hr v = ? v = d t v = (3391 km) (83.6 hr) v = 40.6 km/hr 23

24. C H 3-4 R ATIO AND P ROPORTION Vocabulary Continued: Proportion: is an equation that states that two ratios are equal, written as: a = c bd And you read it as, “a is to b as c is to d” What is the difference between a set of ratios and a proportion????? THE = SIGN IS IN THE PROPORTION ONLY!!!!!! 24

25. C H 3-4 R ATIO AND P ROPORTION Ex.3 Solve for x. 1:16 = ? : 36 1= x 16 36 What should we do now? Yep, cross multiply and start flexing your algebra muscles! x = 2.25 25

26. C H 3-4 R ATIO AND P ROPORTION The proportions can get really big and have variables….no problemo! Ex. 4 Solve each proportion. 2X-2= 2X-4 14 6 6(2X-2) = 14(2X-4) 12x – 12 = 28x – 56 -28x -16x – 12 = -56 + 12 = +12 -16x = -44 x = 2.75 26 Recipe to Solve Equations Step1: Get x term(s) alone on one side. Step2: Combine Like Terms. Step3: Isolate x using opposite functions. Step4: Plug x value back in to original question and check answer.

27. C H 3-4 R ATIO AND P ROPORTION Are we done? Nope, go back in and check your answer…. 2(2.75)-2= 2(2.75)-4 14 6 0.25 = 0.25 YES!!!!! 27

28. C H 3-4 R ATIO AND P ROPORTION Now, you do ODDS 1- 45 28

29. 3-5 P ROPORTIONS AND S IMILAR S HAPES Background: Proportions can help determine the dimensions for certain objects, by comparing two similar shapes. This method is used by architects, designers, and computer aided drafting (CAD) software. Vocabulary: Corresponding side: a side of one object that can be compared to a side of another object in the same location/dimension. 29

30. Ex. 8 corresponds to ______ 12 10 corresponds to _______ 22 30 3-5 P ROPORTIONS AND S IMILAR S HAPES 8 10 12 22

31. How to Use It: Ex.1 Each pair of figures is similar. Find the length of x. 2.5= 5 x 3 2.5(3) = 5(x) 7.5 = 5x 5 5 1.5 = x 31 3-5 P ROPORTIONS AND S IMILAR S HAPES 5 2.5 x 3

32. Now, you do 1-8 ALL PROBLEMS!!!!! 32 3-5 P ROPORTIONS AND S IMILAR S HAPES

33. C H 3-6 E QUATIONS AND P ROBLEM S OLVING You do story problems 1-13 ODDS 33

34. CH 3-7 P ERCENT OF C HANGE Background: Describing relationships using percents can be seen in shopping/pricing, engine efficiencies and grades. Vocabulary: Percent of change: the ratio of the amount of change over the original amount, or % change = amount of change * 100% original amount % error = estimated – actual * 100% actual 34

35. CH 3-7 P ERCENT OF C HANGE How To Use It: Ex.1 Find the percent of change. Describe it as an increase or decrease. 40 cm to 100 cm Amount of change = 100 cm – 40 cm = 60 cm % change = amount of change * 100% original amount % change = 60 cm * 100% 40cm % change = 1.5 * 100% %change = 150% INCREASE 35

36. CH 3-7 P ERCENT OF C HANGE How To Use It: Ex.2 A student estimated the mass of the Physics textbook to be 1750 grams. After measuring the mass on a triple beam balance, the actual textbook mass was 2450 grams. Find the student’s percent error. % error = estimated – actual * 100% actual % error = 1750 – 2450* 100% 2450 %error = 28.6% Do you want a high % error or a low % error? LOW! < 10% in industry is EXCELLENT! 36

37. CH 3-7 P ERCENT OF C HANGE Now you do EVENS 2-26 37

38. CH 3-8 F INDING AND E STIMATING S QUARE R OOTS Background: Square roots and perfect squares are used in many problems including construction and science lab work. Vocabulary: Square root: a special case for a number, that when multiplied by itself, or squared, it gives that number. Ex. 4 2 = 16, so 4 and -4 are square roots of 16 Symbol = √ with a number under this symbol YOU CAN’T HAVE A – SIGN UNDER √ Perfect Square: squares of integers 38 Integer12345678910 Square149162536496481100

39. How To Use It: Ex.1 Use a calculator to find the value of each square root. a. +/- √38 +6.2 or -6.2 b. √19.38 4.4 c. √400 20 39 CH 3-8 F INDING AND E STIMATING S QUARE R OOTS

40. Now you do ODDS 1-19 in 10 minutes! 40

41. You can also simplify a radical instead of solving it. To simplify, look for perfect squares to be taken out of square root sign. How To Use It: Ex.1 Simplify the expression. √32 √8∙4 √2∙4 ∙4 2∙2√2 4√2 41 CH 3-8 F INDING AND E STIMATING S QUARE R OOTS

42. Now you do ODDS 21,27,31-43 42

43. Background: This dude, Pythagoreas lived around 500 B.C. He developed a way to determine the lengths of sides of a right triangle & applied it in the construction industry. Vocabulary: Right triangle: A with one 90° angle Hypotenuse: The longest leg of a Right and IS ALWAYS ACROSS FROM THE RIGHT ANGLE BOX Symbol: c Legs: The other two sides of a Right Symbols: a and b C H 3-9 T HE P YTHAGOREAN T HEOREM 43

44. Let’s label the sides of this triangle: C H 3-9 T HE P YTHAGOREAN T HEOREM 44

45. Pythagorean Theorem: c 2 = a 2 + b 2 NOTE: Must be a Right Triangle! C H 3-9 T HE P YTHAGOREAN T HEOREM c a b 45

46. How to Use It: Ex.1 Find the missing side of the right triangle. a=12 b=? c=35 c 2 =a 2 +b 2 35 2 =12 2 +b 2 1225=144+b 2 1081=b 2 √1081=√ b 2 32.9 = b C H 3-9 T HE P YTHAGOREAN T HEOREM 46

47. How to Use It: Ex.2 Determine whether the given lengths are sides of a right triangle. 20,21,29 c 2 =a 2 +b 2 29 2 =20 2 +21 2 841=400+441 841 = 841 YES C H 3-9 T HE P YTHAGOREAN T HEOREM 47

48. NOW, YOU DO EVENS 2-30 C H 3-9 T HE P YTHAGOREAN T HEOREM 48

49. Yeah! We are done with CHAPTER 3!!!!!!! C H 3 49