ࡱ> ,. )*+nq ֕bjbjt+t+ AAA]    8D 4x  m0 :$!dvNBlyO0  vWorksheet 5 (2.1) Chapter 2 Equations and Inequalities 2.1 Solving First-Degree Equations Summary 1: Important Terminology Used when Solving Equations An equation is a statement that two symbols or groups of symbols are names for the same number. An algebraic equation contains one or more variables. It is called an open sentence because it is neither true nor false. Solving an equation is the process used to find the number or numbers that make an algebraic equation a true numerical statement. A solution or root of an equation is the number that satisfies a given equation. The solution set is the set of all solutions of an equation. First-degree equations in one variable are equations with one variable which has an exponent of one. Equivalent equations are equations that have the same solution. Warm-up 1. a) Is 5 a solution of 2x + 4 = 19 - x? Verify this by showing that the numerical statement is true or false. EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 True or False? Answer: _________ yes or no Problem 1. Is 0 a solution of 3(x - 1) = 2x + 5? Verify this by showing that the numerical statement is true or false. Worksheet 5 (2.1) Summary 2: Further Properties of Equality Used to Solve Equations 1. Addition Property of Equality For all real numbers a, b, and c, a = b if and only if a + c = b + c. Note: Subtraction can be rewritten as addition. Therefore, this property also allows us to subtract the same number on both sides. 2. Multiplication Property of Equality For all real numbers a, b, and c, where cEMBED Equation.30, a = b if and only if ac = bc. Note: Division can be rewritten as multiplication. Therefore, this property also allows us to divide the same number on both sides. Warm-up 2. Solve: a) EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 The solution set is . b) EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 Worksheet 5 (2.1) EMBED Equation.3 EMBED Equation.3 The solution set is . Problems - Solve: 2. -5x - 3 = 12 3. 2(2x - 3) = 5x - 11 Summary 3: Solving First-Degree Equations in One Variable 1. Simplify each side of the equation. 2. Apply the addition property of equality to isolate the variable term on one side and constant on the other side. 3. Apply the multiplication property of equality to obtain the coefficient of 1 for the variable. 4. Write an appropriate solution set and check the solution to verify that the resulting numerical statement is true. Warm-up 3. Solve and check: a) 5n - 4 - 2n = 4n + 12 + 3n - 4 = + 12 3n - 4 + = 7n + 12 + -4n - 4 = 12 -4n - 4 + = 12 + -4n = ( )(-4n) = ( )(16) n = The solution set is . Worksheet 5 (2.1) Check: 5( ) - 4 - 2( ) = 4( ) + 12 + 3( ) ____ - 4 - ____ = ____ + 12 + ____ ____ = ____ True or False? b) 4(m - 1) - (m + 6) = 5(m + 2) 4m - 4 - m - ____ = 5m + 10 ____ - 10 = 5m + 10 3m - 10 + ____ = 5m + 10 + ____ ____ - 10 = 10 -2m - 10 + ____ = 10 + ____ -2m = ( )(-2m) = ( )(20) m = ____ The solution set is _______. Problems - Solve and check: 4. 3n - 1 - 4n = 6n - 7 - 4n 5. -(n - 5) + 3(n + 2) = 4(n - 3) - 1 Summary 4: Using Equations to Solve Word Problems 1. Carefully read the word problem. Reread to get an overview of the situation. Determine known facts and identify unknown quantities. 2. Declare the variable by choosing a letter to use as a variable and tell in words what it will represent. 3. Express other unknown quantities using this variable. 4. Translate the word problem or use a guideline from the word problem to write an algebraic equation to solve. 5. Check results to determine whether or not they satisfy the conditions stated in the original problem. 6. Express answer as a complete sentence. Worksheet 5 (2.1) Warm-up 4. Set up and write an algebraic equation, then solve: a) The sum of two numbers is 55. The larger number is eleven more than the smaller number. Find the two numbers. Declare the variable: Let n = the smaller number = the larger number Write an algebraic equation and solve: n + _______ = 55 2n + ____ = 55 2n + 11 + ____ = 55 + ____ 2n = ____ ( )(2n) = ( )(44) n = ____ ; n + 11 = ____ The numbers are _______ and _______. Problems - Set up and write an algebraic equation, then solve: 6. In a class containing 58 students, the number of women students is one more than twice the number of men students. How many women students are in the class? 7. An air-conditioner repair bill is $87. This included $30 for parts and an amount for 3 hours of labor. What was the hourly rate that was charged for labor? Worksheet 6 (2.2) 2.2 Equations Involving Fractional Forms Summary 1: Using the LCD to Clear Fractions in an Equation The least common denominator, LCD, refers to the least common multiple of a set of denominators. Clearing the equation of all fractions is accomplished by applying the multiplication property of equality. Both sides of the equation are to be multiplied by the LCD in the equation to produce an equivalent equation. When an equation has more than one term on one or both sides of the equation, the distributive property is used to multiply each term by the LCD. Warm-up 1. Find the LCD and multiply both sides of the equation by the LCD to clear the fractions. Give the resulting equivalent equation. a) EMBED Equation.3 The LCD is _____. EMBED Equation.3 EMBED Equation.3 -9 + _____ = _____ The resulting equivalent equation is __________________________. b) EMBED Equation.3 The LCD is _____. EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 The resulting equivalent equation is ___________________________. Worksheet 6 (2.2) Problems - Find the LCD and multiply both sides of the equation by the LCD to clear the fractions. Give the resulting equivalent equation. 1. EMBED Equation.3 2. EMBED Equation.3 Summary 2: Solving Equations Involving Fractional Forms 1. Find the LCD in the equation. 2. Apply the multiplication property of equality to clear the equation of all fractions. This is done by multiplying each term in the equation by the LCD. (See summary 1 above.) 