Algebra Problems… Solutions

Algebra Problems… Solutions
next Algebra Problems… Solutions Set 15 By Herbert I. Gross and Richard A. Medeiros © Herbert I. Gross

For what value of y is it true that
next next Problem #1 For what value of y is it true that 7(y + 4) – 3(y – 5) = 6y + 35 ? Answer: y = 4 © Herbert I. Gross 2

game” to transform the left-hand side of the equation…
Answer: y = 4 Solution: The right-hand side of the equation already has the mx + b (actually, the my + b) form. So we begin by using our “rules of the game” to transform the left-hand side of the equation… next next next 7(y + 4) – 3(y – 5) = 6y + 35 and rewrite it by replacing each subtraction by the “add the opposite” rule. That is… 7(y + 4) – 3(y – 5) + - + - © Herbert I. Gross

next next next next next next Solution: By using the distributive property, we may rewrite 7(y + 4) + -3(y + -5) as… 7(y) + 7(4) + -3(y) + -3(-5) or… 7y y + 15 or… 7y + - 3y or… (7y + - 3y) + ( ) or… 4y + 43 © Herbert I. Gross

We may now replace the left-hand side of
Solution: We may now replace the left-hand side of 7(y + 4) – 3(y – 5) = 6y + 35 by its value 4y + 43 to obtain… next next next next next 4y + 43 = 6y + 35 Next, we might subtract 4y from both sides in the above equation to obtain… 43 = 2y + 35 We may then subtract 35 from both sides above to obtain… 8 = 2y, from which we see that… y = 4 © Herbert I. Gross

next next Notes on #1 Technically speaking we have not as yet proved that y = 4 is a solution of the equation 7(y + 4) – 3(y – 5) = 6y + 35. Rather what we have shown is that if our equation does have a solution, it must be y = 4. As a check that y = 4 is a solution; we should get a true statement, if we replace y by 4 in our equation © Herbert I. Gross 6

Replacing y by 4 in the equation below,
next next next next next Notes on #1 Replacing y by 4 in the equation below, we see that… 4 4 (4) 7( y + 4 ) + -3( y + -5 ) = 6 y or… 7(8) + -3(-1) = or… = 59 and… 59 = 59 © Herbert I. Gross

Notice that in working with the equation
next Notes on #1 Notice that in working with the equation 7(y + 4) – 3(y – 5) = 6y + 35, we began by rewriting the left-hand side into the form my + b. We did this because we already knew how to solve equations that are in the form my + b = ny + c. The point is that in situations that involve logical thought, we often try to reduce unsolved problems to a series of one or more previously solved problems. © Herbert I. Gross 8

next next Notes on #1 Problem # 1is an good example of how we develop new techniques in mathematics, one step at a time. In contrast, “cramming” rarely works in learning to do mathematics. Namely, since each step follows logically from a previous step, it is important to be aware of the significance of each step as it occurs. © Herbert I. Gross 9

already know how to solve.
next Notes on #1 In summary… A major objective in the “game” of algebra is to use the “rules of the game” to paraphrase equations we haven’t yet learned to solve into equations that we already know how to solve. © Herbert I. Gross 10

For what value of p is it true that
next next Problem #2 For what value of p is it true that 2 [3p – 2(4 – p)] = 7p + 32 ? Answer: p = 16 © Herbert I. Gross 11

Answer: p = 16 Solution Again the basic strategy will be to paraphrase the left side of the equation into the mp + b form. As in the previous problem, one way to begin is by replacing each subtraction by the “add the opposite” rule. That is, we rewrite the left side of the equation… 2 [3p – 2(4 – p)] = 7p + 32 as… next next 2 [3p + -2(4 + -p)] = 7p + 32 © Herbert I. Gross

