1. XEI: E XPRESSIONS, EQUATIONS, AND INEQUALITIES

2. EXPRESSIONS What is an expression? An expression is a mathematical statement that consists of terms and represents a value The difference between expressions and equations is that equations have an equal sign and expressions do not. If we see an expression with an equal sign, then there is no value on the other side of the equal sign. What are terms? Terms are parts of an expression separated by a “+” or “-” sign.

3. EXAMPLES OF EXPRESSIONS Example 1 – 1 + 1 + 1 – 5 is an expression Example 2 – (6 + 7 – 10) ∙ 2² is an expression Example 3 – z – (7 – 6 + x) is an expression Example 4 – -4x² - 3x + 2 is an expression Example 5 – x is an expression

4. EVALUATING EXPRESSIONS When we evaluate expressions, we determine the numerical value of the expression. We must use the correct order of operations when we evaluate expressions. Order of Operations: 1. Parentheses 2. Exponents 3. Multiply/Divide (whichever comes first from left to right) 4. Add/Subtract (whichever comes first from left to right)

5. E XAMPLES OF E VALUATING E XPRESSIONS Example 1 – 1 + 1 + 1 – 5 = ? -2 Example 2 – (6 + 7 – 10) ∙ 2² = ? 3 · 2² 3 · 4 12

6. M ORE E XAMPLES OF E VALUATING E XPRESSIONS

7. Y OU T RY

8. S IMPLIFYING E XPRESSIONS What are like terms? Like terms are terms that either have the same variable and power or have no variable at all. Example of like terms – 7x + 2 – 4x – 1 In the expression above, 7x and -4x are like terms, and 2 and -1 are like terms. What does it mean to “combine” like terms? When like terms are “combined,” they are added together.

9. C OMBINING L IKE T ERMS Example 1 – Simplify the following expression by combining like terms. x + 10 – 4 x + 10 – 4 simplifies to x + 6 since 10 and -4 are like terms Example 2 – Simplify the following expression by combining like terms. -2b -10 + b – 1 -2b -10 + b – 1 simplifies to –b – 11 since -2b and b are like terms, and -10 and -1 are like terms.

10. Y OU T RY 1. Simplify x + 3 + 2x + 10 2. Simplify 7x + 4x 3. Simplify 6p – 9p

11. D ISTRIBUTIVE P ROPERTY What is the Distributive Property? a(b + c) = a ∙ b + a ∙ c Examples of Applying the Distributive Property 1) 5(9 + 9n)  5 ∙ 9 + 5 ∙ 9n  45 + 45n 2) 9(-9m + 1)  9 ∙ -9m + 9 ∙ 1  -81m + 9 3) -7(6 – 7n)  (-7) ∙ 6 – (-7) ∙ 7n  -42 – (-49n)  -42 + 49n

12. Y OU T RY 1. Simplify the following expression: 2(-7v – 1) 2. Simplify the following expression: -10(n – 1)

13. S IMPLIFYING E XPRESSIONS Example 1 – Simplify the following expression by using the Distributive Property and combining like terms: -3(5x + 10) + 7 (-3)∙ 5x + (-3) ∙ 10 + 7 distributive property -15x + (-30) + 7 -15x + (-23) combine like terms -15x - 23

14. M ORE Y OU T RY 1. Simplify the following expression by using the Distributive Property and combining like terms: -10(1 – x) + 8x 2. Simplify the following expression by using the Distributive Property and combining like terms: 2m + 9(6m + 10)

15. S OLVING L INEAR E QUATIONS When we solve linear equations, we are trying to find the numerical value of the variable. In trying to find the numerical value of the variable, we are isolating the variable by performing inverse operations. Sometimes, this requires us to combine like terms, use the Distributive Property, or do both.