3. Solve the resulting equivalent equation. 4. Check when directed to do so. Warm-up 2. a) Solve and check: EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 -9 + _____ = _____ _____ = 17 x = _____ The solution set is ______. Worksheet 6 (2.2) Check: EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 True or False? b) Solve: EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 The solution set is ______. Problems - Solve: 3. EMBED Equation.3 4. EMBED Equation.3 Worksheet 6 (2.2) Summary 3: Suggestions for Solving Word Problems 1. Carefully read the word problem. Reread to get overview of given situation. Determine known facts and identify unknown quantities. 2. Utilize a chart, figure, or diagram to clarify the problem. 3. Declare the variable and express unknown quantities in terms of the variable. 4. Find a guideline in the word problem. 5. Translate into an equation using the declared variable. 6. Solve the equation and use solution to determine all facts asked for in the problem. 7. Check results in the original problem. Express answer in a complete sentence. Warm-up 3. Set up and write an algebraic equation, then solve: a) Find a number such that one-half of the number is three more than one-fourth of the number. Declare the variable: Let ____ = a number Write an algebraic equation and solve: EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 The number is _______. Worksheet 6 (2.2) b) In triangle ABC, angle A is one-third angle C and angle B is 40 degrees more than two-thirds of angle A. Find the measures of the three angles in the triangle. Draw a figure: Declare the variable: Let x = measure of angle C _________ = measure of angle A ____________ = measure of angle B Write an algebraic equation and solve: EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 The angle measures are _____, _____, and _____. Worksheet 6 (2.2) Problems - Set up and write an algebraic equation, then solve: 5. John is paid 2 times his normal hourly rate for each hour he works on a holiday. Last week he worked 45 hours, five hours were on a holiday. Find his normal hourly rate if he earned $420. 6. Maria took three college algebra exams and had an average score of 84. Her second exam was eight points better than her first and her third exam was eleven points better than her second exam. Find the three exam scores. Worksheet 7 (2.3) 2.3 Equations Involving Decimals and Problem Solving Summary 1: Using 10n, a Power of 10, to Clear Decimals in an Equation Multiplying a decimal by 10n will result in moving the decimal point n places to the right. Clearing the equation of all decimals is accomplished by applying the multiplication property of equality. Both sides of the equation are to be multiplied by an appropriate power of 10 to produce an equivalent equation. This power of 10 is equal to the most number of decimal places for any given decimal in the equation. When an equation has more than one term on one or both sides of the equation, the distributive property is used to multiply each term by the power of 10. Warm-up 1. Find an appropriate power of 10 to clear all decimals. Give the resulting equivalent equation: a) 2y = .6y + .21 The power of 10 is _______. _____(2y) = _____(.6y + .21) _____(2y) = _____(.6y) + _____(.21) The resulting equivalent equation is ____________________. Problem - Find an appropriate power of 10 to clear all decimals. Give the resulting equivalent equation: 1. 0.05x + 0.08(10000 - x) = 620 Worksheet 7 (2.3) Summary 2: Solving Equations Involving Decimals 1. Find an appropriate power of 10 to clear all decimals using the multiplication property of equality. (See summary 1 above.) 2. Solve the resulting equivalent equation. 3. Check when directed to do so. Warm-up 2. a) Solve and check: .07x + 160 = 152 + .08x _____(.07x + 160) = _____(152 + .08x) _____(.07x) + _____(160) = _____(152) + _____(.08x) _____ + 16000 = _______ + 8x 7x + 16000 + _____ = 15200 + 8x + _____ _____ + 16000 = 15200 -1x + 16000 + _______ = 15200 + ________ -1x = _____ _____(-1x) = _____(-800) x = _____ The solution set is _______. Check: .07( ) + 160 = 152 + .08( ) _____ + 160 = 152 + _____ _____ = _____ True or False? b) Solve: .11(7000 - x) - .12x = 310 _____[.11(7000 - x) - .12x] = _____[310] _____[.11(7000 - x)] - _____[.12x] = _____[310] 11(7000 - x) - _____ = ________ 77000 - _____ - 12x = 31000 77000 - _____ = 31000 77000 - 23x + ________ = 31000 + ________ -23x = __________ _____(-23x) = _____(-46000) x = __________ The solution set is __________. Worksheet 7 (2.3) Problems - Solve: 2. 0.222x = 0.2 - 0.22x , round to nearest hundredth. 3. 0.06x + 0.05(8000 - x) = 450 Summary 3: Suggestions for Solving Word Problems 1. Carefully read the word problem. Reread to get overview of given situation. Determine known facts and identify unknown quantities. 2. Utilize a chart, figure, or diagram to clarify the problem. 3. Declare the variable and express unknown quantities in terms of the variable. 4. Find a guideline in the word problem. 5. Translate into an equation using the declared variable. 6. Solve the equation and use solution to determine all facts asked for in the problem. 