Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 5 By Herbert I. Gross and Richard A. Medeiros 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 next © 2007 Herbert I. Gross

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. next © 2007 Herbert I. Gross 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 next © 2007 Herbert I. Gross Careful In this case, the answer would have been given by… [ - 3( - 4)] 2 = [ + 12] 2 = + 144 next

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. © 2007 Herbert I. Gross PEMDAS (Review) And the agreement used most generally, and which we will use, goes under the acronym PEMDAS.

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. next © 2007 Herbert I. Gross

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 and by our rule for adding two numbers that have different signs, we see that the answer to our problem is - 1. next © 2007 Herbert I. Gross = - 1 next

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 Notes on # 2

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. next © 2007 Herbert I. Gross 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. next Notes on #2

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 + - 3 = - 18 next © 2007 Herbert I. Gross

next © 2007 Herbert I. Gross 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 ÷ + 2 = - 3

next © 2007 Herbert I. Gross Solution: If we now substitute the results of our two equations into the expression… - 18 × - 3 ( - 15 + - 3) × ( - 6 ÷ + 2) we obtain the equivalent expression which by our rule for multiplying two numbers that have the same sign is equal to + 54 = + 54 next

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. next © 2007 Herbert I. Gross Careful

Answer: - 100 Solution: Using PEMDAS we first work inside the parentheses and replace by t by - 5. Thus t 2 becomes ( - 5) 2 ; and ( - 5) 2 means ( - 5) ( - 5). By our rule for multiplying two numbers that have the same sign ( - 5) ( - 5) = + 25. next © 2007 Herbert I. Gross (t 2 – 5)t = + 25 = ( - 5) ( - 5)t 2 = ( - 5) 2 next

Solution: Thus when t = - 5. next © 2007 Herbert I. Gross = + 20 = + 25 – + 5 t 2 – 5 next replacing t by - 5, (t 2 – 5)t becomes... - 100(20)( - 5) which by our rule for multiplying signed numbers is equal to - 100. next

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 © 2007 Herbert I. Gross Note on # 4a (t 2 – 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.

Thus, in the “plain English” format; the expression (t 2 – 5)t becomes… next © 2007 Herbert I. Gross Note on # 4a Step 1Start with any number t -5-5 Step 2Multiply it by itself t 2 ( - 5) = + 25 Step 3 Subtract 5 t 2 – 5 + 25 – + 5 = + 20 Step 4Multiply this by t t 2 – 5t(20)( - 5) = - 100 (t 2 – 5)t

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. next © 2007 Herbert I. Gross Note on # 4a

Solution: Since no grouping symbols appear we use our PEMDAS agreement to rewrite t 2 – 5t… next © 2007 Herbert I. Gross t 2 – 5t next ()() in the form… In words the expression would read… Start with any number (t) Square it (t 2 ) Subtract 5 times the number you chose (t 2 – 5t)

Solution: So we start with t = - 5… next © 2007 Herbert I. Gross (t) 2 – (5t) next We multiply it by itself; (t) 2 next Start with any number t -5-5 Multiply it by itself t 2 ( - 5) = + 25 Multiply t by 5 5t - 25 We multiply t by 5… We subtract 5t from t 2 to obtain… Subtract 5t from t 2 t 2 – 5t + 25 – - 25 next Using the add the opposite rule we obtain… (Or by adding the opposite) + 25 + + 25 = + 50 next

Once you are comfortable working with the algebraic notation, simply replace t by ( - 5) wherever it appears in the expression. © 2007 Herbert I. Gross Note on # 4b + 25 + + 25 + 25 – - 25 ( - 5) 2 – 5 ( - 5)t 2 – 5 t = + 50 next

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 (t 2 – 5)t with t 2 – 5t. © 2007 Herbert I. Gross Note on # 4b

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 t 2 – 5t when t = - 6 means that… next © 2007 Herbert I. Gross Start with any number (t) Square it (t 2 ) Subtract 5 times the number you chose (t 2 – 5t) next

Solution: So we start with t = - 6… next © 2007 Herbert I. Gross (t) 2 – (5t) next We multiply it by itself; (t) 2 next Start with any number t -6-6 Multiply it by itself t 2 ( - 6) = + 36 Multiply t by 5 5t - 30 We multiply t by 5… We subtract 5t from t 2 to obtain… Subtract 5t from t 2 t 2 – 5t + 36 – - 30 next Using the add the opposite rule we obtain… (Or by adding the opposite) + 36 + + 30 = + 66 next

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 t 2 – 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 t 2 – 5t = n. In #4d, we chose 55 as the value of n. © 2007 Herbert I. Gross Looking Ahead

Solution: In #4b, we saw that if we chose t to be - 5 the value of t 2 – 5t would be 50; while in #4c, we saw that if we chose t to be - 6, the value of t 2 – 5t would have been 66. next © 2007 Herbert I. Gross 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 t 2 – 5t = 55. 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 next © 2007 Herbert I. Gross

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… next © 2007 Herbert I. Gross Note on # 4d tt2t2 5t5tt 2 – 5t - 5.0 + 25 - 25 + 25 – - 25 - 5.1 + 26.01 - 25.5 + 26.01 – - 25.5 = + 50 - 5.2 + 27.04 - 26 + 27.04 – - 26 - 5.3 + 28.09 - 26.5 + 28.09 – - 26.5 - 5.4 + 29.16 - 27 + 29.16 – - 27 = + 51.5 = + 53.04 = + 54.59 = + 56.16 next

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… Note on # 4d next © 2007 Herbert I. Gross tt2t2 5t5tt 2 – 5t - 5.0 + 25 - 25 + 25 – - 25 - 5.1 + 26.01 - 25.5 + 26.01 – - 25.5 = + 50 - 5.2 + 27.04 - 26 + 27.04 – - 26 - 5.3 + 28.09 - 26.5 + 28.09 – - 26.5 ??? = + 51.5 = + 53.04 = + 54.59 - 5.4 + 29.16 - 27 + 29.16 – - 27 = + 56.16 + 55

If we wanted an even better approximation, we could evaluate the expression t 2 – 5t, say when t = - 5.31, - 5.32, - 5.33, - 5.34, etc. For example, if we let t = - 5.33, we obtain… next © 2007 Herbert I. Gross Note on #4d tt2t2 5t5tt 2 – 5t - 5.33 + 28.4089 - 26.65 + 28.4089 – - 26.65 = + 55.0589 next which indicates that the required value of t is very nearly equal to - 5.33.

