The Basic Ideas of Algebra - COPYRIGHTED MATERIAL

Chapter One

The Basic Ideas ofAlgebra

Algebra is about reasoning with numbers andcalculations, using symbols, expressions, andequations.

Expressions provides a way of representingcalculations and communicating them toothers. By putting algebraic expressions indifferent but equivalent forms, we canunderstand different aspects of the calculationsthey represent.

Equations describe conditions on and relationsbetween quantities—such as time, distance,cost, or profit—that arise in science, business,or everyday life. By transforming equations tofind solutions, we can answer complexquestions about these quantities.

In this chapter we construct, read, and analyzeexpressions and equations, and we learn thebasic underlying principles for transformingthem.

COPYRIG

HTED M

ATERIAL

2 Chapter One THE BASIC IDEAS OF ALGEBRA

1.1 NUMBERS AND EXPRESSIONS

In our everyday life we develop shortcuts and rules of thumb for calculations that come up repeat-edly.

Example 1 (a) Describe a method for calculating a 20% tip on a restaurant bill, and use it to calculate the tipon a bill of $8.95 and a bill of $23.70.

(b) Choosing the letter B to stand for the bill amount, represent your method in symbols.

Solution (a) Taking 20% of a number is the same as multiplying it by 0.2, so

Tip on $8.95 = 0.2× 8.95 = 1.79 dollarsTip on $23.70 = 0.2× 23.70 = 4.74 dollars.

(b) The tip on a bill ofB dollars is 0.2×B dollars. Usually in algebra we leave out the multiplicationsign or represent it with a dot, so we write 0.2B or 0.2 ·B.

Algebraic ExpressionsAn algebraic expression is a way of representing a calculation with numbers, using letters to standfor some of the quantities in the calculation. For example, the expression 0.2B is formed by a singleoperation, multiplication, between the number 0.2 and the quantity B.

Example 2 (a) Pick two numbers, find their sum and product, and then the average of their sum and product.(b) Write an algebraic expression that represents this calculation.

Solution (a) Take 3 and 5, for example. Their sum is 8 and their product is 15, so the average of the sum andproduct is

8 + 15

2=

23

2.

(b) Let x and y stand for any two numbers. The sum is x+ y and the product is xy, so

Average of sum and product =(x+ y) + xy

2.

Variables and ConstantsWe call the symbols in an algebraic expression variables if we want to consider different possiblevalues for them, and we call the symbols constants if they are fixed quantities or if we want toregard them as fixed quantities. For example, we call the letter B in Example 1 a variable becausewe apply the expression to meals with different prices, even though the method of calculating thetip is always the same. On the other hand, the 0.2 is a constant, representing the tip as a percentageof the bill.

Example 3 A circle of radius r has area πr2. Which symbols in the expression πr2 are variables and which areconstants?

Solution Since the same formula applies to different circles of different radii, it makes sense to think of theletter r as a variable. The numbers π and 2 are constants.

We can use letters as well as numerals to stand for constants.

1.1 NUMBERS AND EXPRESSIONS 3

Example 4 A tip of r percent on a bill of B dollars is given in dollars by the expressionr

100B.

Which letters in this expression would you call variables and which would you call constants if youwere considering(a) Tips by the same person at different meals(b) Tips by different people on the same meal(c) Meal prices and tipping habits in different cities?

Solution (a) Since the same person probably gives the same percentage tip, but the meal price varies, wewould consider r as a constant and B as a variable.

(b) Since the meal price is the same but the people might have different tipping rates, we wouldconsider r as a variable and B as a constant.

(c) Since both meal prices and tipping rates vary from city to city, we would consider both r and Bas variables.

For now we usually consider letters as representing variables, unless it is stated otherwise.Variables are also sometimes called unknowns.

Evaluating ExpressionsIf we give particular values to the variables in an expression, then we can calculate the correspondingvalue of the expression itself. We call this process evaluating the expression.

Example 5 Evaluate the expression 0.2B when B = 16.50. What does your answer represent in the context ofExample 1?

Solution When B = 16.50, we have 0.2B = (0.2)(16.50) = 3.30. This value represents a 20% tip on a billof $16.50.

Example 6 Evaluate the expression using the given value of each variable.(a) 3x− 4y, where x = 2, y = −5(b) 4x2 + 9x+ 7y, where x = −2, y = 3.