7. Check results in the original problem. Express answer as a complete sentence. Warm-up 3. Set up and write an algebraic equation, then solve: a) Kayla bought an opal necklace at a 15% discount sale for $1150. Find the original price of the necklace, rounded to the nearest cent. Declare the variable: Let x = original price _______ = amount of discount Write an algebraic equation and solve: x - _______ = _______ _____(x - .15x) = _____(1150) _____(x) - _____(.15x) = _____(1150) 100x - _____ = 115000 _____ = 115000 x = _______ The original price was______. Worksheet 7 (2.3) b) A total of $12,000 was invested, part at 6% and the remainder at 8%. Find how much was invested at each rate, if the total yearly interest earned was $820. Declare the variable: Let x = amount invested at 6% ___________ = amount invested at 8% ___________ = interest earned at 6% ___________ = interest earned at 8% Write an algebraic equation and solve: .06x + .08(__________) = _____ _____[.06x + .08(12000 - x)] = _____(820) _____[.06x] + _____[.08(12000 - x)] = _____(820) 6x + _____(12000 - x) = 82000 6x + ________ - 8x = 82000 _____ + 96000 = 82000 -2x = ________ x = ________ 12000 - x = ________ There was __________ invested at 6% and __________ invested at 8%. Problems - Set and write an algebraic equation, then solve: 4. Higinio bought a car with 6.5% sales tax included for $17,944. Find the price of the car, rounded to the nearest cent, before tax. 5. Xurry invests a certain amount of money at 7% interest and $1500 less than that amount at 4.5%. His yearly interest is $392.50, find how much he invested at each rate. Worksheet 8 (2.4) 2.4 Formulas Summary 1: Literal equations are equations that contain more than one variable. Formulas are usually literal equations that state a rule in symbolic form. Warm-up 1. a) Solve i = Prt for i, given that P = $800, r =EMBED Equation.3, and t = 3 years. i = Prt i = (800)( )(3) i = _______ b) Solve 3(x - 2y) = 4 for y, given that x = -1. 3(x - 2y) = 4 3( - 2y) = 4 3( ) - 3( ) = 4 -3 - _____ = 4 -3 - 6y + _____ = 4 + _____ -6y = _____ _____(-6y) = _____(7) y = _____ Problems 1. Solve A = P + Prt for r, given that A = $1080, P = $800, and t = 7 years. Express r as a percent. 2. Solve 2x - 5y = -10 for x, given that y = 0. Worksheet 8 (2.4) Summary 2: Changing the Form of a Literal Equation or Formula 1. If necessary, clear all fractions using the LCD and multiplication property of equality. 2. Apply distributive property to clear parentheses when appropriate. 3. Use the addition property of equality to collect terms containing the desired variable on one side and all other terms on the opposite side. 4. If two or more terms contain the desired variable on one side, use the distributive property to rewrite the expression. 5. Apply the multiplication property of equality to obtain a coefficient of 1 for the desired variable. 6. If needed, apply the symmetric property of equality so the desired variable is on the left side of the equation. Warm-up 2. a) Solve EMBED Equation.3 for b2. _____[A] = _____EMBED Equation.3 _____ = h(b1 + b2) 2A = _____ + _____ 2A + _____ = hb1 + hb2 + _____ 2A - hb1 = _____ _____(2A - hb1) = _____(hb2) EMBED Equation.3 EMBED Equation.3 b) Solve -3x - y = 5 + 4y for y. -3x - y + _____ = 5 + 4y + _____ -3x - _____ = 5 -3x - 5y + _____ = 5 + _____ -5y = _____ + _____ _____(-5y) = _____(3x + 5) y = EMBED Equation.3 or y = EMBED Equation.3 Worksheet 8 (2.4) Problems - Solve: 3. EMBED Equation.3, for w. 4. EMBED Equation.3, for x. Summary 3: Suggestions for Solving Word Problems Using Formulas 1. Carefully read the word problem. Reread to get overview of given situation. Determine known facts and identify unknown quantities. 2. Utilize a chart, figure, or diagram to clarify the problem. 3. Declare the variable and express unknown quantities in terms of the variable. 4. Use a formula as a guideline in the word problem. 5. Translate into an equation using the declared variable. 6. Solve the equation and use the solution to determine all facts asked for in the problem. 7. Check results in the original problem. Express answer as a complete sentence. Warm-up 3. Set up and write an algebraic equation, then solve: a) The length of a rectangular lot is 16 feet. If the total distance around the lot is 56 feet, find the width of the lot. Draw a figure: Worksheet 8 (2.4) Declare the variable: Let x = width Write an algebraic equation and solve: P = 2l + 2w _____ = 2( ) + 2( ) _____ = 32 + _____ _____ = 2x _____ = x The width is __________. b) After 3 hours, two cars traveling in opposite directions on I-24 are 360 miles apart. One car is traveling 10 mph faster than the other car. Find the rate of speed for both cars. Declare the variable: Let x = speed of slower car _______= speed of faster car Chart: Diagram: d=rt r t ------------------------------------ fast x + 10 3 3x 3( ) ------------------------------------ <-------+----------> slow x 3 distance traveled ------------------------------------ 360 miles Write an algebraic equation and solve: 3x + ___________ = _____ 3x + 3x + _____ = 360 _____ + 30 = 360 6x = _____ x = _____ ; x + 10 = _____ The speed of the slower car is _______ and the speed of the faster car is _______. Worksheet 8 (2.4) Problems - Set up and write an algebraic equation, then solve: 5. How many milliliters of 15%-salt solution must be added to 200 ml of 25%-salt solution to obtain a 20%-salt solution? 6. In a given triangle, the largest angle is three times the smallest angle. The third angle is ten more than the smallest angle. Find the measures of all three angles. Worksheet 9 (2.5) 2.5 Inequalities Summary 1: Statements of inequality express one of the following: 1. EMBED Equation.3 means a is less than b. 2. EMBED Equation.3 means a is less than or equal to b. 3. EMBED Equation.3 means a is greater than b. 4. EMBED Equation.3 means a is greater than or equal to b. Algebraic inequalities contain one or more variables. They are open sentences which are neither true nor false. Solving an inequality is the process used to find the number or numbers that make an algebraic inequality a true numerical statement. These numbers are solutions of the inequality. Warm-up 1. a) Is -3 a solution of EMBED Equation.3? Verify this by showing that the numerical statement is true or false. EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 True or False? Answer: _________ yes or no Problem 1. Is 2 a solution of EMBED Equation.3? Verify this by showing that the numerical statement is true or false. Worksheet 9 (2.5) Summary 2: Properties of Inequalities 1. Addition Property of Inequality For all real numbers a, b, and c, a > b if and only if a + c > b + c. Note: When adding the same number on both sides of the inequality, the sense of the inequality remains the same. 2. Multiplication