16. E XAMPLES OF S OLVING L INEAR E QUATIONS Example 1 – Solve the following equation: -5n + 3 + 5 = 13 -5n + 8 = 13 combine like terms - 8 - 8 inverse operations -5n = 5 ÷ -5 ÷ -5 inverse operations n = -1

17. M ORE E XAMPLES OF S OLVING L INEAR E QUATIONS Example 2 – Solve the following equation: 3(5a + 5) = 75 15a + 15 = 75 Distributive Property - 15 - 15 Inverse operations 15a = 60 ÷ 15 ÷ 15 Inverse operations a = 4

18. M ORE E XAMPLES OF S OLVING L INEAR E QUATIONS Example 3 – Solve the following linear equation: 6x – 2 = 3x + 7 - 3x -3x Inverse operations to move variable to left side 3x – 2 = 7 + 2 + 2 Inverse operations 3x = 9 ÷ 3÷ 3 Inverse operations x = 3

19. M ORE E XAMPLES OF S OLVING L INEAR E QUATIONS Example 4 – Solve the following linear equation: 5 – 4(2 – 4a) = 6a + 27 5 + (-4)(2 – 4a) = 6a + 27 Change subtraction to “adding a negative” 5 + (-4) ∙ 2 – (-4) ∙ 4a = 6a + 27 Distributive Property 5 + (-8) – (-16a) = 6a + 27 5 + (-8) + 16a = 6a + 27 Change “double negative” to addition -3 + 16a = 6a + 27 Combine like terms 16a – 3 = 6a + 27 Commutative Property to put variable first - 6a - 6a Inverse operations to move variable to left side 10a – 3 = 27 + 3 + 3 Inverse operations 10a = 30 ÷ 10 ÷ 10 Inverse operations a = 3

20. Y OU T RY 1. Solve the following equation: -2 – 6(x + 2) = -5x -16 2. Solve the following equation: 6 – 2(3 – 3m) = -12 – 6m

21. S ETTING UP AND S OLVING E QUATIONS When we set up equations from word problems, we must first identify all of our variables (known and unknown). Next we must translate the words in the problem to math symbols so that we can derive an equation. Then we replace all known variables with their numerical value and solve for the unknown variable. Finally, answer the question in a complete sentence.

22. E XAMPLE OF S ETTING UP AND S OLVING E QUATIONS Example 1 – A skating rink charges $125 per hour and $6 per guest in order to rent the rink for a party. If we rented the rink for two hours and 200 people attended the party, then how much did the rink charge us? Solution – Let C be the total amount that the rink charges. Let h be the number of hours the rink is rented. Let p be the number of people who attend the party C = 125 h + 6 p C = 125(2) + 6(200) C = 250 + 1200 C = 1450 The rink charged $1450.

23. M ORE E XAMPLES OF S ETTING UP AND S OLVING E QUATIONS Example 2 – A large bookshelf holds 50 books, and a small bookshelf holds 20 books. How many books are there if 4 large bookshelves and 6 small bookshelves are filled? Solution – Let B be the total amount of books. Let l be the number of large bookshelves Let s be the number of small bookshelves B = 50 l + 20 s B = 50(4) + 20(6) B = 200 + 120 B = 320 There are a total of 320 books.

24. A NOTHER E XAMPLE OF S ETTING UP AND S OLVING E QUATIONS Example 3 – A hotel charges $80 per night and $10 per family member per night to rent a room. How much would a family of four be charged to rent a room for five nights? Solution – Let C be the total cost. Let n be the number of nights for which the room is rented. Let f be the number of family members C = 80 n + 10( f)(n) C = 80(5) + 10(4)(5) C = 400 + 200 C = 600 The hotel charged $600 for a family of four to rent a room for five nights.

25. Y OU T RY Lebron James scored 27 points last night. He made six 2-point field goals and nine free throws. How many 3-point field goals did he make?

26. S OLVING “W ORK ” W ORD P ROBLEMS When solving “work” word problems, we are trying to find out how long it takes two or more people to do a job together if we know how long it takes them to do the job individually. Also, we can find out how long it takes a person to do a job individually if we know how long it takes another person to do the job individually and if we know how long it takes them to do the job together.

27. E XAMPLE OF S OLVING “W ORK ” W ORD P ROBLEMS Example 1 – Ijahanna takes 3 hours to do her homework. It takes Emmanuel 4 hours to do his homework. How long would it take them to do their homework if they worked together?

28. A NOTHER E XAMPLE OF S OLVING “W ORK ” W ORD P ROBLEMS Example 2 – It takes Adam eight hours to paint a fence. Stephanie can paint the same fence in ten hours. If they worked together how long would it take them? Solution – Let t be the amount of time it takes them to paint the fence together.