Solution (a) If x = 2 and y = −5, then

3x− 4y = 3 · 2− 4 · (−5) = 6 + 4 · 5 = 26.

(b) If x = −2 and y = 3, then

4x2 + 9x+ 7y = 4 · (−2)2 + 9 · (−2) + 7 · 3 = 4 · 4− 9 · 2 + 7 · 3 = 19.

Example 7 Juan wants to buy chips and soda for a party. He estimates that he will need 5 bags of chips and 10bottles of soda. His total cost, T , for these items is

T = 5c+ 10s,

where c is the price of a bag of chips and s is the price of a bottle of soda. If chips cost $2.99 perbag and soda costs $1.29 per bottle, find the total cost.

Solution We have c = 2.99 and s = 1.29, so

T = 5 · c+ 10 · s = 5 · 2.99 + 10 · 1.29 = 27.85 dollars.


Reading ExpressionsIt is important to be able to read an expression and determine the calculation it represents. One wayis to start with the variables and see how the expression is built up.

Example 8 The expressions in each pair represent different calculations. Explain how they are different.(a) 2x2 and (2x)2

(b) 2l + w and 2(l + w)(c) 3− x+ y and 3− (x+ y)

Solution To interpret an expression, we pay attention to the order of operations and to the placement ofparentheses.(a) In the expression 2x2, we square the number x and multiply the result by 2. In (2x)2, we

multiply the number x by 2 then square the product.(b) In 2l + w, we begin with two numbers, l and w. We double l and add w to it. In 2(l + w), we

add l to w, then double the sum.(c) In 3− x+ y, we subtract x from 3, then add y. In 3− (x+ y) we add x and y, then subtract the

sum from 3.

Longer expressions can be built up from simpler ones. To read a complicated expression, wecan sometimes break the expression down as a sum or product of parts and analyze each part. Theparts of a sum are called terms, and the parts of a product are called factors.

Example 9 Describe how each expression breaks down into component parts.

(a)1

2h(a+ b) (b) 3(x− y) + 4(x+ y)

Solution (a) This expression is the product of three factors, 1/2, h, and a+ b. The last factor is a sum of twoterms, a and b.

(b) This expression is the sum of two terms, 3(x − y) and 4(x + y). The first term is a product oftwo factors, 3 and x−y, and the second is a product of 4 and x+y. The factors x+y and x−yare sums of two terms.

Example 10 The surface area of a cylinder of radius r and height h is given by the formula

Surface area = 2πr2 + 2πrh.

(a) Describe in words how the surface area is computed.(b) Find the surface area of the cylinder of radius 3 inches and height 4 inches.

Solution (a) Take the square of the radius and multiply the result by 2π, then take the product of the radiusand the height and multiply the result by 2π, then add the results of these two calculations.

(b) If r = 3 and h = 4 then

Surface area = 2π(3)2 + 2π(3)(4) = 18π + 24π = 42π in2.

Writing ExpressionsIt is important to be able to write an algebraic expression from a verbal description. One way to dothis is to identify the variables and the mathematical operations associated with them.

1.1 NUMBERS AND EXPRESSIONS 5

Example 11 You buy a bond for p dollars, an amount called the face value. The interest each year is 5% of theprincipal, and after t years the total interest is the product of the number of years and the interest.The balance is the sum of the face value and the interest.(a) Write an expression for the total interest after t years.(b) Write an expression for the balance after t years.

Solution (a) The variables are the number of years, t, and the face value, p. Because 5% =1

20, the interest

is given by

(Number of years) · 1

20· Principal = t

1

20p =

tp

20.

(b) The balance is the sum of the face value and the interest, and is given by

p+tp

20.

Example 12 A student’s grade in a course depends on a homework grade, h, three test grades, t1, t2, and t3, anda final exam grade f . The course grade is the sum of the homework grade, the average of the threetest grades, and twice the final exam grade. Write an expression for the course grade.

Solution The primary mathematical operation is the addition of three terms. The first term is the homeworkgrade. The second term is the average of the three test grades, which is their sum divided by three.The third term is twice the final grade. The course grade is given by

Course grade = h+t1 + t2 + t3

3+ 2f.

Exercises and Problems for Section 1.1Exercises

1. The surface area of a right circular cylinder with radius rand height h is 2πr2 + 2πrh. Identify the constants andthe variables in this expression.

In Exercises 2–5, describe the sequence of operations that pro-duces the expression.