Property of Inequality a) For all real numbers a, b, and c with c > 0, a > b if and only if ac > bc. Note: When multiplying the same positive number on both sides of the inequality, the sense of the inequality remains the same. b) For all real numbers a, b, and c with c < 0, a > b if and only if ac < bc. Note: When multiplying the same negative number on both sides of the inequality, the sense of the inequality changes; therefore, the inequality sign must be reversed. In general, the process for solving inequalities closely parallels that for solving equations. Warm-up 2. Solve: a) 2x - 5 > 7 2x - 5 + _____ > 7 + _____ 2x ___ 12 _____(2x) > _____(12) x ___ 6 b) EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 Worksheet 9 (2.5) EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 Problems - Solve: 2. EMBED Equation.3 3. EMBED Equation.3 Summary 3: Appropriate Ways to Express the Solution of an Inequality SET GRAPH INTERVAL NOTATION EMBED Equation.3 < (===> EMBED Equation.3 a EMBED Equation.3 < [===> EMBED Equation.3 a EMBED Equation.3 <====) > EMBED Equation.3 b EMBED Equation.3 <====] > EMBED Equation.3 b Notes: 1. Set notation likeEMBED Equation.3is translated as "the set of all x such that x is less than -4." 2. The symbol ( or ) is used to exclude an endpoint or with the infinity symbol, . 3. The symbol [ or ] is used to include an endpoint. Worksheet 9 (2.5) Warm-up 3. a) Sketch a graph of x < 3 and express in interval notation. Graph: <%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%> -1 0 1 2 3 4 5 Interval notation: b) Express (-, 8] in set-builder notation. Set: EMBED Equation.3 Problems 4. Express the graph below both in set notation and interval notation: < [========> -6 -5 -4 5. Sketch a graph for the solution expressed as (3 , ). Summary 4: Key Factors to Consider when Solving Inequalities 1. The solving process for inequalities closely parallels that of solving equations. 2. The inequality sign reverses when multiplying both sides of the inequality by a negative number. 3. Use an appropriate format to express the solution of an inequality - set, graph, or interval notation. 4. One solution can be checked to possibly catch a mistake. It is not possible to check all solutions of an inequality. Worksheet 9 (2.5) Warm-up 4. a) Solve and graph the solution set on a number line: 2(3 - x) < 14 6 - _____ < 14 6 - 2x + _____ < 14 + _____ -2x < _____ _____(-2x) < _____(8) x ___ -4 Graph: b) Solve and express the solution set using interval notation: 2(x + 4) - (x - 2) EMBED Equation.3 -3(x - 1) 2x + 8 - _____ + _____ EMBED Equation.3 -3x + _____ x + _____ EMBED Equation.3 -3x + 3 x + 10 + _____ EMBED Equation.3 -3x + 3 + _____ _____ + 10 EMBED Equation.3 3 4x + 10 + _____ EMBED Equation.3 3 + _____ 4x EMBED Equation.3 _____ _____(4x) EMBED Equation.3 _____(-7) x EMBED Equation.3 _____ Interval notation: Problems 6. Solve and graph the solution set on a number line: -3 + 6x > -15 7. Solve and express the solution set using interval notation: -3(2x + 1) - 2EMBED Equation.3 -2(x + 5) Worksheet 10 (2.6) 2.6 More on Inequalities and Problem Solving Summary 1: Compound Inequalities Compound statements are formed when mathematical statements are joined by the words and or or. Conjunctions are compound statements that use and. The solution set for a given conjunction is the intersection of the solution sets for the inequalities joined by the word and. Expressing the Solution of a Conjunction: SET GRAPH INTERVAL NOTATION EMBED Equation.3 < (===) > (a, b) a b EMBED Equation.3 < [===) > [a, b) a b EMBED Equation.3 < (===] > (a, b] a b EMBED Equation.3 < [===] > [a, b] a b Note: a < x < b is the compact form for a < x and x < b. The compact form can only be used for conjunctions where the solution is between two endpoints. Disjunctions are compound statements that use or. The solution set for a given disjunction is the union of the solution sets for the inequalities joined by the word or. Expressing the Solution of a Disjunction: SET GRAPH INTERVAL NOTATION EMBED Equation.3 <===) (===> EMBED Equation.3 a b Note: The endpoints a and b will be included in the solution for , . In this case, use [ or ]. Worksheet 10 (2.6) Warm-up 1. Graph the compound inequality and express in interval notation: a) x > -2 and x < 3 Give the compact form of this statement: __________ Graph: <%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%> Interval notation: b) EMBED Equation.3 Graph: <%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%> Interval notation: c) x > -2 or x > 3 Graph: <%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%> Interval notation: Problems - Graph the compound inequality and express in interval notation: 1. x > 1 and x < 3 (What is the compact form of this statement?) 2. x < -5 or x > 1 3. x -3 or x < 2 Worksheet 10 (2.6) Summary 2: Key Factors to Consider when Solving Compound Statements 1. For disjunctions, solve each inequality separately in the compound sentence. The solution set is the union of these solutions. 2. For conjunctions, solve each inequality separately in the compound sentence. The solution set is the intersection of these solutions. Note: If the conjunction is in compact form, it can be solved by isolating the variable in the middle. The properties of inequality will be applied simultaneously on the middle, left, and right of the compact form. Warm-up 2. Solve, then express the solution set as a graph and in interval notation: a) x - 5 < -3 or x - 3 > 3 x - 5 + _____ < -3 + _____ or x - 3 + _____ > 3 + _____ x < _____ x > _____ The solution set is {x________or________}. Graph: <%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%> Interval notation: __________EMBED Equation.3 __________ b)EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 Worksheet 10 (2.6) EMBED Equation.3 The solution set is {x_______________}. Graph: <%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%%%%%%,%%> Interval notation: Problems - Solve, then express the solution set as a graph