29. Y OU T RY Working alone, it takes Asanji 14 minutes to sweep a porch. Julio can sweep the same porch in 8 minutes. Find how long it would take them if they worked together. Solution –

30. S OLVING “W ORK ” W ORD P ROBLEMS W HEN T RYING T O F IND A N I NDIVIDUAL ’ S T IME Example 1 – John can pick forty bushels of apples in 14 hours. One day his friend Pranav helped him and it only took 7.24 hours. Find how long it would take Pranav to do it alone. Solution – Let x be the amount of time it takes Pranav to do the job alone.

31. A NOTHER E XAMPLE OF S OLVING “W ORK ” W ORD P ROBLEMS W HEN T RYING T O F IND A N I NDIVIDUAL ’ S T IME Example 2 – Working alone, Perry can tar a roof in nine hours. One day his friend Imani helped him and it only took 4.24 hours. How long would it take Imani to do it alone? Solution – Let x be the time it takes Imani to tar a roof alone.

32. Y OU T RY Working together, Amanda and Willie can sweep a porch in 5.32 minutes. Had he done it alone it would have taken Willie 13 minutes. Find how long it would take Amanda to do it alone. Solution –

33. S OLVING D ISTANCE -R ATE - T IME W ORD P ROBLEMS When solving distance-rate-time word problems, we must first know that distance = rate x time or d = r ∙ t Another word for rate is speed. Also, we must know what we are trying to find. Then we must identify what we know already from the problem. Finally, we must set up and solve the equation.

34. E XAMPLE OF S OLVING D ISTANCE - R ATE -T IME W ORD P ROBLEM Example 1 – Joe left the hospital and traveled toward the recycling plant at an average speed of 27 km/h. Kristin left some time later traveling in the same direction at an average speed of 45 km/h. After traveling for three hours, Kristin caught up with Joe. How long did Joe travel before Kristin caught up? Solution – What are we trying to find? We are trying to find out how many hours Joe traveled before Kristin caught up with him. What do we already know? We know Joe’s speed was 27 km/h. We know Kristin’s speed was 45 km/h. We know Kristin’s travel time was 3 hours. We know they both traveled the same distance. We know d = r ∙ t How do we set up an equation? Since they both traveled the same distance, Joe’s “rate x time” must equal Kristin’s “rate x time” Since we don’t know Joe’s travel time, we’ll let it be t. 27 t = 45(3) 27 t = 135 t = 5 So, Joe traveled for 5 hours before Kristin caught up with him.

35. W E T RY Lea left the airport and drove toward the lake at an average speed of 30 mph. Nadia left some time later driving in the same direction at an average speed of 75 mph. After driving for two hours Nadia caught up with Lea. How long did Lea drive before Nadia caught up? Solution –

36. Y OU T RY Kayla left the White House and drove toward Georgetown at an average speed of 20 mph. Jennifer left some time later driving in the same direction at an average speed of 25 mph. After driving for four hours, Jennifer caught up with Kayla. Find the number of hours Kayla drove before Jennifer caught up with her. Solution -

37. A NOTHER E XAMPLE OF S OLVING D ISTANCE -R ATE -T IME W ORD P ROBLEM Example 2 – Gabriella left James’ house and drove toward Uptown. Three hours later, Shayna left driving at 70 mph in an effort to catch up to Gabriella. After driving for two hours, Shayna finally caught up. What was Gabriella’s average speed? Solution – What are we trying to find? We are trying to find Gabriella’s average speed or her rate. What do we already know? We know Shayna’s speed was 70 mph. We know Shayna drove for 2 hours. We know Gabriella drove for 5 hours. We know that d = r ∙ t We know that they went the same distance since Shayna caught up with Gabriella. How do we set up an equation? Since they both traveled the same distance, Gabriella’s “rate x time” must equal Shayna’s “rate x time.” Since we don’t know Gabriella’s rate, we will let it be r. r ∙ 5 = 70(2) 5 r = 140 r = 28 So, Gabriella drove at an average speed of 28 mph.