2. 2(u+ 1) 3. 2u+ 1

4. 1− 3(B/2 + 4) 5. 3− 2(s+ 5)

In Exercises 6–9, write an expression for the sequence of op-erations.

6. Subtract x from 1, double, add 3.

7. Subtract 1 from x, double, add 3.

8. Add 3 to x, subtract the result from 1, double.

9. Add 3 to x, double, subtract 1 from the result.

In Exercises 10–16, write an expression for the quantity de-scribed.

10. Twelve x’s

11. Six more than a number, n

12. Six less than a number, n

13. Seven less than twice the radius, r

14. Four years later than the present year, t

15. Four years earlier than the present year, t

16. Two more than five q’s

In Exercises 17–23, evaluate the expression using the valuesgiven.

17. 3x2 − 2y3, x = 3, y = −1

18. −16t2 + 64t+ 128, t = 3

19. (0.5z + 0.1w)/t, z = 10, w = −100, t = −10

20. (1/4)(x+ 3)2 − 1, x = −7

21. (a+ b)2, a = −5, b = 3

22. (1/2)h(B + b), h = 10, B = 6, b = 8

23. ((b− x)2 + 3y)/2− 6/(x− 1), b = −1, x = 2, y = 1


24. A caterer for a party buys 75 cans of soda and 15 bags ofchips. Write an expression for the total cost if soda costss dollars per can and chips cost c dollars per bag.

25. Apples are 99 cents a pound, and pears are $1.25 apound. Write an expression that represents the cost, C,

in dollars, of a pounds of apples and p pounds of pears.

26. A person’s body mass index is their weight in pounds, w,multiplied by 704.5, and divided by the square of theirheight in inches, h. Write an expression for the bodymass index.

Problems

In Problems 27–33, both a and x are positive. What is the ef-fect of increasing a on the value of the expression? Does thevalue increase, decrease, or remain unchanged?

27. ax+ 1 28. x+ a 29. x− a

30. x

a+ 1 31. x+

1

a32. ax− 1

a

33. a+x−(2+a)

34. The children in a family contribute equally for a gift totheir mother. Each child contributes

c

5, where c is the to-

tal cost of the gift.

(a) How many children are there?(b) If the gift cost $200, how much would each child

contribute?(c) Find the value of c if each child contributes $50.(d) Write an expression for the amount contributed by

each child if there are 3 children.

35. The number of people who attend a concert is 160 − pwhen the price of a ticket is $p.

(a) What is the practical interpretation of the 160?(b) Why is it reasonable that the p term has a negative

sign?(c) The number of people who attend a movie at ticket

price $p is 175−p. If tickets are the same price, doesthe concert or the movie draw the larger audience?

(d) The number of people who attend a dance perfor-mance at ticket price $p is 160 − 2p. If tickets arethe same price, does the concert or the dance perfor-mance draw the larger audience?

The surface area, S, of a cylinder with radius r feet and heighth feet is given by S = 2πr2 + 2πrh square feet. In Prob-lems 36–39, find the surface area of the cylinder with the givenradius and height.

36. Radius 5 feet and height 10 feet.

37. Radius 10 feet and height 5 feet.

38. Radius 6 feet and height half the radius.

39. Radius b feet and height half the radius.

40. A car travels from Tucson to San Francisco, a distance of870 miles. It has rest stops totaling 5 hours. While driv-ing, it maintains a speed of v mph. Give an expressionfor the time it takes. What is the difference in time takenbetween a car that travels 65 mph and a car that travels75 mph?

41. A mathematics professor uses the following scheme todetermine the final grade: he takes the sum of the foursemester grades g1, g2, g3, and g4, then adds twice thefinal grade, f , then divides the total by 6. Write an ex-pression for calculating final grade.

42. The number of registered voters in Town A is x and thenumber in Town B is y. If 42% of the voters in Town Aand 53% of those in Town B are registered Republicans,write an expression for the total number of Republicanvoters in both towns.

43. Write an expression that is the sum of two terms, the firstbeing the quotient of x and z, the second being the prod-uct of L+ 1 with the sum of y and 2k.

44. A triangle with base b and height h has area (1/2)bh.Write a brief explanation of where the 1/2 in this formulacomes from.

45. The perimeter of a rectangle of length l and width w isgiven by the formula P = 2l + 2w. Write a brief expla-nation of where the constants 2 in this expression comefrom.