2 [3p + -2(4) + -2(-p)] or… 2 [3p + -8 + 2p]
Solution: Recalling that we simplify an algebraic expression by starting with the innermost set of grouping symbols (in this case, the parentheses), we next use the distributive property to rewrite 2 [3p + -2(4 + -p)] as… next next next 2 [3p + -2(4) + -2(-p)] or… 2 [3p p] © Herbert I. Gross

…which by the distributive property can be rewritten as…
Solution: Then by using the commutative and associative properties of addition, the expression 2 [3p p] within the brackets can be rewritten as… next next next 2 [5p + -8] …which by the distributive property can be rewritten as… 10p (or 10p – 16) © Herbert I. Gross

by its value 10p + -16 to obtain the equivalent equation…
Solution: We may now replace the left side of the equation 2 [3p + -2(4 + -p)] = 7p + 32 by its value 10p to obtain the equivalent equation… next next next 10p + -16 2 [3p + -2(4 + -p)] = 7p + 32 Subtracting 7p from both sides of the above equation, and then adding 16 to both sides of the new equation, we see that… 3p = 48 or p =16 © Herbert I. Gross

Subtract this result from 3 times p.
next next next next next next Notes on #2 In terms of a “program”, 2 [3p – 2(4 – p)] means… Program #1 2 3p 2 4 p Start with p. Subtract it from 4. Multiply the result by 2. Subtract this result from 3 times p. Multiply the result by 2. © Herbert I. Gross 16

10p – 16 told us was that this was equivalent to the simpler program…
next next next next Notes on #2 What the expression 10p – 16 told us was that this was equivalent to the simpler program… Program #2 10 p 16 Start with p. Multiply it by 10. Subtract 16. © Herbert I. Gross 17

Notes on #2 In terms of a chart… p 4 – p 3p 2(4 – p) 3p – 2(4– p)
next next Notes on #2 In terms of a chart… p 4 – p 3p 2(4 – p) 3p – 2(4– p) 2[3p –2(4 –p)] 10p 10p – 16 1 3 6 -3 -6 10 2 6 4 20 3 1 9 2 7 14 30 4 12 24 40 5 -1 15 -2 17 34 50 6 -2 18 -4 22 44 60 © Herbert I. Gross 18

next next next next Notes on #2 In terms of a “program”, the right-hand side of the equation 2 [3p + -2(4 + -p)] = 7p + 32 becomes… 7 p 32 Program #3 Start with p. Multiply it by 7. Add 32. © Herbert I. Gross 19

next Notes on #2 And what we showed in solving this problem is that the only time Program #1 (or, equivalently, Program #2) and Program #3 yield the same output for a given input is when p = 16 © Herbert I. Gross 20

Notes on #2 As a check, we see that… p 10p 10p – 16 7p 7p + 32 1 10 -6
next Notes on #2 As a check, we see that… p 10p 10p – 16 7p 7p + 32 1 10 -6 7 39 2 20 4 14 46 … 16 160 144 112 17 170 154 119 151 18 180 164 126 158 etc. © Herbert I. Gross 21

Notes on #2 In essence, p = 16 is the equilibrium point.
next next next next Notes on #2 In essence, p = 16 is the equilibrium point. p 10p 10p – 16 7p 7p + 32 15 150 134 105 137 16 160 144 112 17 170 154 119 151 15 134 137 16 144 144 17 154 151 As the chart indicates; if p is less than 16, 10p – 16 is less than 7p + 32. However, when p is greater than 16, 10p – 16 is greater than 7p + 32. © Herbert I. Gross 22

For what value of q is it true that
next next Problem #3 For what value of q is it true that 1/7 (q + 3) + 6 = 1/2(q + 5) ? Answer: q = 16 © Herbert I. Gross 23

Since 14 is a common multiple of 7 and 2, we can paraphrase…
Answer: q = 11 Solution Since 14 is a common multiple of 7 and 2, we can paraphrase… 1/7(q + 3) + 6 = 1/2(q + 5) into an equation involving only whole numbers by multiplying both sides of the equation by 14. Doing this we obtain… next next 14 [1/7 (q + 3) + 6] = 14[1/2(q + 5)] © Herbert I. Gross