and in interval notation: 4. x - 3 > -2 or x + 1 < 3 5. EMBED Equation.3 Summary 3: Solving Word Problems Involving Inequalities 1. The strategies previously outlined for equations hold true for inequalities as well. 2. Analyze the situation to determine which inequality symbol is appropriate. Worksheet 10 (2.6) Warm-up 3. Set up and write an algebraic inequality, then solve: a) In general chemistry, Kate scored 95, 83, 89, and 85 on four exams this semester. If she is to earn an A, 90 or higher, for the semester, what must she score on the fifth and final test for the semester? Declare the variable: Let x = score on the fifth test Write an algebraic inequality and solve: EMBED Equation.3 _____EMBED Equation.3_____(90) 95 + 83 + 89 + 85 + xEMBED Equation.3 _____ _____ + xEMBED Equation.3 450 352 + x + _____EMBED Equation.3 450 + _____ xEMBED Equation.3 _____ She must score at least ______. Problem - Set up and write an algebraic inequality, then solve: 6. Darla has $2000 to invest. If she invests $1000 at 9% interest, at what rate must she invest the remaining amount so that the two investments earn more than $130 of combined yearly interest? Worksheet 11 (2.7) 2.7 Equations and Inequalities Involving Absolute Value Summary 1: Solving Absolute Value Statements when k > 0 Absolute value equations and inequalities must be converted to their corresponding equivalent compound statements when k > 0. The compound statement can then be solved and the solution set can be graphed or expressed in interval notation. 1. EMBED Equation.3 is equivalent to EMBED Equation.3 2. EMBED Equation.3 is equivalent to EMBED Equation.3 orEMBED Equation.3 3. EMBED Equation.3 is equivalent to EMBED Equation.3 Warm-up 1. a) Solve and give the solution set in set notation: EMBED Equation.3 EMBED Equation.3 or EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 or EMBED Equation.3 The solution set is { , } b) Solve and graph the solution set: EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 Worksheet 11 (2.7) EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 Graph: Problem - Solve and write solution in interval notation: 1. EMBED Equation.3 Summary 2: Solving Absolute Value Statements when k 0 Absolute value equations and inequalities are solved by inspection when k < 0. 1. ForEMBED Equation.3, the solution set is . 2. ForEMBED Equation.3, the solution set is . 3. ForEMBED Equation.3, the solution set is (-, ). Absolute value equations have exactly one solution when k = 0. 1.EMBED Equation.3is equivalent toEMBED Equation.3 Worksheet 11 (2.7) Warm-up 2. 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eY=ȀLQR(Xbo2iϼdo-j<h~ǪU^$4G]pr'Qq97%_zTWWx _8gN=UeZI[|ʊeC³-|P][%'Vw 'p؏J;!WutFt}tkw:L4w3[ZUADd  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      "#$%&'(-0Q123456879:;<=>?@ACBEDFGIHJKLMNOPRSsUTVWXYZ[\]^`_badcefghijklmnopqrutvwyx{z}|~Root Entry F8fg0/Data !QEWordDocumentAObjectPool~(0g0_1060589686F(0"20Ole CompObjfOle10Native  !#$%*,-.04678:;?ABCEIKLMOPTVWXZ^`abdhjklnrtuvx|~ FMicrosoft Equation 3.0 DS Equation Equation.39q 2(    ) + 4 = 19 -    h@yII 2(    )Equation Native ObjInfo _1060589687 F"20"20Ole   + 4 = 19 -     FMicrosoft Equation 3.0 DS Equation Equation.39q     + 4 =    CompObj fOle10NativeEquation Native hObjInfoL@yII     + 4 =     FMicrosoft Equation 3.0 DS Equation Equation.39q_1060589688F"20"20Ole CompObjfOle10Native 14 =    0@yII 14 =    Equation Native LObjInfo_1060589689 "F"20"20Ole CompObj fOle10Native"Equation Native &,ObjInfo' FMicrosoft Equation 3.0 DS Equation Equation.39q @yII `"_1060589690F"20"20Ole (CompObj)fOle10Native! + FMicrosoft Equation 3.0 DS Equation Equation.39q -11 = 3x + 1<@yII -11 = 3Equation Native /XObjInfo1_1060589691($F"20"20Ole 2x + 1 FMicrosoft Equation 3.0 DS Equation Equation.39q -11 +     = 3x + 1 +    CompObj#%3fOle10Native'&5Equation Native 9ObjInfo<x@yII -11 +     = 3x + 1 +     FMicrosoft Equation 3.0 DS Equation Equation.39q_1060589693*F"20"20Ole =CompObj)+>fOle10Native-,@ -12 = 3x,@yII -12 = 3xEquation Native DHObjInfoF_1060589694F0F"20"20Ole GCompObj/1HfOle10Native32JEquation Native NObjInfoQ FMicrosoft Equation 3.0 DS Equation Equation.39q (    )(-12 ) = (    ) ( 3 x)|@yII (    )(-12 ) = (    ) ( 3 x) FMicrosoft Equation 3.0 DS Equation Equation.39q     = x_10605896956F"2090Ole RCompObj57SfOle10Native98U,@yII     = x FMicrosoft Equation 3.0 DS Equation Equation.39q x =     Equation Native YHObjInfo[_10605896964@<F9090Ole \CompObj;=]fOle10Native?>_Equation Native cLObjInfoe0@yII x =      FMicrosoft Equation 3.0 DS Eq_1060589697BF9090Ole fCompObjACgfOle10NativeEDiuation Equation.39q 2(x - 1) = 5x + 7P@yII 2(x - 1) = 5x + 7Equation Native mlObjInfoo_1060589698:RHF90B0Ole p FMicrosoft Equation 3.0 DS Equation Equation.39q 2x -     = 5x + 7T@yII 2x -  CompObjGIqfOle10NativeKJsEquation Native wpObjInfoy   = 5x + 7 FMicrosoft Equation 3.0 DS Equation Equation.39q 2x - 2 +      = 5x + 7 +     _1060589699NFB0B0Ole zCompObjMO{fOle10NativeQP}ߌ@yII 2x - 2 +      = 5x + 7 +      FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native ObjInfo_1060589700LXTFB0B0Ole CompObjSUfOle10NativeWVEquation Native TObjInfo -3x - 2 = 78@yII -3x - 2 = 7_1060589702ZFB0B0Ole CompObjY[fOle10Native]\ FMicrosoft Equation 3.0 DS Equation Equation.39q -3x - 2 +     = 7 +    t@yII -3x - 2 +     = 7 + Equation Native ObjInfo_1060589703.