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next Algebra Problems… Solutions Set 5 © 2007 Herbert I. Gross By Herbert I. Gross and Richard A. Medeiros next Problem #1 What signed number is 2 named by 3( 4) ? Answer: -48 © 2007 Herbert I. Gross next Answer: -3(-4)2 = -48 Solution: We do exponents before we multiply. What's being raised to the second power (squared) is -4. Since the product of two negative numbers is positive, we see that (-4)2 = (-4)(-4) = +16 Replacing (-4)2 by +16, we obtain… -3(-4)2 = -3(+16) And since the product of two numbers that have opposite signs is negative, -3(+16) = -48 © 2007 Herbert I. Gross next Careful Don't overlook the exponent. That is, if we look at the expression… -3(-4)2 too quickly, it might seem that we are multiplying two negative numbers. We have to observe that -3 is not multiplying -4, but rather (-4)2, which is +16. © 2007 Herbert I. Gross next Careful Since we raise to a power before we multiply, the exponent refers only to -4. If instead we had wanted to multiply -4 by -3 before we raised the expression to the second power, we would have had to write the expression as… [-3(-4)]2 In this case, the answer would have been given by… [-3(-4)]2 = [+12]2 = +144 © 2007 Herbert I. Gross next PEMDAS (Review) Once again, we have to stress that there is ambiguity when grouping symbols are omitted; for that reason, we need some general agreement concerning the order of operations. And the agreement used most generally, and which we will use, goes under the acronym PEMDAS. © 2007 Herbert I. Gross next PEMDAS Review In PEMDAS the agreed Order of Operations that we act on is… Parentheses Exponents Multiplication Division Addition and Subtraction © 2007 Herbert I. Gross next Problem #2 What signed number is 2 named by 8 – ( 3) ? Answer: -1 © 2007 Herbert I. Gross next Answer: 8 – (-3)2 = -1 Solution: By our order of operations (PEMDAS agreement) we read 8 – (-3)2 as 8 – [(-3)2]. That is, we do what's inside the parentheses first, then raise to powers before we subtract. We know that (-3)2 means (-3)(-3) and that (-3)(-3) = +9. © 2007 Herbert I. Gross next Solution (cont.): Therefore we may rewrite the expression 8 – [(-3)2] in the equivalent form… +8 – +9 By the add-the-opposite rule we may rewrite the expression +8 – +9 in the equivalent form… +8 + -9 = -1 and by our rule for adding two numbers that have different signs, we see that the answer to our problem is -1. © 2007 Herbert I. Gross next Notes on #2 Again, notice the importance of the grouping symbols. We multiply -3 by itself before we subtract it from +8. Yet if we had read the expression 8 – (-3)2 too quickly, we might have confused it with the expression 8 – -(3)2. © 2007 Herbert I. Gross next Notes on #2 Then in calculating the expression 8 – -(3)2, we would have used the add-the-opposite rule and our answer would have been… 8 + +(3)2 = 17. In short, 17 is the correct answer to 8 + (-3)2, BUT is the incorrect answer to the given problem, which was… 8 – (-3)2. © 2007 Herbert I. Gross next Problem #3 What signed number is named by (-15 + -3) × (-6 ÷ +2)? Answer: +54 © 2007 Herbert I. Gross next Answer: (-15 + -3) × (-6 ÷ +2) = +54 Solution: Since there are no exponents, and we are using the order of operations (PEMDAS), we begin by working inside the parentheses first. We know from the previous lessons that… -15 © 2007 Herbert I. Gross + -3 = -18 next Solution: …and from the present lesson we know that when we divide two signed numbers we divide their magnitudes and choose the positive sign for the quotient if the two numbers have the same sign and the negative sign otherwise. This means that… -6 © 2007 Herbert I. Gross ÷ + 2 = -3 next Solution: If we now substitute the results of our two equations into the expression… (-15-18 + -3) × (-6 -÷3 +2) = +54 we obtain the equivalent expression which by our rule for multiplying two numbers that have the same sign is equal to +54 © 2007 Herbert I. Gross next Solution: In more compact form, what we did was… ( 15 + -3) -18 × × (6 ÷ -3 = +54 © 2007 Herbert I. Gross +2) = next Careful In this problem we had to perform three different arithmetic operations with signed numbers. When we have to perform several different types of operations, we have to be careful not to confuse one rule with another. Unless we have internalized the rules, it is all too easy for example to be confused about why the sum of two negative numbers is always negative, but the product of two negative numbers is always positive. © 2007 Herbert I. Gross next Problem #4a Evaluate (t2 – 5)t when t = 5. Answer: -100 © 2007 Herbert I. Gross next Answer: -100 Solution: (t2 – 5)t Using PEMDAS we first work inside the parentheses and replace by t by -5. Thus t2 becomes (-5)2; and (-5)2 means (-5) (-5). By our rule for multiplying two numbers that have the same sign (-5) (-5) = +25. t2 = (-5)2(-5) = +25 © 2007 Herbert I. Gross next Solution: Thus when t = -5. 