Since the left side of the equation
Solution: Since the left side of the equation 14 [1/7(q + 3) + 6] = 14[1/2(q + 5)] has the form a(b + c), where a = 14, b = 1/7(q + 3), and c = 6; we may use the distributive property to rewrite our equation as… next next 14 [1/7(q + 3)] + 14(6) = 14[1/2(q + 5)] © Herbert I. Gross

14 [1/7(q + 3)] + 14(6) = 14[1/2(q + 5)] as…
Solution: Since 14(1/7) = 2, 14 × 6 = 84 and 14 (1/2) = 7, it is not difficult to see from the properties of multiplication that we may rewrite the equation… 14 [1/7(q + 3)] + 14(6) = 14[1/2(q + 5)] as… next next next next next next next 2(q + 3) + 84 = 7(q + 5) or… 2q = 7q + 35 or… 2q + 90 = 7q + 35 or… 90 = 5q + 35 or… 55 = 5q or… q = 11 © Herbert I. Gross

next next Notes on #3 Some students find it easier to first use the distributive property and then multiply both sides by the least common denominator. That is, we could have solved the equation 1/7(q + 3) + 6 = 1/2(q + 5) by first using the distributive property to rewrite it as… q/7 + 3/7 + 6 = q/2 + 5 © Herbert I. Gross 27

next next Notes on #3 q/7 + 3/7 + 6 = q/2 + 5 We could then have multiplied each term on both sides of the above equation by 14 to obtain… 2q = 7q + 35 …whereupon we could have then solved the equation as we did in our previous solution. © Herbert I. Gross 28

next next next next Notes on #3 Remember that you can (and should) check your answer by replacing it in the original equation and seeing whether it becomes a true statement. Replacing q by 11 in equation 1/7(q + 3) + 6 = 1/2(q + 5) gives us… 1/7( q ) + 6 = 1/2( q ) 11 11 1/7( 14 ) + 6 = 1/2( 16 ) = 8 © Herbert I. Gross 29

The main reason for choosing the least
next next Notes on #3 In simplifying q/7 + 3/7 + 6 = q/2 + 5, it was not necessary to multiply by the least common multiple of 7 and 2. Any other common multiple would have worked as well. The main reason for choosing the least common multiple is to keep the arithmetic as simple as possible. However, if it is not obvious what the least common multiple is, it is often easier (especially with the help of a calculator) to use any common denominator and then solve the resulting equation. © Herbert I. Gross 30

Answer: 300 (miles) Problem #4
next next next Problem #4 A car leaves from point O moving at a constant speed of 50 miles per hour. One hour later, a second car, traveling in the same direction as the first car, leaves point O and travels at a constant speed of 60miles per hour. How many miles must the second car travel before it catches up to the first car? Answer: 300 (miles) © Herbert I. Gross 31

Answer: 300 miles Solution:
We may let t stand for the number of hours the second car has to travel to overtake the first car, and we may let d denote the number of miles the second car is from the point O. Since the speed of the second car is 60 miles per hour, the relationship for the second car between d (distance) and t (time) for the second car is given by… next next d = 60t © Herbert I. Gross

second car traveled, t + 1 denotes the time the first car traveled.
Solution: Since the first car started an hour earlier, it has traveled one hour longer than the second car, and since t denotes the time the second car traveled, t + 1 denotes the time the first car traveled. next next Therefore, the relationship between d and t for the first car is given by… d = 50(t + 1) © Herbert I. Gross

Since the two cars would have traveled the
Solution: Since the two cars would have traveled the same distance d from O when they meet; we may equate the value of d in the equation d = 60t to the value of d in the equation d = 50(t + 1) to obtain… next next 60t = 50(t + 1) …which, by the distributive property may be rewritten as… 60t = 50t + 50 © Herbert I. Gross