`FB0B0Ole     FMicrosoft Equation 3.0 DS Equation Equation.39q -3x =     CompObj_afOle10NativecbEquation Native TObjInfo8@yII -3x =      FMicrosoft Equation 3.0 DS Equation Equation.39q (    )(-3 x) = ( _1060589704fFB0B0Ole CompObjegfOle10Nativeih   )( 9 )t@yII (    )(-3 x) = (    )( 9 )Equation Native ObjInfo_1060589705dplFB0B0Ole CompObjkmfOle10NativeonEquation Native LObjInfo FMicrosoft Equation 3.0 DS Equation Equation.39q x =     0@yII x =      FMicrosoft Equation 3.0 DS Equation Equation.39q -34 + 12x = 23_1060589706rFB0B0Ole CompObjqsfOle10NativeutEquation Native xObjInfo_1060589707jxFB0J0Ole \@yII -34 + 12x = 23 FMicrosoft Equation 3.0 DS Equation Equation.39q      -34 + 12x() =      CompObjwyfOle10Native{zEquation Native ObjInfo23()߰@yII      -34 + 12x() =      23()_1060589708~FJ0J0Ole CompObj}fOle10Native FMicrosoft Equation 3.0 DS Equation Equation.39q      -34() +      12x() =      23()@yII      -34() +    Equation Native ObjInfo_1060589710|FJ0J0Ole   12x() =      23() FMicrosoft Equation 3.0 DS Equation Equation.39q 25(4x - 1) + 1 = -14(5x + 2)CompObjfOle10NativeEquation Native ObjInfoߌ@yII 25(4x - 1) + 1 = -14(5x + 2) FMicrosoft Equation 3.0 DS Eq_1060589711FJ0J0Ole CompObjfOle10Nativeuation Equation.39q      254x - 1() + 1[] =      -145x + 2()[]@yII      254x - 1() + 1[] =      Equation Native  ObjInfo_1060589712vFJ0J0Ole -145x + 2()[] FMicrosoft Equation 3.0 DS Equation Equation.39q       25(4x - 1)[] +       CompObjfOle10NativeEquation Native 4ObjInfo1[] =       -14(5x + 2)[]@yII       25(4x - 1)[] +       1[]   !"&()*,0234678<>?@BCDHJKLNOSUVWYZ^`abdhjklnosuvwyz{ =       -14(5x + 2)[] FMicrosoft Equation 3.0 DS Equation Equation.39q    (4x - 1) +      =     (5x + 2_1060589713FJ0J0Ole CompObjfOle10Native)ߤ@yII    (4x - 1) +      =     (5x + 2) FMicrosoft Equation 3.0 DS EqEquation Native  ObjInfo _1060589714FJ0S0Ole CompObjfOle10NativeEquation Native ObjInfouation Equation.39q 4x - 19 = 3x + 13 - x + 4߀@yII 4x - 19 = 3x + 13 - x + 4 FMicrosoft Equation 3.0 DS Equation Equation.39q 12(6x - 1) + 12 = -2x - 5_1060589715FS0S0Ole CompObjfOle10NativeEquation Native  ObjInfo#_1060589716FS0S0Ole $߀@yII 12(6x - 1) + 12 = -2x - 5 FMicrosoft Equation 3.0 DS Equation Equation.39q -34 + 12x =CompObj%fOle10Native'Equation Native +xObjInfo- 23\@yII -34 + 12x = 23 FMicrosoft Equation 3.0 DS Eq9_1060589718FS0S0Ole .CompObj/fOle10Native1uation Equation.39q      -34 + 12x() =      23()߰@yII      -34 + 12x() =      23Equation Native 5ObjInfo9_1060589719FS0S0Ole :() FMicrosoft Equation 3.0 DS Equation Equation.39q      -34() +      12x() =      23()CompObj;fOle10Native=Equation Native AObjInfoE@yII      -34() +      12x() =      23() FMicrosoft Equation 3.0 DS Eq_1060589720FS0S0Ole FCompObjGfOle10NativeIuation Equation.39q -34 + 12(    ) = 23p@yII -34 + 12(    ) = 23Equation Native MObjInfoP_1060589721^FS0U[0Ole Q FMicrosoft Equation 3.0 DS Equation Equation.39q -34 +          = 23CompObjRfOle10NativeTEquation Native XObjInfo[p@yII -34 +          = 23 FMicrosoft Equation 3.0 DS Equation Equation.39q          = 2_1060589722FU[0U[0Ole \CompObj]fOle10Native_3P@yII          = 23 FMicrosoft Equation 3.0 DS EqEquation Native clObjInfoe_1060589723FU[0U[0Ole fCompObjgfOle10NativeiEquation Native mObjInfopuation Equation.39q 4x - 19 = 3x + 13 - x + 4߀@yII 4x - 19 = 3x + 13 - x + 4 FMicrosoft Equation 3.0 DS Equation Equation.39q       4x - 19() =       3x + 13 - x + 4()_1060589724FU[0U[0Ole qCompObjrfOle10NativetEquation Native xObjInfo|_1060589726FU[0U[0Ole }@yII       4x - 19() =       3x + 13 - x + 4() FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj~fOle10NativeEquation Native 0ObjInfo      4x - 19() =      3x + 13() -     (x) +     (4)@yII      4x - 19() =      3x + 13() -     (x) +     (4) FMicrosoft Equation 3.0 DS Equation Equation.39q_1060589727FU[0`}d0Ole CompObjfOle10Native 4x - 1 =    (3x + 1) - 9x + 36߈@yII 4x - 1 =    (3x + 1) - 9x + 36Equation Native ObjInfo_1060589728F`}d0`}d0Ole  FMicrosoft Equation 3.0 DS Equation Equation.39q 4x - 1 = 9x +     - 9x + 36|@yII 4x - 1 CompObjfOle10NativeEquation Native ObjInfo= 9x +     - 9x + 36 FMicrosoft Equation 3.0 DS Equation Equation.39q 4x - 1 =    _1060589729F`}d0`}d0Ole CompObjfOle10Native@@yII 4x - 1 =     FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native \ObjInfo_1060589730F`}d0`}d0Ole CompObjfOle10NativeEquation Native LObjInfo 4x =    0@yII 4x =    _1060589731F`}d0`}d0Ole CompObjfOle10Native FMicrosoft Equation 3.0 DS Equation Equation.39q x =    ,@yII x =    Equation Native HObjInfo_1060589732F`}d0l0Ole  FMicrosoft Equation 3.0 DS Equation Equation.39q 3a - 14 + a - 23 = 2120 + a - 15CompObjfOle10NativeEquation Native ObjInfo߰@yII 3a - 14 + a - 23 = 2120 + a - 15 FMicrosoft Equation 3.0 DS Equation Equation.39q_1060589734Fl0l0Ole CompObjfOle10Native 12(2x - 1) = 3 + 13(5x + 2)߈@yII 12(2x - 1) = 3 + 13(5x + 2)Equation Native ObjInfo_1060589739Fl0l0Ole  FMicrosoft Equation 3.0 DS Equation Equation.39q       =       + 3\@yII      CompObj fOle10Native  Equation Native xObjInfo =       + 3 FMicrosoft Equation 3.0 DS Equation Equation.39q 4(     ) = 4(     ) + 4(3)_1060589740Fl0l0Ole CompObj fOle10Nativet@yII 4(     ) = 4(     ) + 4(3) FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native ObjInfo_1060589741 Fl0l0Ole CompObjfOle10NativeEquation Native |ObjInfo       =       + 12`@yII       =       + 12_1060589742Fl0l0Ole CompObjfOle10Native FMicrosoft Equation 3.0 DS Equation Equation.39q       = 128@yII       = 12Equation Native TObjInfo_1060589743N Fl0@Fu0Ole    !%'()+/12359;<=?CEFGIJNPQRTUVZ\]^cefgijnpqrtuy{|} FMicrosoft Equation 3.0 DS Equation Equation.39q x +         +               = 180CompObj!fOle10Native#"Equation Native ObjInfo ߜ@yII x +         +               = 180 FMicrosoft Equation 3.0 DS Equation Equation.39q 9(   ) + 9(     )_1060589744&F@Fu0@Fu0Ole  CompObj%' fOle10Native)( + 9(        ) = 9(180)߬@yII 9(   ) + 9(     ) + 9(        ) = 9(180)Equation Native ObjInfo_1060589746$0,F@Fu0@Fu0Ole  FMicrosoft Equation 3.0 DS Equation Equation.39q 9x + 3x +              = 1620߄@yII 9x + 3xCompObj+-fOle10Native/.Equation Native ObjInfo" +              = 1620 FMicrosoft Equation 3.0 DS Equation Equation.39q 14x + 360 = 1620_10605897472F@Fu0@Fu0Ole #CompObj13$fOle10Native54&L@yII 14x + 360 = 1620 FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native *hObjInfo,_1060589748*B8F@Fu0@Fu0Ole -CompObj79.fOle10Native;:0Equation Native 4PObjInfo6 14x = 12604@yII 14x = 1260_1060589749>F@Fu0`|0Ole 7CompObj=?8fOle10NativeA@: FMicrosoft Equation 3.0 DS Equation Equation.39q x =     0@yII x =     Equation Native >LObjInfo@_1060589750<HDF`|0`|0Ole A FMicrosoft Equation 3.0 DS Equation