2 – +5 = +t25 5 = +20 replacing t by -5, (t2 – 5)t becomes... which by our rule for multiplying signed numbers is equal to -100. © 2007 Herbert I. Gross -100 -5) (20)( next Note on #4a Until you have internalized the arithmetic of signed numbers and the language of algebra, a good strategy is to translate the expression into “English” first. More specifically, given the algebraic expression (t2 – 5)t we work within the parentheses first and we raise to the second power before we subtract. After that, we do the indicated multiplication. © 2007 Herbert I. Gross next (t2 – 5)t Note on #4a Thus, in the “plain English” format; the expression (t2 – 5)t becomes… Step 1 Step 2 Start with any number t Multiply it by itself t2 Step 3 Step 4 Subtract 5 t2 – 5 Multiply this by t t2 – 5t © 2007 Herbert I. Gross -5 (-5) (- 5) – +5 (20)(-5) + 25 = = = +25 +20 -100 next Note on #4a The table/recipe shows each step of the calculation, and shows -5 replacing t. Better yet, the order in which the steps are written already indicates the order of operations, in advance. This lets us concentrate on the arithmetic, without being distracted by such questions as “Do I change the sign first or do I multiply first”, etc. © 2007 Herbert I. Gross next Problem #4b Evaluate t2 – 5t when t = 5. Answer: +50 © 2007 Herbert I. Gross next Answer: +50 Solution: Since no grouping symbols appear we use our PEMDAS agreement to rewrite t2 – 5t… in the form… (t )2 – (5t ) In words the expression would read… Start with any number (t) Square it (t2) Subtract 5 times the number you chose (t2 – 5t) © 2007 Herbert I. Gross next Solution: (t)2 – (5t) So we start with t = -5… We multiply it by itself; (t)2 We multiply t by 5… We subtract 5t from t2 to obtain… Using the add the opposite rule we obtain… Start with any number t Multiply it by itself t2 Multiply t by 5 5t Subtract 5t from t2 t2 – 5t (Or by adding the opposite) © 2007 Herbert I. Gross -5 (-5) (- 5) -25 +25 – -25 +25 + +25 = +25 = +50 next Note on #4b Once you are comfortable working with the algebraic notation, simply replace t by (-5) wherever it appears in the expression. (-5)t22 –– 55 t(-5) +25 – -25 +25 + +25 = +50 © 2007 Herbert I. Gross next Note on #4b However, until you feel comfortable with manipulating the algebraic symbols, practice translating the algebraic expressions into “plain English”. Otherwise, for example, it might be easy to confuse (t2 – 5)t with t2 – 5t. © 2007 Herbert I. Gross next Problem #4c Evaluate t2 – 5t when t = 6? Answer: +66 © 2007 Herbert I. Gross next Answer: +66 Solution: This is word-for-word the same as Exercise 4b except that we replace -5 by -6. More specifically, in verbal form evaluating the expression t2 – 5t when t = -6 means that… Start with any number (t) Square it (t2) Subtract 5 times the number you chose (t2 – 5t) © 2007 Herbert I. Gross next Solution: (t)2 – (5t) So we start with t = -6… We multiply it by itself; (t)2 We multiply t by 5… We subtract 5t from t2 to obtain… Using the add the opposite rule we obtain… Start with any number t Multiply it by itself t2 Multiply t by 5 5t Subtract 5t from t2 t2 – 5t (Or by adding the opposite) © 2007 Herbert I. Gross -6 (-6) (- 6) -30 +36 – -30 +36 + +30 = +36 = +66 next Looking Ahead The reason for giving two such closely connected problems such as #4b and #4c will become more transparent when we present the solution for problem #4d. The key point in #4d is that if t2 – 5t = +50 when t = -5, and it equals 66 when t = -6 then if n represents any number that is between 50 and 66, there is at least one value for t between -5 and -6 for which t2 – 5t = n. In #4d, we chose 55 as the value of n. © 2007 Herbert I. Gross