38. W E T RY A passenger plane flew to Jakarta and back. The trip there took seven hours, and the trip back took five hours. The plane averaged 490 mph on the return trip. Find the average speed of the trip there. Solution –

39. Y OU T RY An aircraft carrier made a trip to Tahiti and back. The trip there took ten hours and the trip back took 13 hours. The aircraft carrier averaged 20 km/h on the return trip. Find the average speed of the trip there. Solution –

40. A NOTHER E XAMPLE OF S OLVING D ISTANCE -R ATE -T IME W ORD P ROBLEM Example 3 – Allan left school two hours before Tony. They drove in opposite directions. Tony drove at 25 mph for one hour. After this time, they were 160 miles apart. What was Allan’s speed?

41. S OLUTION T O P REVIOUS E XAMPLE What are we trying to find? We are trying to find Allan’s speed or rate. What do we already know? We know that Tony and Allan drove in opposite directions from school. We know Tony drove for 1 hour. We know Allan drove for 3 hours, since he left 2 hours before Tony. We know Tony’s speed was 25 mph. We know that after 3 hours of driving for Allan and 1 hour of driving for Tony, they were 160 miles apart. How do we solve? First, we find the distance that Tony drove by multiplying his rate by his time. d = r ∙ t d = 25(1) d = 25 So, Tony drove 25 miles. Next, we find the distance Allan drove by subtracting the distance Tony drove from the distance they were apart. 160 – 25 = 135 Finally, we find Allan’s rate by using the d = r ∙ t formula. d = r ∙ t 135 = r ∙ 3 135 = 3r r = 35 So, Allan drove at a speed of 45 mph.

42. W E T RY A fishing boat left Port 52 and traveled toward Madagascar at an average speed of 10 mph. An aircraft carrier left some time later traveling in the opposite direction with an average speed of 20 mph. After the fishing boat had traveled for four hours, the ships were 120 miles apart. How long did the aircraft carrier travel? Solution –

43. Y OU T RY Ryan left Julio’s house at the same time as Mark. They traveled in opposite directions. Mark traveled at a speed of 65 mph. After one hour, they were 110 miles apart. How fast did Ryan travel? Solution –

44. F UNCTIONS A function is how something works. A function in math is a relation between a set of inputs and a set of outputs. The output of a function f corresponding to an input x is denoted by f ( x ) (read as “ f ” of “ x ”). For example, in the function f ( x) = x ² the output will be the square of the input. So, f( 4) = 4² = 16. Four is the input and 16 is the output.

45. E VALUATING F UNCTIONS Example 1 – f(n) = n + 3 Find f( 0 ) Solution – f( 0 ) = 0 + 3 = 3 Example 2 – k(a) = a ² + 2 a Find k (-7) Solution – k (-7) = (-7)² + 2(-7) = 49 + (-14) = 35

46. Y OU T RY g(n) = 3 n – 4 Find g (9) f(x) = x ² − 4 Find f (-10)

47. P ROPERTIES OF E XPONENTS

48. M ORE P ROPERTIES OF E XPONENTS

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50. P OLYNOMIALS What are polynomials? Polynomials are algebraic expressions defined by their degree and number of terms. What is the degree of a polynomial? The degree of a polynomial is the highest power of the variable in the polynomial. What are terms? Terms are parts of an expression separated by a “+” or “-” sign.

51. E XAMPLES OF POLYNOMIALS Example 1 – 2x² - 3x + 4 is a polynomial 2x², -3x, and 4 are the terms So, this polynomial has 3 terms Example 2 – x – 9 is a polynomial x and -9 are the terms So, this polynomial has 2 terms

52. D EGREES OF P OLYNOMIALS What is the degree of a polynomial? The degree of a polynomial is the highest power of the variable in the polynomial. * If there is no variable in the polynomial, then the degree is 0. Example 1 – 2x² - 3x + 4 is a polynomial x² is the exponent with the highest power in the polynomial So, the degree of the polynomial is 2

53. M ORE E XAMPLES OF D EGREES OF P OLYNOMIALS Example 3 – 4x³ - 2x² + 5x – 7 is a polynomial x³ is the exponent with highest power So, the degree of the polynomial is 3 Example 4 – 10 is a polynomial It has no variable, so its degree is 0

54. N AMING P OLYNOMIALS How are polynomials named? Polynomials are named according to their degree and number of terms.

55. P OLYNOMIAL N AMING C HART DegreeTerms 0Constantn/a 1LinearMonomial 2QuadraticBinomial 3CubicTrinomial 4QuarticPolynomial w/ 4 terms 5QuinticPolynomial w/ 5 terms

