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Math 117 Lecture 14 notes The question “What is algebra?” has several answers. For example, “Algebra is generalized arithmetic” and “Algebra is a language” are frequently offered. Others describe algebra as “the mathematics of patterns.” Each of these descriptions alludes to the key idea that algebra is an extension of arithmetical thinking. In particular, algebra plays these three roles in mathematical reasoning: Algebra describes generality. For example, saying that 5 + 8 = 8 + 5 is a fact about the commutativity of addition for a particular pair of whole numbers, but the fundamental idea is that 5 and 8 can be replaced with any two whole numbers. This can be said in words, but the idea is conveyed most effectively with the algebraic statement “m + n = n + m for any two whole numbers m and n.” Algebra solves problems. Consider this problem: “in two more years, Angie will be twice the age of her three-year-old brother; how old is Angie?” To solve this problem algebraically, let the symbol x represent Angie’s unknown age, a placeholder for a number we wish to determine. The problem stated in words is then equivalent to the symbolic statement x + 2 = 22(3 + 2). That is, x + 2 = 10 and so x = 10 – 2 = 8. Thus, the equation has been solved to reveal that Angie is currently eight years old. Here the letter x played the role of an unknown. Algebra reveals and explains patterns. The figure below shows the first three “trapezoid” trains constructed from matchsticks. Consider the question: What is the relationship between the number of trapezoids in the train and the number of matchsticks required to form the train? This relationship can be expressed the equation m = 1 + 4t, where the variable t denotes the number of trapezoids and the variable m denotes the number of matchsticks. Algebra Reform = Algebraic Thinking • Numerical • Symbolic • Graphical • Application to “real life” = Understanding Algebraic Connections Multiply 23 x 54 Solving an equation means “undoing” what’s happened to x. Generally speaking, the “order of operations” is reversed to solve equations. 2 Solve: 2x + 3 = 7 Solve: x + 4 = 13 Math 117 Lecture 13 page 2 Factor: 2 x + 5x + 6 Multiply: (x + 3)(x + 1) Simplify: 2 x + 1 + x + 3x + 1 + x 2 +x 2 The Ancient Greeks were able to solve equations such as x + because they could apply Geometry to find a solution. 10x = 39 Bottle and Cork Problem: A bottle and a cork together cost $1.00. If the bottle costs 90 cents more than the cork, how much does the cork cost? In a. b. c. d. understanding the problem, each of the following is true except: the combined cost of the bottle and cork is $1 the cork cost 10 cents the bottle costs 90 cents more than the cork we are asked to find the cost of the cork You decide to try to strategy of using a variable to solve the problem. You let c represent the cost of the cork in cents. The cost of the bottle in cents is represented by: a. 100 + c b. 90 c. c + 90 d. none of these Proceeding to solve the problem, you find the cost of the bottle is: a. 95¢ b. 90¢ c. 10¢ d. 5¢ Looking back, what other strategies might you have used? Given: 3x + 4 = 19 Problem-solving strategy: Guess and test. As the name of this method suggests, one guesses values for the variable and substituttes to see if a true equation results. Problem-solving strategy: Work backwards. The left side of the equation shows that x was ultiplied by 3 and then 4 was added to botain 19. Thus, if we subtract 4 from 19 and divide by 3, we can work backward to the value of x. Here 19 – 4 = 15 and 15 ÷ 3 = 5, so x = 5. The diagram summarizes this. x3 5 +4 19 15 5 ÷3 –4 Math 117 Lecture 13 page 3 Problem-solving strategy: Draw a Picture Take away 4 from each side Divide the ones into three equal groups (one group for each square). Therefore, = or, x =5 Research tells us that the use of objects and containers as analogues for variables help students solve linear equations. This analogue positively influenced not only student’s achievement, but also their attitude toward the topic. We say that two algebraic expressions are equivalent to each other if, for each assignment of numberical values to the variables in the expressions, the values of the two expressions are equal. In other words, functions that yield exactly the same set of data pairs are called equivalent. Two equations are equivalent if they have the same solution set. Ex: 0.5(a + b) is equivalent to a/2 + b/2 Ex: 3x + 1 = 7 is equivalent to x = 2 What permits you to create equivalent expressions and equations, and what prevents it? When transforming expressions into equivalent ones, consider: • commutative law for addition and multiplication 3 2 2 – 5 does not equal 5 – 2 and 2 does not equal 3 • associative law for addition and multiplication (2x + 3) + 7 is equivalent to 2x + (3 + 7) Math 117 Lecture 13 page 4 • opposite of a sum and opposite of a product –(a + b) is equivalent to (–a) + (–b) –(ab) is equivalent to (–a)•b and a(–b) • division is “multiplying by the reciprocal” 6x ÷ 2 is equivalent to 1/2(6x) • distributive property for expressions, or order of operation powers & roots multiplication & division addition & subtraction adding “like terms” is really using the distributive property Ex: 2x + 3x is really (2+ 3)x or 5x • the “Power of One” for simplifying complex fractions Ex: 4 3 3 9 for solving equations Ex: 3x = 18 Rules for transforming equations into equivalent ones: • additive property of equality • multiplicative property of equality • zero product property Ex: (x – 2)(x + 3) = 0 but not (x – 2)(x + 3) = a non-zero Is there a property for squaring both sides of an equation? Raising both sides to any power? Finding roots of both sides? When will this operation work, and when will it fail? Two equations are equivalent if they have the same solution set. Unfortunately, there are transformations which you can perform on an equation which seem perfectly correct, but which change the solution set. Since it is the original equion you are trying to solve, you must be aware of the types of transformations which might not produce an equivalent equation. 