46. The formula for the volume of a cylinder is the product ofthe area of a circle and another factor. What is the otherfactor?

1.2 TRANSFORMING EXPRESSIONS

A number can be expressed in many different ways. For example, 1/4 and 0.25 are two differentways of expressing the same number. Similarly, we can have two or more different ways of express-ing the same calculation.

1.2 TRANSFORMING EXPRESSIONS 7

Example 1 Pares says she has an easy way to figure out a 20% tip on a bill: she moves the decimal point in thebill one place to the left then doubles the answer.(a) Check that Pares’ method gives the same answer on bill amounts of $8.95 and $23.70 as the

method in Example 1 on page 2.(b) Does Pares’ method always work? Explain your answer using algebraic expressions.

Solution (a) For $8.95, first we move the decimal point to the left to get 0.895, then double to get a tip of$1.79. For $23.70, we move the decimal point to the left to get 2.370, then double to get a tip ofget $4.74. Both answers are the same as in Example 1.

(b) Moving the decimal point to the left is the same as multiplying by 0.1. So first Pares multipliesthe bill by 0.1, then multiplies the result by 2. Her calculation of the tip is

2(0.1B).

We can simplify this expression by regrouping the multiplications:

2(0.1B) = (2 · 0.1)B = 0.2B.

This last expression for the tip is the same one we found in Example 1 on page 2.

For each value of B the two expressions, 2(0.1B) and 0.2B, give equal values for the tip. Wesay that the two expressions are equivalent.

Equivalent ExpressionsTwo expressions are equivalent if, for every value of the variables in them, they have equalvalues.

Because the variables in an expression represent numbers, we can try a few values to help usdecide whether two expressions are equivalent. If the resulting calculations give the same answer,we have some evidence that the expressions might be equivalent.

Example 2 Is the expression a/2 equivalent to the expression (1/2)a?

Solution Choose a value for a, say a = 10, and evaluate both expressions:a

2=

10

2= 5

(1

2

)a =

(1

2

)10 = 5.

Table 1.1 shows that a/2 and (1/2)a have the same value for four other values of a. In fact a/2 and(1/2)a have the same value for any value of a, since dividing by 2 is the same as multiplying by1/2. Because the two expressions produce equal results for all values of a, they are equivalent.

Table 1.1 Table comparing a/2 and(1/2)a for different values of a

a 6 9 14 36

a/2 3 4.5 7 18

(1/2)a 3 4.5 7 18


Checking that two expressions are equal at one or two values of the variable does not guaranteethat they are equal at all values. We cannot show that two expressions are equivalent just by checkingvalues. On the other hand, we can show that two expressions are not equivalent if we find values ofthe variables that give them different values.

Example 3 Is the expression (x+ y)2 equivalent to x2 + y2?

Solution If we choose, for example, x = 4 and y = 3 and evaluate the two expressions, we get

(x+ y)2 : (4 + 3)2 = 72 = 49

x2 + y2 : 42 + 32 = 16 + 9 = 25.

Since 25 is not equal to 49, it follows that (x+ y)2 is not equivalent to x2 + y2.

What Changes Can we Make to an Expression?In Example 1 we transformed 2(0.1B) into the equivalent expression 0.2B by regrouping the mul-tiplication. What actions may we perform on an expression to produce an equivalent one? The basicrule is:

We can do anything to an expression that does not change its value.

This means, among other things, that we can reorder and regroup addition, and reorder andregroup multiplication, because these actions do not change the value of a numerical expression.For example, reordering allows us to write

3 · 5 = 5 · 3 and 3 + 5 = 5 + 3,

while regrouping allows us to write

2 · (3 · 5) = (2 · 3) · 5 and 2 + (3 + 5) = (2 + 3) + 5.

These same principles are valid for algebraic expressions:

ab = ba and a+ b = b+ a

a(bc) = (ab)c and a+ (b+ c) = (a+ b) + c.

Of course, we cannot reorder the numbers in a subtraction or a division, since, for instance,3− 5 6= 5− 3, and 3/5 6= 5/3.

Example 4 During a normal month a bike shop sells q bicycles at a price of p dollars, and its gross revenue indollars is given by the expression qp. During a sale month it halves the selling price and the numberof bicycles sold triples. Write an expression for the revenue during a sale month. Does the revenuechange, and if so, how?