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 t 2 – 5t = 55. © 2007 Herbert I. Gross 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. next

© 2007 Herbert I. Gross Careful Be careful not to assume that there is only one value of t for which t 2 – 5t. For example, when t = 10, t 2 – 5t = 50. And when t = 11, t 2 – 5t = 66. Hence there is another value of t that is between 10 and 11 for which t 2 – 5t = 55. How can this be, and are there still other values of t for which t 2 – 5t = 55? Questions such as these will be answered as our course continues.

next The formula that relates the Fahrenheit temperature (F) to the Celsius temperature (C) is… C = 5/9 (F – 32). Problem #5a © 2007 Herbert I. Gross What is the Celsius temperature when the Fahrenheit temperature is 14°? 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… next © 2007 Herbert I. Gross = 5 / 9 ( + 14 – + 32) C = 5 / 9 ( F – + 32) = 5 / 9 ( - 18) = 5 / 9 ( + 14 + - 32) = - 10 next

To review the logic behind each step, we start with… next © 2007 Herbert I. Gross next We first replaced F by + 14… We then used the “add the opposite rule” to obtain… C = 5 / 9 And finally, we used our rules of arithmetic to obtain… ( - 18) C = 5 / 9 ( + 14 + - 32) C = 5 / 9 ( + 14 – + 32) C = 5 / 9 ( F – + 32)

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… next © 2007 Herbert I. Gross next C = - 10 C = 5 / 9 ( - 18)

next The formula that relates the Fahrenheit temperature (F) to the Celsius temperature (C) is… C = 5/9 (F – 32). Problem #5b © 2007 Herbert I. Gross What is the Fahrenheit temperature when the Celsius temperature is - 20°? next Answer: - 4°F

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… next © 2007 Herbert I. Gross - 20 = 5 / 9 ( F – + 32) but now we replace C by - 20 to obtain the equation. C = 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… next © 2007 Herbert I. Gross 9 / 5 × - 20 = 9 / 5 × 5 / 9 ( F – + 32) - 20 = 5 / 9 ( F – + 32) next

Solution: Since 9 / 5 × - 20 = - 36 and 9 / 5 × 5 / 9 = 1, we may rewrite the equation as... next © 2007 Herbert I. Gross next - 36 = F – + 32 - 36 = 1 (F – + 32) 9 / 5 × - 20 = 9 / 5 × 5 / 9 ( F – + 32) next ()

Solution: By the “add the opposite” rule, we may rewrite… next © 2007 Herbert I. Gross next and we now add + 32 to both sides of the equation to obtain… - 4 = F + 0 - 36 + + 32 = F + - 32 + + 32 - 36 = F + - 32 Since - 36 + + 32 = - 4 and - 32 + + 32 = 0, we may rewrite the equation as… next - 36 = F – + 32 ()

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. next © 2007 Herbert I. Gross Note on # 5

In terms of our “program”… next © 2007 Herbert I. Gross “Recipe” next Converting Fahrenheit to Celsius Start with Fahrenheit Temp.F14 Subtract 32F – 32 14 – 32 = - 18 Multiply by 5 / 9 5 / 9 (F – 32) 5 / 9 ( - 18) Answer is Celsius Temp. C = 5 / 9 (F – 32) - 10 problem 5a becomes…

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… next © 2007 Herbert I. Gross Calculator next 14 –32÷9×5=

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. next © 2007 Herbert I. Gross Recipe

In terms of a “plain English” chart… next © 2007 Herbert I. Gross Doing & Undoing Doing RecipeUndoing Recipe Start with FAnswer is F Subtract 32Add 32 Multiply by 5 / 9 Divide by 5 / 9 i.e. Multiply by 9 / 5 Answer is CStart with C next

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 © 2007 Herbert I. Gross Note on # 5 C = 5 / 9 ( F – + 32)

To “unblock” the parentheses (that is, to “undo” the last step in the formula) we multiply both sides by 9 / 5 to obtain… next © 2007 Herbert I. Gross Note on # 5 9 / 5 C = 1(F – + 32) 9 / 5 C = ( 9 / 5 5 / 9 )(F – + 32) 9 / 5 C = 9 / 5 [ 5 / 9 (F – + 32)] C = 5 / 9 (F – + 32) next

In other words, the formula now becomes… © 2007 Herbert I. Gross Note on #5 F = 9 / 5 C + 32 9 / 5 C + 32 = F and if we now add 32 to both sides of the above formula we see that… next 9 / 5 C = F – 32 or in more traditional format…

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 / 5 C + 32 if we were given the value of C, and we wanted to find the corresponding value of F. next © 2007 Herbert I. Gross 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. next © 2007 Herbert I. Gross 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. next © 2007 Herbert I. Gross Note on # 5

Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 5 By Herbert I. Gross and Richard A. Medeiros next.

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Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 5 By Herbert I. Gross and Richard A. Medeiros next.

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