We then subtract 50t from both sides of the equation below to give us…
Solution: We then subtract 50t from both sides of the equation below to give us… next next next 60t = 50t + 50 50t 50t 10t = 50 t = 5 And replacing t by 5 in either equation d = 60t or equation d = 50(t + 1) we see that d = 300. © Herbert I. Gross

next next Notes on #4 As a check, notice that at the instant the second car catches up to the first car, both cars have traveled the same number of miles. Since the first car started 1 hour before the second car; if the second car traveled for 5 hours, the first car must have traveled for 6 hours. At 50 miles per hour for 6 hours, we see that the second car also travels 300 miles. © Herbert I. Gross 36

50 miles per hour and that the second car had an average speed of
next next Notes on #4 From a “real world” point of view, the “weakness” of this exercise is that it is very unlikely that a car would travel at a constant speed for 5 or 6 hours. A more realistic version would be to say that the first car had an average speed of 50 miles per hour and that the second car had an average speed of 60 miles per hour. © Herbert I. Gross 37

next next next Notes on #4 Note that, to find the average speed we would have to know how far the car traveled and how long that took. That is, we define the average speed to be the quotient we get when we divide the distance by the time. If the speed happens to be constant, then the average speed is the same as the constant speed. © Herbert I. Gross 38

next next Notes on #4 A common strategy for solving algebra problems is to let x (or any other letter) stand for the quantity we are asked to find. Hence, as a first step, we read the problem and see what it asks for. Since this problem asks for how far the second car traveled in order to catch up to the first car; we might let d stand for the distance from the point O to the point P at which the second car catches up to the first car. © Herbert I. Gross 39

Notes on #4 From the relationship… distance = speed × time
next next next Notes on #4 From the relationship… distance = speed × time …we see that it is also true that… time = distance speed Therefore, the time it took the first car to get to the point P is given by d/50 and the time it took the second car to get to the point P is d/60. © Herbert I. Gross 40

next next Notes on #4 Since the first car traveled for one hour more than the second car, the equation that relates their times is given by… The number of hours that the first car traveled The number of hours that the second car traveled + 1 = © Herbert I. Gross 41

next next Notes on #4 Since we know that the number of hours the first car travels is d/50 and the number of hours the second car travels is d/60, the equation… The number of hours that the first car traveled = The number of hours that the second car traveled + 1 becomes… d/50 = d/60 + 1 © Herbert I. Gross 42

d/50 = d/60 + 1, we had to add 1 to the
next next next Notes on #4 Aside: Since 60 is more than 50, d/60 is less than d/50, and that’s why in the equation d/50 = d/60 + 1, we had to add 1 to the right-hand side. Since 300 is the least common multiple of 50 and 60, we may multiply both sides of our equation by 300 to obtain… 300(d/50) = 300(d/60 + 1) 6d = 5d + 300 d = 300 © Herbert I. Gross 43

one gets to be either inadequate or too cumbersome.
next next Notes on #4 In general, we seek the simplest solution to any problem we face, and we look for a more complicated solution only when the simpler one gets to be either inadequate or too cumbersome. In this sense, we prefer to use algebra for solving a word problem only in the event that we cannot see an easier way to solve the problem. © Herbert I. Gross 44

next next Notes on #4 Very often so-called “word problems” in algebra can be done by using arithmetic. In fact, we usually elect to use algebra only when it is not easy for us to see what the arithmetic method is. For example, one non-algebraic way to do this problem is to notice that the first car has a 50 miles head start on the second car. © Herbert I. Gross 45

The second car gains 10 miles on the first car each hour it travels.
next next next Notes on #4 That is, the first car has a 1 hour head start, and during that hour, it travels 50 miles. The second car gains 10 miles on the first car each hour it travels. In other words, each hour after the first hour, the first car travels 50 miles and the second car travels 60 miles. © Herbert I. Gross 46

next next Notes on #4 Since the second car gains 10 miles on the first car for each hour that the second car travels, it will take the second car 5 hours to make up the 50 mile head start the first car has. And at a constant speed of 60 miles per hour, the second car travels 300 miles in those 5 hours. © Herbert I. Gross 47