1. Multiplying by an expression which can equal zero (an extraneous solution is a number which satisfies a transformed equation by not the original equation) Ex: x = 5 (x – 2) x = (x – 2) 5 Multiplying both members of an equaion by an expression which Math 117 Lecture 13 page 5 can equal zero is called an irreversable step. You cannot go backwards and divide both members by that expression since division by zero is undefined. 2. Dividing by a variable can lose a valid solution. 2 Ex: x = 4x Consider the equation: 42(x – 4) + 210 = 252 Below are two different solution paths for this equation. Path 1 Path 2 42(x – 4) + 210 = 252 42(x – 4) + 210 = 252 (x – 4) + 5 = 6 42x – 168 + 210 = 252 x+1 = 6 42x + 42 = 252 x=5 42x = 210 x = 5 A good teacher will guide students toward looking at various strategies for solving problems. Ex: For each of the following equations or inequalities, i. Solve using the properties of equations or inequalities ii. Solve again, but this time use a different solution path – one starting with a different first step iii. Find still another solution path – one with a different number of steps than the first iv. Finally, compare the three solution parths. What are the advantaes and disadvantages of each? 1. 2. 3. 4. 3x 5 3x " 4= 3x + 30 15 " 12 +3> x "2 8 0.75(52x + 28) = 2.25(44x + 20) 8123 – 4.2x < 7673 – 3.7x Beberman text, 1962: “The unofficial score of a nation in the Olympic Games can be found by the following formula. S = 10x + 5y + 4z + 3w + 2r + t In this formula x = the number of first places y = the number of second places z = the number of third places w = the number os fourth places r = the number of fifth places t = the number of sixth places S = the total score of the nation The right member of this formula is an example of a polynomial. Bebermans’ text contains two history sections entitled “The Story of Algebra,” Part 1 and Part 2, which give in brief form some ideas of how algebra has developed over the centuries. For example, Math 117 Lecture 13 page 6 “An algebra problem appeared in a book written in Egypt 5000 years ago and was copied into another book about 3500 years ago by a scribe named Ahmes. Ahmes wrote on a coarse paper called papyrus, and part of the book that he copied is still in existence. One of the problems in his book as we might state it is: If one seventh of a number is added to the number, the sum is 19. As Ahmes wrote it: Aha (heap), 1/7 of it added to it, becomes it: 19. It is interesting to see how the Egyptians solved this problem … And, of course, the problem that Beberman made famous: “If a cricket chirps 100 times a minute, how fast does an ant run? Metric System of Measruement, SI (Le Systeme International d’Unites) Base units in SI (Metric System) length – meter (m) mass – kilogram (kg) temperature – degrees Celsius (°C) or Kelvin time – second (s) luminous intensity – candella (cd) amount of substance – mole (mol) electrical current – ampere (A) A related unit (not a base unit): volume – liter (l or L) SI prefixes (1991): 10 10 24 21 18 10 15 10 12 10 10 10 10 10 9 6 3 2 1 10 yotta zetta exa peta tera giga mega kilo hecto deka Y Z E P T G M k h da –1 10 –2 10 –3 10 –6 10 –9 10 –12 10 –15 10 –18 10 –21 10 –24 10 deci d centi c milli m micro µ nano n pico p femto f atto a zepto z yocto y Rules for using metric prefixes: • Prefix symbols are printed in roman type (upright) without spacing between the prefix symbol and the unit symbol. Math 117 Lecture 13 page 7 • • • The grouping formed by the prefix symbol attached to the unit symbol constitutes a new inseparable symbol which can be raised to a positive or negative power. Compound prefixes are not to be used. A prefix should never be used alone. “Metric sense” means being able to select the best unit of measure for a situation and being able to estimate the measurement. Examples: A newborn baby might weigh 3 kg An average adult might weigh 60 kg A bedroom might measure 3x4 m A paperclip is about 1 cm wide and weighs 1 g Room temp is about 22 °C Your shoe size might be 26 cm A football field is about 100 m A bath might use 250 L of water To convert between one metric unit and another, move the decimal point the correct number of places in the direction of the desired unit. Ex: 240 cm = m 1 dm = mm 0.8 L = ml 230 mg = g 3500 mm = cm 10 ml = L 0.03 hm = dm 0.93 cm = m 0.08 kg = µg 57 cg = kg 31 cm3 = ml 1200 cc = L 1 cubic decimeter (a cube 10 cm x 10 cm x 10 cm) = 1 liter 3 and 10 x 10 x 10 cm = 1000 cubic cm (cc or cm ) and 1/1000 of a liter is a milliliter, 3 then 1 cc = 1 cm = 1 ml = A nice feature of the metric system is the relationship between length, volume, and mass. 1 liter of water at 4° C has a mass of 1 kg 1 cc of water at 4° C has a mass of 1 g Metric Program National Institute of Standards and Technology Technology Administration U.S. Department of Commerce www.nist.gov/metric (email) [email protected]