next Problem #4d Use the answers to parts (b) and (c) to find a value of t for which 2 t – 5t = 55 (Round off your answer to the nearest integer). Answer: t = -5 © 2007 Herbert I. Gross next Answer: t = -5 Solution: In #4b, we saw that if we chose t to be -5 the value of t2 – 5t would be 50; while in #4c, we saw that if we chose t to be -6, the value of t2 – 5t would have been 66. Since 55 is less than 66 but greater than 50, we may conclude that there must be at least one value of t between -5 and -6 such that t2 – 5t = 55. © 2007 Herbert I. Gross next Solution: Moreover because 50 is closer in value to 55 than 66 is, we assume that the required value of t is closer to -5 than to -6 which leads us to believe that the value of t, rounded off to the nearest integer, is -5 © 2007 Herbert I. Gross next Note on #4d To verify our assumption (and we might use a calculator to make the necessary computations less tedious to obtain), we could make a table such as… t t2 5t t2 – 5t -5.0 +25 -25 +25 – -25 -5.1 +26.01 -25.5 -5.2 +27.04 -5.3 +28.09 -26.5 -5.4 +29.16 © 2007 Herbert I. Gross -26 -27 +26.01 – -25.5 = +50 = +51.5 +27.04 – -26 = +53.04 +28.09 – -26.5 = +54.59 +29.16 – -27 = +56.16 next Note on #4d Since 55 is between 54.59 and 56.16, we know that the required value of t is, in fact, between -5.3 and -5.4… t t2 5t t2 – 5t -5.0 +25 -25 +25 – -25 -5.1 +26.01 -25.5 -5.2 +27.04 -5.3 +28.09 -26.5 -26 ? ? ? -5.4 +29.16 -27 © 2007 Herbert I. Gross +26.01 – -25.5 = +50 = +51.5 +27.04 – -26 = +53.04 +28.09 – -26.5 = +54.59 +55 +29.16 – -27 = +56.16 next Note on #4d If we wanted an even better approximation, we could evaluate the expression t2 – 5t, say when t = -5.31, -5.32, -5.33, -5.34, etc. For example, if we let t = -5.33, we obtain… t t2 5t -5.33 +28.4089 -26.65 t2 – 5t +28.4089 – -26.65 = +55.0589 which indicates that the required value of t is very nearly equal to -5.33. © 2007 Herbert I. Gross next This problem illustrates an important technique for solving equations that may be beyond the scope of our algebra knowledge. Later in the course, we will learn how to obtain exact algebraic solutions to equations such as t2 – 5t = 55. Meantime, there is nothing wrong with making guesses (sometimes referred to as trial-and-error or guess-and-check). By repetitive guessing and checking, we can correctly find the value of t accurate to any required number of decimal places. © 2007 Herbert I. Gross next Careful Be careful not to assume that there is only one value of t for which t2 – 5t. For example, when t = 10, t2 – 5t = 50. And when t = 11, t2 – 5t = 66. Hence there is another value of t that is between 10 and 11 for which t2 – 5t = 55. How can this be, and are there still other values of t for which t2 – 5t = 55? Questions such as these will be answered as our course continues. © 2007 Herbert I. Gross next Problem #5a The formula that relates the Fahrenheit temperature (F) to the Celsius temperature (C) is… C = 5/9 (F – 32). What is the Celsius temperature when the Fahrenheit temperature is 14°? © 2007 Herbert I. Gross Answer: -10°C next Answer: -10°C Solution: Starting with the formula C = 5/9 (F – 32), we replace F by 14 and then obtain the following sequence of steps… C = 5/9( F – +32) = 5/9(+14 – +32) = 5/9(+14 + -32) = 5/9(-18) = -10 © 2007 Herbert I. Gross next To review the logic behind each step, we start with… We first replaced F by +14… We then used the “add the opposite rule” to obtain… And finally, we used our rules of arithmetic to obtain… F – +32) C = 5/9(+14 C = 5/9(+14 + -32) C = 5/9 ( -18) © 2007 Herbert I. Gross next Answer: -10°C Solution: Since we are multiplying a positive number (5/9) by a negative number (-18) the product will be negative and since 5/9 × 18 = 10, we see that the equation can be written in the form… -10 5 C = C = /9(-18) © 2007 Herbert I. Gross next Problem #5b The formula that relates the Fahrenheit temperature (F) to the Celsius temperature (C) is… C = 5/9 (F – 32). What is the Fahrenheit temperature when the Celsius temperature is -20°? © 2007 Herbert I. Gross Answer: -4°F next Answer -4°F Solution: In part 5a we were faced with a direct computation (arithmetic). In this part we are faced with an indirect computation (algebra). That is, we are still working with the formula… but now we replace C by -20 