Solution Since the bike shop sells q bicycles during a normal month, and since that number triples duringa sale month, it sells 3q bicycles during a sale month. Similarly, the price is (1/2)p during a salemonth. Thus, the revenue during a sale month is

(3q)

(1

2p

).

By regrouping and reordering the factors in this product we get

(3q)

(1

2p

)= 3 · 1

2qp =

3

2qp.

Since the original revenue was qp the new revenue is 3/2 or 1.5 times the original revenue.


Example 5 If x+ y + z = 1 find the value of (x+ 10) + (y − 8) + (z + 3).

Solution We can remove the parentheses and reorder the terms in the expression to get

x+ 10 + y − 8 + z + 3 = x+ y + z + 10− 8 + 3.

Then, by regrouping, we find

(x+ y + z) + (10− 8 + 3) = (x+ y + z) + 5.

Since x+ y + z = 1, it follows that (x+ 10) + (y − 8) + (z + 3) = 6.

In summary

Reordering and regroupingFor addition and multiplication, terms and factors in an expression may be reordered or re-grouped without changing the value of the expression.

The Distributive Law

Example 6 If Abby’s bill at a restaurant is B dollars and Renato’s is b dollars, write two expressions for a 20%tip on the total.

Solution The total bill is B + b, so the total tip is 0.2(B + b). Alternatively, Abby’s tip is 0.2B and Renato’stip is 0.2b, so the total tip is 0.2B + 0.2b.

The two expressions in Example 6 are equivalent, that is,

0.2(B + b) = 0.2B + 0.2b for all values of B and b.

This is an instance of the

Distributive Law

a(b+ c) = ab+ ac

and

a(b− c) = ab− ac.In particular, if a = −1,

−1(b+ c) = −(b+ c) = −b− cand

−1(b− c) = −(b− c) = −b+ c.


Example 7 Use the distributive law to write two equivalent expressions for the perimeter of a rectangle, andinterpret each expression in words.

Solution Let l be the length and w the width of the rectangle. The perimeter is the sum of the lengths of the 4sides. Since there are 2 sides of length l and 2 of length w, the perimeter is given by the expression2l+ 2w. By the distributive law, this is equivalent to 2(l+w). The expression 2l+ 2w says “doublethe length, double the width, and add.” The expression 2(l + w) says “add the length to the widthand double.”

Example 8 Show that the following expressions are equivalent:x+ y

3and

x

3+y

3.

Solution Choose two values for x and y, say x = 4 and y = 11. Thenx+ y

3=

4 + 11

3=

15

3= 5

x

3+y

3=

4

3+

11

3=

15

3= 5.

Since the two expressions have the same value, it is possible that they are equivalent. Since dividingby 3 is the same as multiplying by 1/3, we can write

x+ y

3=

1

3(x+ y)

x

3+y

3=

1

3x+

1

3y.

Then the distributive law tells us that1

3(x+ y) =

1

3x+

1

3y.

Example 9 An even integer is an integer that can be written in the form 2 × integer. Show that the sum of twoeven integers is an even integer.

Solution Suppose you have two even integers, say 2m and 2n, where m and n are both integers. Their sumis 2m+ 2n. But

2m+ 2n = 2(m+ n) = 2× integer,

which has the form of an even integer. So the sum of two even integers is even.

Like terms

If two or more terms have exactly the same variables raised to exactly the same powers, they aresaid to be like terms. We use the distributive law to combine like terms into a single term:

2x2 + 5x2 = (2 + 5)x2 = 7x2.

(Think “two x2 plus five x2 is seven x2.”) Unlike terms may not be combined. For example, theterms in 3x+ 3x2 may not be combined because the variables are raised to different powers.


Example 10 Combine like terms in each of the following expressions:(a) 3x2 + 6x+ 9x− x2 (b) ax+ bx− cx2.

Solution (a) We can regroup the terms in this sum as 3x2 − x2 + 6x + 9x. Then we combine x2 terms,using 3x2 − x2 = 2x2 (think “three x2 take away one x2 is two x2”) and the x terms, using6x+ 9x = 15x. Therefore,

3x2 + 6x+ 9x− x2 = 2x2 + 15x.

(b) We can rewrite ax+ bx as (a+ b)x. Then ax+ bx− cx2 is equivalent to (a+ b)x− cx2.