56. E XAMPLES OF N AMING P OLYNOMIALS Example 1 – 2x² - 3x + 4 has a degree of 2 and has 3 terms So, it is a quadratic trinomial Example 2 – x – 9 has a degree of 1 and has 2 terms So, it is a linear binomial

57. M ORE E XAMPLES Example 3 – 4x³ - 2x² + 5x – 7 has a degree of 3 and has 4 terms So, it is a cubic polynomial with 4 terms Example 4 – 10 has a degree of 0 and has 1 term So, it is a constant monomial or often referred to as just a “constant”

58. Y OU T RY Name the following polynomials: 1. 8x³ - 27 2. 2x² - 7x + 10 3. 5x 4. -3x ⁴ - 2x² + 1

59. A DDING AND S UBTRACTING P OLYNOMIALS When we add/subtract polynomials, we can only add/subtract like terms. *Remember that like terms are terms that have the same variable and power. Align all like terms and then add or subtract. Make sure final answer is arranged from highest power to lowest Example 1 – 4r 4 + 3r³ + (r³ - 5r 4 ) Solution – 4r 4 + 3r³ + (-5r 4 ) r³ -1r 4 + 4r³

60. M ORE E XAMPLES OF A DDING /S UBTRACTING P OLYNOMIALS Example 2 – (5 + 5n²) – (5 – 3n²) Solution – 5 + 5n² − 5 (-3n²) 8n² Example 3 – (5x 2 – 3x – x 4 ) + (x 4 - 5x 3 + 5x 2 ) Solution – 5x 2 – 3x – x 4 + 5x 2 x 4 (-5x 3 ) 10x 2 – 3x -5x 3 -5x 3 + 10x 2 – 3x

61. Y OU T RY Simplify the following expression. (4x + x³ − 3) + (4 – 5x – 5x³) Simplify the following expression. (4r² − 3r – 3r 4 – 3r³) – (4r³ − 3r² − r 4 + 3r)

62. M ULTIPLYING P OLYNOMIALS / H OW T O U SE A N E XPANSION B OX Monomial x Binomial Example 1 – a(b + c) 1. Put the first factor on the side and the other factor on the top 2. Add what is in the boxes. Combine all like terms if possible. So, a(b + c) = ab + ac b+c aa ∙ ba ∙ c

63. M ORE E XAMPLES O F U SING A N E XPANSION B OX Binomial x Binomial Example 2 – (a + b)(c + d) 1. Put first factor on side and other factor on top. 2. Add the boxes and combine like terms if possible. (a + b)(c + d) = ac + ad + bc + bd c+d aa ∙ ca ∙ d +bb ∙ cb ∙ d

64. R EAL E XAMPLES Example 3 – 7(x + 2) = 7x + 14 x+2 77x14

65. M ORE R EAL E XAMPLES Example 4 – (x + 6)(x – 4) x² -4x + 6x – 24 = x² + 2x – 24 x-4 xx²-4x +66x-24

66. Y OU T RY 1. 2x(3x + 7) 2. (x – 5)(4x + 2) 3. 3x²(2x² - 3x + 4)

67. Q UADRATIC E XPRESSIONS What is a quadratic expression? A quadratic expression is a single variable, degree 2 polynomial Examples of quadratic expressions: 1. x² + 10x + 21 2. x² - 81 3. 4x² - 25 4. x² - 7x 5. -6x² - 22x -20

68. S IMPLE Q UADRATIC E XPRESSIONS What are simple quadratic expressions? Simple quadratic expressions are in the form x² + bx + c How do we factor simple quadratic expressions? 1. Find the pair of numbers that multiply to produce c and add up to b 2. Place those two numbers in the following binomial factors: (x + __)(x + __)

69. E XAMPLES OF F ACTORING S IMPLE Q UADRATICS Example 1 – Factor x² + 10x + 21 1. Find the pair of factors of 21 that add up to 10: 1, 21  1 + 21 = 22  NO -1, -21  -1 + -21 = -22  NO 3, 7  3 + 7 = 10  YES -3, -7  -3 + -7 = -10  NO 2. So, x² + 10x + 21 = (x + 3)(x + 7)