to obtain the equation. -20 C © 2007 Herbert I. Gross = 5/9( F – +32) next Solution: Since 5/9 is multiplying what's inside the parentheses, we “unblock” the parentheses by dividing both sides of the equation by 5/9. Dividing by 5/9 is equivalent to multiplying by 9/5. Hence, we may rewrite the equation as… 9/ © 2007 Herbert I. Gross -20 = 5 9/ (× F 5/ – +F +32) × ( 32) – 5 5 9 9 next Solution: Since 9/5 × -20 = -36 and 9/5 × 5/9 = 1, we may rewrite the equation as... 9/ -20 -36== 91 5/– )+(32) +32) × ( / (F × F – 5 5 9 -36 © 2007 Herbert I. Gross = F – +32 next Solution: By the “add the opposite” rule, we may rewrite… and we now add +32 to both sides of the equation to obtain… Since -36 + +32 = -4 and -32 + +32 = 0, we may rewrite the equation as… = F –++-32 --36 36 ++ ++32 32 == FF ++ --32 32 ++ ++32 32 -4 =(F + 0 ) -36 © 2007 Herbert I. Gross next Note on #5 Using the recipe model that we introduced in Lesson 1 we can illustrate the connection between #5a and #5b. Namely, in words, the formula C = 5/9(F – 32) tells us how to find the Celsius temperature once we know the Fahrenheit temperature. © 2007 Herbert I. Gross next “Recipe” In terms of our “program”… problem 5a becomes… Converting Fahrenheit to Celsius Start with Fahrenheit Temp. F 14 Subtract 32 F – 32 14 – 32 = -18 5/ (F – 32) 5/ (-18) Multiply by 5/9 9 9 -10 Answer is Celsius Temp. C = 5/9(F – 32) © 2007 Herbert I. Gross next Calculator Notice that if your calculator doesn't have a fraction mode, multiplying by 5/9 is the same as dividing by 9 and then multiplying by 5 (or first multiplying by 5 and then dividing by 9). Therefore, the sequence of key strokes might be… 14 © 2007 Herbert I. Gross – 32 ÷ 9 × 5 = next Recipe • For problem #5b we would want to use the “undoing” process. That is, if the last step was multiplying by 5, the first step in the “undoing” program would be to start with the Celsius temperature and then divide by 5. We would next “undo” dividing by 9 which means multiplying by 9. Finally, we would “undo” subtracting 32 by adding 32. © 2007 Herbert I. Gross next Doing & Undoing • In terms of a “plain English” chart… Doing Recipe Undoing Recipe Start with F Answer is F Subtract 32 Add 32 Multiply by 5/9 Divide by 5/9 i.e. Multiply by 9/5 Answer is C Start with C © 2007 Herbert I. Gross next Note on #5 Once we have used a “plain English” chart long enough, it becomes more natural for us to work with the algebraic format. For example, starting with C = 5/9( F – +32) © 2007 Herbert I. Gross next Note on #5 To “unblock” the parentheses (that is, to “undo” the last step in the formula) we multiply both sides by 9/5 to obtain… 9/ 9/ 59/ (F +32) 5/ – +32)] C = [ (F – 95 5 9 9/ • 5/ )(F – +32) C = ( 5 5 9 9/ © 2007 Herbert I. Gross +32) C = 1(F – 5 next Note on #5 In other words, the formula now becomes… and if we now add 32 to both sides of the above formula we see that… 9/ C 5 + = 32 F –=32 F F = 9/5C + 32 or in more traditional format… © 2007 Herbert I. Gross next Note on #5 The two formulas look quite different, but they are equivalent ways to express the relationship between F and C. Most likely if we wanted to do only direct computations, we would use formula C = 5/9(F – 32) if we were told the value of F, and we wanted to find the corresponding value of C; and we would use formula F = 9/5C + 32 if we were given the value of C, and we wanted to find the corresponding value of F. © 2007 Herbert I. Gross next Note on #5 Recall that 0°C represents the freezing point of water and 100°C represents the boiling point of water. Thus, while we don't often think of it in those terms, the fact is that the Celsius scale is analogous to percents. For example, if the temperature is 20°C it means that it is 20% of the way between the freezing point of water and the boiling point of water. That's one nice advantage that Celsius has over Fahrenheit. © 2007 Herbert I. Gross next Note on #5 On the other hand, a degree on the Celsius scale represents a greater change in temperature than a degree on the Fahrenheit scale. More specifically, the rate is 5°C per 9°F. Thus, a Celsius degree is almost double a Fahrenheit degree. Therefore, being off by 1 degree on the Celsius scale represents a greater error than by being off 1 degree on the Fahrenheit scale. © 2007 Herbert I. Gross