Knowing When Expressions are not EquivalentSo far we have considered three operations for producing equivalent expressions: reordering, re-grouping, and using the distributive law. Knowing what you cannot do with expressions is often asimportant as knowing what you can do. In Example 3 we saw that we could not replace (x + y)2

by x2 + y2. If you are not sure whether two expressions are equivalent, you can try some specificvalues for the variables. If you do not get equal values for both expressions, they are not equivalent.

Example 11 Say whether the expression1

x+

1

y

is equivalent to

(a)2

x+ y(b)

y + x

xy

Solution (a) Putting x = 1 and y = 1 in each expression, we get

1

x+

1

y=

1

1+

1

1= 1 + 1 = 2,

2

x+ y=

2

1 + 1=

2

2= 1.

Since 2 6= 1, the expression 1/x+ 1/y is not equivalent to 2/(x+ y).(b) With x = 1 and y = 1, we get

y + x

xy=

1 + 1

1 · 1 =2

1= 2,

so it is possible that 1/x+ 1/y is equivalent to (y + x)/xy. To see that this is indeed the case,combine the fractions 1/x and 1/y over a common denominator.

1

x+

1

y=y

y· 1

x+x

x· 1

y=

y

xy+

x

xy=y + x

xy.



In Exercises 1–3,(a) Write an algebraic expression representing each of the

given operations on a number b.(b) Say whether the expressions are equivalent, and explain

what this tells you.

1. “Multiply by one fifth”“Divide by five”

2. “Multiply by 0.4”“Divide by five-halves”

3. “Multiply by eighty percent”“Divide by eight-tenths”

For Exercises 4–9, say whether the two expressions are equiv-alent.

4. 3(z + w) and 3z + 3w 5. (a− b)2 and a2 − b2

6.√a2 + b2 and a+ b 7. −a+ 2 and −(a+ 2)

8. (3− 4t)/2 and 3− 2t 9. bc− cd and c(b− d)

10. Which of the following expressions is equivalent to3(x2 + 2)− 3x(1− x)?

(a) 6 + 3x (b) −3x+ 6x2 + 6

(c) 3x2 + 6− 3x− 3x2 (d) 3x2 + 6− 3x

Simplify and then combine like terms in each of the expres-sions in Exercises 11–16.

11. 3p2 − 2q2 + 6pq − p2

12. y3 + 2xy − 4y3 + x− 2xy

13. (1/2)A+ (1/4)A− (1/3)A

14. (a+ 4)/3 + (2a− 4)/3

15. 3(2t− 4)− t(3t− 2) + 16

16. z3 + 5z3 − 3z

For Exercises 17–21, say whether the the attempt to combinelike terms is correct or incorrect.

17. 2x2 + 3x3 = 5x5

18. 2AB2 + 3A2B = 5A3B3

19. 3h2 + 2h2 = 5h2

20. 3b+ 2b2 = 5b3

21. Write a sentence explaining what it means for two ex-pressions to be equivalent.

Problems

22. To convert from miles to kilometers, Abby takes the num-ber of miles, m, doubles it, then subtracts 20% from theresult. Renato first divides the number of miles by 5, andthen multiplies the result by 8.

(a) Write an algebraic expression for each method.(b) Use your answer to part (a) to decide if the two meth-

ods give the same answer.

23. The formula for the area of a triangle is often expressedasA = (1/2)bh. Is the expression bh/2 equivalent to theexpression (1/2)bh?

24. On Figure 1.1, indicate an interval of length 1 − x, andthen use this to indicate an interval of length 1− (1−x).What two expressions does this suggest are equivalent?

-�x -�

1

Figure 1.1

25. On Figure 1.2, indicate intervals of length (a) x + 1(b) 2(x+ 1) (c) 2x (d) 2x+ 1

What does your answer tell you about whether 2x+1 and 2(x+ 1) are equivalent?

-�1

-�x

-�x

-�1

Figure 1.2

In Problems 26–28, each row of the table is obtained by per-forming the same operation on all the entries of the previousrow. Fill in the blanks in the table.

26.−2 −1 0 1 2 a

−1 0 1 2 3

−1/2 0 1/2 1 3/2

1/2 0 −1/2 −1 −3/2

3/2 1 1/2 0 −1/2

1.3 EQUATIONS IN ONE VARIABLE 13

27.−2 −1 0 1 2 a

−1 −1/2 0 1/2 1

1 1/2 0 −1/2 −1

3/2 1 1/2 0 −1/2

28.1 2 3 4 5 x x− 1

2 6 8 10 B

3 5 7 11

29. Consider the following sequence of operations on a num-ber n: “Add four, double the result, add the original num-ber, subtract five, divide by three.”

