All work must be shown or no credit will be awarded. Box all answers!! Order of Operations

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Steps: All work must be shown or no credit will be awarded. Box all answers!! Order of Operations 1. Do operations that occur within grouping symbols. If there is more than one set of symbols, work from the inside out.. Evaluate powers. 3. Multiplication and division from left to right. 4. Addition and subtraction from left to right. Non-Calculator Procedures A. Calculator Keystrokes 6 4 3 ( 6) 6 x 4 ^ + (-3) ^ + - 6 + =64 496 = 4 496 = 0 96 = 9 6 = 3 = 30 This is a screen shot of what the screen on a TI-84 Plus calculator would look like. B. 4 5 3 6 3 5 4 5 3 4 = 6 5 5 = 4 5 1 6 35 5 = 4 1 9 5 = 68 34 ( 4 (-5 3 x (-) ^ )) (-6 x (-3 ) + 5) This is a screen shot of what the screen on a TI-84 Plus calculator would look like. = C. 9 6 1 8 4-9 (6 ( 1 (8 )) 4 ) 9 6 1 4 = = = 9 6 38 4 9 6 19 4 = 9 3 4 = 9 36 = This is a screen shot of what the screen on a TI-84 Plus calculator would look like.

Use the order of operations to simplify each problem. Show all work! 1. 83 3 ( 4) 5. 8 3 4 ( 8) 5 3 3. 0 4 ( 14) 5 4 13 4. 98 3 5 4 1 Evaluating Expressions Steps: 1. Substitute the given numbers in for the variables.. Simplify using the order of operations. Let x =, y = and z = 3 4 Non-Calculator Procedures Calculator Keystrokes A. x + y + z To start these problems on the calculator, you need to store = 3 4 the values for x, y and z into the calculator. = 4 8 1 1 1 1 3 1 = 1 1 Storing the Values for the Variables: - STO ALPHA STO ENTER - 3 STO ALPHA 1 ENTER 4 STO ALPHA ENTER After these inital steps, the = 11 1 values will be stored for those variables on your calculator. So, for Example A, type in ALPA STO + ALPHA 1 + ALPHA ENTER To convert the decimal you get to a fraction, press MATH and ENTER twice.

Let x =, y = and z = 3 4 B. xy + z Calculator Keystrokes : = ALPHA STO * ALPHA 1 + ALPHA ENTER 3 4 MATH ENTER twice to change to a fraction = 4 3 4 16 1 = 1 1 = 3 1 C. 16 xz 16 * ALPHA STO * ALPHA = 16 4 14 = 16 4 8 = 16 4 64 8 = 4 4 = 9 4 = 3 Evaluate each expression. Show all work! Let a = 10, b = 5 1, c = 4 and d = 3 5. 4b ad + ab 6. 50 abc. 3abc 4bc + 6ad 1 5 8. ac bd Combining Like Terms Like terms are terms that have the same variable which are raised to the same power. Examples include: x 1 and 6x, 4 y and 1y or x and x. When combining like terms, you combine the coefficients which are the numbers in front of the variables and keep the variables the same.

5x x 1 1x 8x 4 1 B. x 8x x 5 3 = x 11x 8x 1 5 = x x 8x =14x8x 1 = 8 x x = 6x 1 A. Simplify each expression by combining like terms. Show all work! 4x 5y 3 9x y 10. 9. 3u t 9t u 11. x y 1 x 1. 3 3 4 y 3x y x 4 5 Using the Distributive Property a( b c) ab ac ( b c) a or a( b c) ab ac ( b c) a A. 5(3x ) 5 3x 5 B. 3x x 5y 3x x 3x 5y x 10 3x xy C. 6(x 4) 6 x 6 4 x 3y 5x 5x x 5x 3y D. Distribute. 1x 4 10x 13. 3x 1 14. 43a. 3x 4 16. y y 4 xy 1. 5y 3 z( x) 18. 5x 3x 3 3 8 10 4 19. x xy y

Distribute then Combine Like Terms A. 6x3x 4 B. 3x 5 x 3x 1 = 6x3x 3 4 = 3x 5x 6x 3 = 6x3x 1 = 3x1x 3 = 3x 1 =3xx 3 = 4x 3 C. 3 x 5 3x 58x D. 6 5 = 3 x519x 1 = x xy x x y x 6x xy x 5xy x 3 x 95x 60 = 6x 4xy x 3 9x 60 = 6x 4xy x = = =39x 60 = 8x 4xy = 9x 63 Distribute then combine like terms. Show all work! 0. 3x 1 3x 9x 1. 1 3x x 8. 10k 3 5 6k 3. 65u1 34u 1 4. 34x6 x 5. 85 9 4x x 6. 3m 5 m 5m. 58 m m 8. 34x 5 4 x 6 9. x x 4 6 3 10 4 3 9 14 30. 9 4 z 3 wz z w z z 3 6r t r t r 5t r 31.