70. M ORE E XAMPLES OF F ACTORING Q UADRATICS Example 2 – Factor x² + 7x – 30 1. Find the pair of factors of -30 that add up to 7: 1, -30  1 + -30 = -29  NO -1, 30  -1 + 30 = 29  NO 5, -6  5 + -6 = -1  NO -5, 6  -5 + 6 = 1  NO 3, -10  3 + -10 = -7  NO -3, 10  -3 + 10 = 7  YES 2. So, x² + 7x – 30 = (x + -3)(x + 10)

71. M ORE E XAMPLES OF F ACTORING Q UADRATICS Example 3 – Factor x² − 14x + 48 1. Find the pair of factors of 48 that add up to -14: 1,48  1 + 48 = 49  NO -1,-48  -1 + -48 = -49  NO 2, 24  2 + 24 = 26  NO -2,-24  -2 + -24 = -26  NO 3, 16  3 + 16 = 19  NO -3,-16  -3 + -16 = -19  NO 4, 12  4 + 12 = 16  NO -4,-12  -4 + -12 = -16  NO 6, 8  6 + 8 = 14  NO -6,-8  -6 + -8 = -14  YES 2. So, x² - 14x + 48 = (x + -6)(x + -8)

72. YOU TRY 1. x² + 5x – 24 2. x² − 8x – 33 3. x² + 15x + 50

73. F ACTOR O THER Q UADRATIC E XPRESSIONS What are perfect square trinomials? Perfect square trinomials are produced by one binomial multiplied by itself. In x² + bx + c, c is the square of a number and b is 2 times that number. Example 1 – x² + 16x + 64 = (x + 8)(x + 8) = (x + 8)² since 64 is 8 ² and 16 is 2 ∙ 8

74. M ORE E XAMPLES OF P ERFECT S QUARE T RINOMIALS Example 2 – x² - 6x + 9 = (x + -3)(x + -3) = (x + -3)² since 9 is ( -3 )² and -6 is 2 ∙ -3 Example 3 – x² + 12x + 36 = (x + 6)(x + 6) = (x + 6)² Example 4 – x² -20x + 100 = (x + -10)(x + -10) = (x + -10)²

75. Y OU T RY Factor the following quadratic expressions: 1. x² - 24x + 144 2. x² + 10x + 25

76. D IFFERENCE OF S QUARES What is difference of squares? Difference of squares is a binomial in which one square is being subtracted from another. Factors of difference of squares are two binomials where one is the sum of the square roots of the squares and the other is the difference of the square roots of the squares.

77. E XAMPLES OF D IFFERENCE OF S QUARES Example 1 – x² − 81 = (x + 9)(x – 9) since x² and 81 are perfect squares Example 2 – x² − 100 = (x + 10)(x – 10) since x² and 100 are perfect squares Example 3 – 4x² − 49 = (2x + 7)(2x – 7) since 4x² and 49 are perfect squares

78. Y OU T RY Factor the following quadratics: 1. x² − 25 2. 9x² − 36

79. S OLVING S IMPLE Q UADRATIC E QUATIONS When we solve quadratic equations, we must first make sure that the right side of the equation is 0 and then factor the quadratic expression on the left side of the equation. Next, we set both factors equal to 0 and then we solve for the variable in both equations. Example 1 – Solve the following equation: x² − 2x – 48 = 0 Solution – (x – 8)(x + 6) = 0 x – 8 = 0 or x + 6 = 0 So, x = 8 or x = -6

80. M ORE E XAMPLES OF S OLVING S IMPLE Q UADRATIC E QUATIONS Example 2 – Solve the following equation: x² + 10x = -16 Solution – x² + 10x + 16 = 0 add 16 to both sides to put 0 on right side (x + 8)(x + 2) = 0 factor left side x + 8 = 0 or x + 2 = 0 set both factors equal to 0 and solve for x So, x = -8 or x = -2

81. M ORE E XAMPLES OF S OLVING S IMPLE Q UADRATIC E QUATIONS Example 3 – Solve the following equation: x² − x = 20 Solution – x² − x – 20 = 0 subtract 20 from both sides to make right side equal 0 (x – 5)(x + 4) = 0 factor quadratic on left x – 5 = 0 or x + 4 = 0 set both factors equal to 0 and then solve for x So, x = 5 or x = -4

82. Y OU T RY Solve the following equation: x² − 6x – 40 = 0 Solve the following equation: x² + 4x = 5