Equation.39q x3 = (    )3 =       CompObjCEBfOle10NativeGFDEquation Native HObjInfoKx@yII x3 = (    )3 =        FMicrosoft Equation 3.0 DS Equation Equation.39q 2x9 + 40 = 2(_1060589751JF`|0`|0Ole LCompObjIKMfOle10NativeMLO    )9 + 40 =       ߨ@yII 2x9 + 40 = 2(    )9 + 40 =       Equation Native SObjInfoW_10605897536fPF`|0`|0Ole X FMicrosoft Equation 3.0 DS Equation Equation.39q 8 12 %$@yII 8 12 %CompObjOQYfOle10NativeSR[Equation Native _@ObjInfo` FMicrosoft Equation 3.0 DS Equation Equation.39q A =  12 h(* 1 b + * 2 b)_1060589754VF`|0`|0Ole aCompObjUWbfOle10NativeYXdEquation Native hObjInfok_1060589755T`\F`|0 0Ole lp@yII A =  12 h(* 1 b + * 2 b) FMicrosoft Equation 3.0 DS Equation Equation.39q  12 h(* 1 b +CompObj[]mfOle10Native_^oEquation Native sObjInfov * 2 b )[]t@yII  12 h(* 1 b + * 2 b )[]_1060589756bF 0 0Ole wCompObjacxfOle10Nativeedz FMicrosoft Equation 3.0 DS Equation Equation.39q            h = * 2 bd@yII            h = * 2 bEquation Native ~ObjInfo_1060589757ZrhF 0 0Ole  FMicrosoft Equation 3.0 DS Equation Equation.39q * 2 b =            h      !"#$%&(')*+-,./012345687;:]=<>?@ABCDEFGHJIKLMNOPQRSUTVWXYZ[\^_a`bcdefghijlkmnopqrstuwvxyz{|}~lT<  C A ? 2\GpH3 .s@-@85`!0GpH3 .s@-@ XJx=PjA}3g j^WPGf ڊoWKlԟ'Ղk }En飼VЀ3/.vMVOi25ɇ齥 tSEт3+&_:pc="ݕTIb(!|:KRXA&Z 5DdT<  C A ?  2WX&Ja@oWk*3!`!+X&Ja@oWk*@ XJxUPAjP}3I[BU>"ݵ`BJ^E.ugoS.qJ,4|~oB(@Zrʬ(Dyn5M0 ) -rB,|#[d{xfr.&Zv=uPƊn˫4OȈi2sd&p |.dJ\TqR%m#o[Ŀq#]ɒĖ+ƕbn,Iwb{=4Dd@<  C A ?  2r )-HY*kN`!F )-HY*k ( xcdd``~ 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infinity, b ]P@yII ( - infinity, b ]_1060589793"F2ə02ə0Ole CompObj!#fOle10Native%$ FMicrosoft Equation 3.0 DS Equation Equation.39q { x | x < -4 }D@yII { x | xEquation Native `ObjInfo_10605897942(F2ə02ə0Ole  < -4 } FMicrosoft Equation 3.0 DS Equation Equation.39q { x |             }CompObj')fOle10Native+*Equation Native tObjInfoX@yII { x |             } FMicrosoft Equation 3.0 DS Equation Equation.39q   -3(x - 1)_1060589795.F2ə02ə0Ole CompObj-/fOle10Native10<@yII d"  -3(x - 1) FMicrosoft Equation 3.0 DS EqEquation Native XObjInfo_1060589796,84F2ə02ə0Ole CompObj35fOle10Native76Equation Native ,ObjInfouation Equation.39q @yII d"_1060589797:F2ə0Й0Ole CompObj9;fOle10Native=< FMicrosoft Equation 3.0 DS Equation Equation.39q @yII d"Equation Native ,ObjInfo_1060589798@FЙ0Й0Ole  FMicrosoft Equation 3.0 DS Equation Equation.39q @yII d"CompObj?AfOle10NativeCBEquation Native ,ObjInfo FMicrosoft Equation 3.0 DS Equation Equation.39q _1060589800FFЙ0Й0Ole CompObjEGfOle10NativeIHEquation Native ,ObjInfo_1060589801DPLFЙ0Й0Ole 0@yII d" FMicrosoft Equation 3.0 DS Equation Equation.39q CompObjKMfOle10NativeONEquation Native ,ObjInfo  "#$)+,-/35679=?@ACGIJKMQSTUWX\^_`bcgijkmqstuz|}~@yII d" FMicrosoft Equation 3.0 DS Equation Equation.39q _1060589802RFЙ0ٙ0Ole CompObjQSfOle10NativeUT@yII d" FMicrosoft Equation 3.0 DS Equation Equation.39q Equation Native  ,ObjInfo _1060589803JbXFٙ0ٙ0Ole  CompObjWY fOle10Native[ZEquation Native ,ObjInfo@yII d" FMicrosoft Equation 3.0 DS Equation Equation.39q_1060589804^Fٙ0ٙ0Ole CompObj]_fOle10Nativea` @yII d" FMicrosoft Equation 3.0 DS EqEquation Native ,ObjInfo_1060589805\hdFٙ0ٙ0Ole CompObjcefOle10Nativegf!Equation Native %,ObjInfo&uation Equation.39q @yII d"_1060589806jFٙ0ٙ0Ole 'CompObjik(fOle10Nativeml* FMicrosoft Equation 3.0 DS Equation Equation.39q { x | a < x < b }P@yII { x | a < x < b }Equation Native .lObjInfo0_1060589808VpFٙ0ᙸ0Ole 1 FMicrosoft Equation 3.0 DS Equation Equation.39q { x | a  x < b }CompObjoq2fOle10Nativesr4Equation Native 8lObjInfo:P@yII { x | a d" x < b } FMicrosoft Equation 3.0 DS Equation Equation.39q { x | a < x  b}_1060589809vFᙸ0ᙸ0Ole ;CompObjuw<fOle10Nativeyx>L@yII { x | a < x d" b} FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native BhObjInfoD_1060589810t|Fᙸ0ᙸ0Ole ECompObj{}FfOle10Native~HEquation Native LhObjInfoN { x | a  x  b}L@yII { x | a d" x d" b}_1060589811Fᙸ0ᙸ0Ole OCompObjPfOle10NativeR FMicrosoft Equation 3.0 DS Equation Equation.39q { x | x < a  or  x > b }l@yII { x | x < a  or  x > b Equation Native VObjInfoY_1060589812zFᙸ0`ꙸ0Ole Z} FMicrosoft Equation 3.0 DS Equation Equation.39q ( - infinity, a) union ( b, infinity )CompObj[fOle10Native]Equation Native aObjInfodߤ@yII ( - infinity, a) union ( b, infinity ) FMicrosoft Equation 3.0 DS Equation Equation.39q_1060589813F`ꙸ0`ꙸ0Ole eCompObjffOle10Nativeh x  0  or  x > 4L@yII x d" 0  or  x > 4Equation Native lhObjInfon_1060589815F`ꙸ0`ꙸ0Ole oCompObjpfOle10NativerEquation Native v<ObjInfow FMicrosoft Equation 3.0 DS Equation Equation.39q union @yII unionI_1060589816F`ꙸ0`ꙸ0Ole xCompObjyfOle10Native{ FMicrosoft Equation 3.0 DS Equation Equation.39q -8  3x - 2  7H@yII -8 d" 3xEquation Native dObjInfo_1060589817nF`ꙸ0`ꙸ0Ole  - 2 d" 7 FMicrosoft Equation 3.0 DS Equation Equation.39q -8 +        3x - 2 +        7 +      CompObjfOle10NativeEquation Native ObjInfo߼@yII -8 +       d" 3x - 2 +       d" 7 +       FMicrosoft Equation 3.0 DS Eq_1060589818F`ꙸ0e0Ole CompObjfOle10Nativeuation Equation.39q          3x         p@yII       d"   3x   d"      Equation Native ObjInfo_1060589819Fe0@0Ole  FMicrosoft Equation 3.0 DS Equation Equation.39q       (-6 )        (3 x)        (9 )CompObjfOle10NativeEquation Native ObjInfo߸@yII       (-6 ) d"       (3 x) d"       (9 ) FMicrosoft Equation 3.0 DS Equation Equation.39q_1060589820F@0@0Ole CompObjfOle10Native -2         3H@yII -2 d"       d" 3Equation Native dObjInfo_1060589821F@0@0Ole CompObjfOle10NativeEquation Native `ObjInfo FMicrosoft Equation 3.0 DS Equation Equation.39q 0  4x + 2 < 8D@yII 0 d" 4x + 2 < 8 FMicrosoft Equation 3.0 DS Equation Equation.39q 95 + 83 + 89 + 85 + (     )(     )  90_1060589823F@0@0Ole CompObjfOle10NativeEquation Native ObjInfo_1060589824F@0@0Ole ߰@yII 95 + 83 + 89 + 85 + (     )(     ) e" 90 FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjfOle10NativeEquation Native ObjInfo 95 + 83 + 89 + 85 + x5()  ߈@yII 95 + 83 + 89 + 85 + x5() e" _1060589825F@0`.0Ole CompObjfOle10Native FMicrosoft Equation 3.0 DS Equation Equation.39q @yII e"Equation Native ,ObjInfo_1060589826F`.0`.0Ole  FMicrosoft Equation 3.0 DS Equation Equation.39q CompObjfOle10NativeEquation Native ,ObjInfo@yII e" FMicrosoft Equation 3.0 DS Equation Equation.39q _1060589827F`.0`.0Ole CompObjfOle10Native@yII e" FMicrosoft Equation 3.0 DS Equation Equation.39q Equation Native ,ObjInfo_1060589828F`.0`.0Ole CompObjfOle10NativeEquation Native ,ObjInfo@yII e" FMicrosoft Equation 3.0 DS Equation Equation.39q | ax + b | = k_1060589830F`.0 V 0Ole CompObjfOle10NativeD@yII | ax + b | = k FMicrosoft Equation 3.0 DS EqEquation Native `ObjInfo_1060589831F V 0 V 0Ole CompObjfOle10NativeEquation Native ObjInfo   "#$&*,-.04678:;?ABCEIKLMOSUVWY]_`acdhjklnosuvwy}uation Equation.39q ax + b = -k  or  ax + b = kx@yII ax + b = -k  or  ax + b = k FMicrosoft Equation 3.0 DS Equation Equation.39q | ax + b | < k     o !"