First combine like terms, then evaluate. Show all work! 3. 5 x ( y) 86x 9y if x = 4 and y = 33. x 4y 6x 3y if x = 1 and y = 8 34. x y x y 5 if x = 3 and y = 6 Translating Expressions, Equations, & Inequalities Words that mean Addition Words that mean Subtraction Words that mean Multiplication Words that mean Division Plus Minus Times Divided by Sum Difference Product Quotient More than Less than Of Increased by Decreased by Multiplied by Subtracted from to = to > to Equals Is greater than Is greater than or equal to Is Is more than Is more than or equal to At least to < Is less than to Is less than or equal to At most Translate each expression, equation or inequality. x A. 1 decreased by the quotient of twice a number and 3. 1 3 B. 5 less than the sum of 5 and twice a number is less than The sum of x and it s cube. (n + 5) 5 < x + x 3 C. more than the sum of half a number and 9 is equal to the product of 4 and the same number plus 3. D. The product of 5 and a number is at most the quotient of the sum of and x, and 9. 1 9 4 3 x x x 5n 9

Translate each expression, equation or inequality. 35. The cube of the sum of x and twice y, decreased by the product of a and b. 36. Ten less than the difference of three and a number n is less than nine. 3. Six more than the quotient of a number and is 4. Steps: Solving Equations & Inequalities 1. Distribute and rid the problem of all grouping symbols.. Clear any fractions or decimals by multiplying every term by the common denominator. 3. Combine like terms on the same side of the equation. 4. Move the variable to the left hand side of the equation. 5. Add or subtract the constant to get the variable alone. 6. Multiply or divide by the coefficient in front of the variable to solve the equation. A. 3a 5 11 B. 1 4x x 3 +5 = +5 x x 3a = 6 1 5x = 3 3 3 1 1 a = 5x = 5 5 x = 3 C. 4x 1 10 D. 3x 4 4x 3 x 3x 4x 1 10 3x 1 4x + 1 = x + 3 x + 180 = 4x 0 x = 5 +0 +0 1 1 30 = 4x x = 5 4 4 = x To check a solution, substitute the answer back into the original equation. 1 4(3) = 3 3 1 1 = 0 0 = 0 Since this is a true statement, x = 3 is a solution.

E. xx 1 36 x 35x 1 F. 3 y 1 4 y 11 y x +1x + 36 = x x + 36 3y 3 4y 8 = 11 y x x y 11 = 11 y 1x + 36 = x + 36 +y + y +x +x 11 = 11 1x + 36 = 36 36 36 (No Solution because this is a false statement) 1x = 0 1 1 x = 0 G. 3a 5 a 3a 6 34 a H. x 9 5 3 14 3a [5 a + 3a 18] = + 1 3a 95x 14 3a [ 13 + a] = 14 3a 4 5x 14 3a + 13 a = 14 3a 4 4 a + 13 = 14 3a 5x 10 +3a +3a 5 5 5a + 13 = 14 x 13 13 When you divide by a negative, you must 5a = 1 reverse the inequality symbol. 5 5 a = 1 5 I. 5a 3 9 4 3a 5a 3 9 4 3a 5a 3 13 3a 3a 3a a 3 > 13 +3 +3 a > 10 a > 5 To check a solution, substitute a number within the conditions of the solution back into the original inequality. Let a = 0 since 0 > -5, so 5(0) 3 > - 9 (4 3(0)) 0 3 > - 9 (4 0) -3 > - 9 4-3 > - 13 Since this is a true statement, a > -5 is the solution. Solve the following equations or inequalities. 38. 3 1 4 8 x 39. 4x 9 19 40. 8 3 5x x 41. 18 0 4. 3 x 1 x 1 43. 6x 10x 3 5 10

45. 9 x 3 4x 3 46. 4x 5x 3 9x 5 44. 9x 1 x 5 4. 6x 3 18 316 3x Literal Equations 48. Solve for x: ax = y 49. Solve for m: m y n x