Notes on #4 As we’ve mentioned before it is sometimes helpful to make
next next Notes on #4 As we’ve mentioned before it is sometimes helpful to make a chart and look for patterns. For example, in the present situation, suppose we denote by “Car 1” the car that started first, and we denote the second car by “Car 2”. © Herbert I. Gross 48

The chart might then look like…
next next next Notes on #4 Distance Traveled Car 1 Car 2 After 1 hour 50 miles 0 miles The chart might then look like… After 2 hour 100 miles 60 miles After 3 hour 150 miles 120 miles After 4 hour 200 miles 180 miles After 5 hour 250 miles 240 miles After 6 hour 300 miles After 6 hour 300 miles After 7 hour 350 miles 360 miles © Herbert I. Gross 49

Notice the need to keep track of what our variable represents.
next next next Notes on #4 Notice the need to keep track of what our variable represents. Distance Traveled Car 1 Car 2 After 5 hour 250 miles 240 miles After 6 hour 300 miles Namely in solving the problem, we found that the two cars were at the same place when t = 5. Yet the chart seems to indicate that it happens when t = 6? How can t equal both 5 and 6? © Herbert I. Gross 50

(t + 1) the second car traveled.
next next Notes on #4 The answer is that there is no contradiction, because in our case, the time was measured relative to Car 2, while in the chart the time is measured relative to Car 1. In more mathematically precise terminology, we might have let t1 represent the time (t) the first car traveled, and t2 represent the time (t + 1) the second car traveled. © Herbert I. Gross 51

next Notes on #4 In this particular example, it didn’t take long for the chart to help us arrive at the correct answer. Notice, however, that even if the problem had been more complex the chart showed us rather quickly that every hour the distance between the two cars decreased by 10 miles. © Herbert I. Gross 52

Answer: -40°C (or equivalently -40°F)
next next next Problem #5 The relationship between the Fahrenheit (F) temperature scale and the Celsius (C) temperature scale is given by the linear relationship… C = 5/9(F – 32) At what temperature will both scales yields the same reading? Answer: -40°C (or equivalently -40°F) © Herbert I. Gross 53

When the scales have the same reading,
Answer: -40°C (or equivalently -40°F) Solution When the scales have the same reading, it means that C = F. Hence, we may replace C by F in the formula C = 5/9(F – 32) to obtain… next next next F = 5/9(F – 32) To eliminate the fraction in the equation above, we may multiply both sides of the equation by 9 to obtain… 9F = 5(F – 32) © Herbert I. Gross

We may then use the distributive property to obtain…
Solution We may then use the distributive property to obtain… next next next next 9F = 5(F – 32) 9F = 5F – 160 …and if we next subtract 5F from both sides of the above equation we obtain… 4F = -160 F = -40 © Herbert I. Gross

next next Notes on #5 In F = 5/9(F – 32) we replaced C by F in formula C = 5/9(F – 32). It would have been just as logical to replace F by C in the formula C = 5/9(F – 32). In this case, the equation would have been… C = 5/9 (C – 32) © Herbert I. Gross 56

Notes on #5 Notice that equations F = 5/9(F – 32) and
next next Notes on #5 Notice that equations F = 5/9(F – 32) and C = 5/9(C – 32) are exactly the same, except for the name of the variable. However, since either equation tells us the temperature at which the two scales would have the same reading… (that is, when C = F); it doesn’t matter which equation we use. © Herbert I. Gross 57

Notes on #5 What we’ve shown in this problem is that
next next Notes on #5 What we’ve shown in this problem is that -40° means exactly the same thing on either scale (-40°C or -40°F). In general, it does make a difference whether the temperature reading is in Celsius degrees or Fahrenheit degrees. © Herbert I. Gross 58

Algebra Problems… Solutions

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