#$%&'(*),+.-/12T346587:9;<=>?A@CBEDFGHIJKLMNOPRQSUVWXYZ[\^]_`abcdefgihjklmpqrstuvwxyz{|}~:2~7\{DN8!uZ`!R7\{DN8!u  xQJA}3wəD#@ \($ 鴷Hiy`aa' ~HΛyfȂ;Q(D!2k#&_ɫ?*ږh=ڄTDOy<k⺐Ղ/vj}{'u%BͷvI'd˲egMv~Uc!l>kk_[\Afx<̟'7f(~wS<0^}ޖ/5g b_,fV;NVwD#Ddl<  C A7? ;2# Yko\ԁ`!T# Yko "xQJ@}3/< WyvZ}@NEN:m,KK?Aﰲ&ҁI;yv h,IX@C\uzi]\}&v-:DTY98m <z C^}UGH~IdpZ0HΊ*e(wT1SqqbMx^y5=9iZ)۹?iJ}սvdz' =ksx6Pk/DDd l<  C A8? <2W^noMZg23`!+^noMZg2x=PNBA=3p ;L4>hqB%@ca Ԛx]dvLΙ9:vh\HY+/D!.:wtm{Ⱦ䓮{-n+MܠTR;%72ˉ&rO<z/F6(LbUm6^~tRlk1a &C x["Qg. 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^x@yII 2 - m +      = -4 +      FMicrosoft Equation 3.0 DS EqEquation Native bObjInfoe_1060589842($F $0 $0Ole fCompObj#%gfOle10Native'&iEquation Native mObjInfopuation Equation.39q 2 - m +      = 4 +     t@yII 2 - m +      = 4 +      FMicrosoft Equation 3.0 DS Equation Equation.39q -m =     _1060589843*F $0 $0Ole qCompObj)+rfOle10Native-,tEquation Native xPObjInfoz_1060589844^0F $0-0Ole {4@yII -m =      FMicrosoft Equation 3.0 DS Equation Equation.39q -m =     CompObj/1|fOle10Native32~Equation Native PObjInfo4@yII -m =      FMicrosoft Equation 3.0 DS Equation Equation.39q_10605898466F-0-0Ole CompObj57fOle10Native98     (-m) =     (-6)d@yII     (-m) =     (-6)Equation Native ObjInfo_10605898474@<F-0-0Ole CompObj;=fOle10Native?>Equation Native |ObjInfo FMicrosoft Equation 3.0 DS Equation Equation.39q     (-m) =     (2)`@yII     (-m) =     (2) FMicrosoft Equation 3.0 DS Equation Equation.39q m = 6_1060589848BF-0-0Ole CompObjACfOle10NativeEDEquation Native <ObjInfo_1060589849:RHF-0-0Ole  @yII m = 6I FMicrosoft Equation 3.0 DS Equation Equation.39q m =     CompObjGIfOle10NativeKJEquation Native LObjInfo0@yII m =      FMicrosoft Equation 3.0 DS Equation Equation.39q | 2x + 1 |  11_1060589850NF-050Ole CompObjMOfOle10NativeQPH@yII | 2x + 1 | d" 11 FMicrosoft Equation 3.0 DS EqEquation Native dObjInfo_1060589851LXTF5050Ole CompObjSUfOle10NativeWVEquation Native ObjInfouation Equation.39q       2x + 1      h@yII      d" 2x + 1 d"      FMicrosoft Equation 3.0 DS Equation Equation.39q -11 +       2x + 1 +       11 +     _1060589853ZF5050Ole CompObjY[fOle10Native]\Equation Native ObjInfo_1060589854Fv`F5050Ole ߸@yII -11 +      d" 2x + 1 +      d" 11 +      FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj_afOle10NativecbEquation Native hObjInfo -12  2x      L@yII -12 d" 2x d"     _1060589855fF5050Ole CompObjegfOle10Nativeih FMicrosoft Equation 3.0 DS Equation Equation.39q     (-12)      (2x)      (10)ߜ@yII     (-12) d"     (2x)Equation Native ObjInfo_1060589856dplF50>0Ole  d"     (10) FMicrosoft Equation 3.0 DS Equation Equation.39q         x        CompObjkmfOle10NativeonEquation Native ObjInfod@yII      d"   x   d"      FMicrosoft Equation 3.0 DS Equation Equation.39q | 6 - 7x | > 22_1060589857rF>0>0Ole CompObjqsfOle10NativeutH@yII | 6 - 7x | > 22 FMicrosoft Equation 3.0 DS EqEquation Native dObjInfo_1060589858jxF>0>0Ole CompObjwyfOle10Native{zEquation Native `ObjInfouation Equation.39q | ax + b | = kD@yII | ax + b | = k_1060589859~F>0>0Ole CompObj}fOle10Native  "#$&*,-.04678:>@ABDHJKLNRTUVX[\]^_`bcdefghj FMicrosoft Equation 3.0 DS Equation Equation.39q | ax + b | < kD@yII | ax + Equation Native `ObjInfo _1060589861|F>0>0Ole  b | < k FMicrosoft Equation 3.0 DS Equation Equation.39q | ax + b | > kCompObj fOle10Native Equation Native `ObjInfoD@yII | ax + b | > k FMicrosoft Equation 3.0 DS Equation Equation.39q | ax + b | = 0_1060589862F>0QF0Ole CompObjfOle10NativeD@yII | ax + b | = 0 FMicrosoft Equation 3.0 DS EqEquation Native `ObjInfo_1060589863FQF0QF0Ole CompObjfOle10Native!Equation Native %TObjInfo'uation Equation.39q ax + b = 0.8@yII ax + b = 0._1060589864FQF0QF0Ole (CompObj)fOle10Native+ FMicrosoft Equation 3.0 DS Equation Equation.39q | 2x + 1 | < -5H@yII | 2x + Equation Native /dObjInfo1_1060589865FQF0QF0Ole 21 | < -5 FMicrosoft Equation 3.0 DS Equation Equation.39q | 2x + 1 | > -5CompObj3fOle10Native5Equation Native 9dObjInfo;H@yII | 2x + 1 | > -5 FMicrosoft Equation 3.0 DS Equation Equation.39q | 2x + 1 | = 0_1060589866FQF0QF0Ole <CompObj=fOle10Native?D@yII | 2x + 1 | = 0 FMicrosoft Equation 3.0 DS EqEquation Native C`ObjInfoE_1060589868FQF0QF0Ole FCompObjGfOle10NativeIEquation Native M`ObjInfoOuation Equation.39q | y - 5 | > -2D@yII | y - 5 | > -2_1060589869FQF0QF0Ole PCompObjQfOle10NativeS FMicrosoft Equation 3.0 DS Equation Equation.39q | 3m + 4 | < -10L@yII | 3m + Equation Native WhObjInfoY1Table!SummaryInformation(Z4 | < -10 Oh+'0t  0 < HT\dlWorksheet 5 (2orkTL UserL UL U Normal.dot TL User2 UMicrosoft Word 8.0@@x0@x <  C Ak? {2̛\Mu:`!m̛\MjH;xcdd``6ed``beV dX,XĐ c A?db~6FnĒʂT ~35;a#&br<FKt@ڈ+6W|3^gEܨR`6 Ud0!_AOc`Ml?XQoOXPŒ}̨\̨Ä*?sQ 8."%v0o8+KRs!Fh V-J}Ddt@<  C Al? |2l::)hٰ\LHc`!@::)hٰ\L  xcdd``~ @bD"L1JE `x؅,56~) M @ ?`c`8 UXRYvo`0L` ZZr.PHq%0f{XA|8 oOcAk|-42f? `Fw6*H0~.hpÎP``CI)$5OZI:.DdP <  C Am? }2nY5U 1ê;z_`!rnY5U 1ê;lH@xcdd``6cd``beV dX,XĐ c A?db 7$# L aA $37X/\!(?71]46J`l͆5|8߅ķPxA$8U^׆pq!?BN(?$ױ  0_‚j#ocFbF&T>È*o4p< 0y{Ĥ\Y\ I9@@f:fIDd<  C An? ~2p;rS1-E8m#(L`!D;rS1-E8m#(@ PVHxcdd`` @bD"L1JE `x0:p Yjl R A@1h]P#7T obIFHeA*0d @Hfnj_jBP~nbC#%@i mĕhR `/l U+o`E?kXPAX_aBj:F41e8,"%v0o8+KRs! 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