(a) Write an expression in n giving the result of the op-erations.

(b) Show that the result is always one more than thenumber you start with.

30. (a) Show the following expression is equivalent to y:

(3(y − 2) + 6)− y2

(b) Use your answer to part (a) to write down a sequenceof operations on a number y that always yields thenumber you started with.

31. Give a sequence of operations on a number x that alwaysyields 10 no matter what number you start with, and ex-plain why your answer works.

32. Consider the following sequence of operations: “Pick anumber, add 3, double the result, subtract 4 from that an-swer and triple what you get.”

(a) If you pick 10, what number do you get?(b) If you pick 5, what number do you get?(c) If you pick a number and end up with 30, what num-

ber did you pick?(d) Explain how to find someone’s number if you are

told what the answer is after doing all the steps.(e) Maris has a shortcut for finding someone’s number.

“Take the answer, divide by 6, and subtract 1.” IsMaris right?

In Problems 33–34, use the fact that an even number is onethat can be written in the form 2n, and an odd number is onethat can be written in the form 2n+ 1, where n is an integer.

33. Is the sum of an odd and an even integer always odd, al-ways even, or sometimes one and sometimes the other?Use algebraic expressions to justify your answer.

34. Is the product of an odd and an even integer always odd,always even, or sometimes one and sometimes the other?Use algebraic expressions to justify your answer.

1.3 EQUATIONS IN ONE VARIABLE

Different algebraic expressions represent different calculations. Often we want to know when twodifferent calculations give the same answer.

Example 1 You have a $10 discount certificate for a pair of pants. When you go the store you discover that thereis a 25% off sale on all pants. For what tag price will you end up paying the same amount with eachdiscount method?

Solution Let p be the tag price, in dollars, of a pair of pants. With the discount certificate you pay p − 10,and with the store discount you pay 75% of the price, or 0.75p. You want to know what value of pmakes the following statement true:

p− 10 = 0.75p.

By trying several different values of p, we find that p = 40 does the job. (See Table 1.2.) You paythe same price with either discount method when the tag price is $40.

Table 1.2 Comparison of two discount methods

Tag price, p (dollars) 20 30 40 50 60

Discount certificate, p− 10 10 20 30 40 50

Store discount, 0.75p 15 22.50 30 37.50 45

Notice that in Example 1 we are using the equal sign in a different way than we did in Sec-tion 1.2. There we used it to indicate that two expressions were equivalent, for example, 2(l+w) =2l + 2w. These expressions have the same value for all values of l and w.


However, here the two expressions in Example 1, p − 10 and 0.75p, are not equivalent. InTable 1.2, we see that these expressions are not equal for p = 20, 30, 50, or 60. Our purpose isto find a value for p for which the two expressions are equal, namely p = 40. A statement likep − 10 = 0.75p, where we want to find a value that makes the statement true, is an example of anequation.

An equation is a statement that two expressions are equal. A value of the variable that makesan equation true is called a solution of the equation, and we say that the solution solves orsatisfies the equation.

The equation p−10 = 0.75p is true for p = 40, so p = 40 is a solution. The equation x2 = −16is not true for any value of x, so it has no solutions.

Example 2 Which of the following are not equations?(a) 3(x− 5) + 6− x (b) 3(x− 5) + 6 = x(c) ax2 + bx+ c (d) at2 + t = bt2 − 2

Solution An equation must include an equal sign, so (a) and (c) are not equations.

Solutions of EquationsWe have seen that an equation may be true for some values of the variable and false for others.To test whether a value of the variable is a solution to an equation, you evaluate each side of theequation and see if the two sides are equal. For the equation in Example 1, p = 40 is a solutionbecause

p− 10 = 0.75p

p = 40 : 40− 10 = 0.75 · 40 are they equal?

30 = 30 yes, statement is true.

On the other hand, trying p = 50, we get

p− 10 = 0.75p

p = 50 : 50− 10 = 0.75 · 50 are they equal?

40 = 37.50 no, statement is false,

so p = 50 is not a solution.

Example 3 For each of the following equations, say which of the given values is a solution.(a) 3− 4t = 5− (2 + t), for the values t = −3, 0(b) 3x2 + 5 = 8, for the values x = −1, 0, 1.

Solution For each equation, we test the given values of the variable to see if the two sides are equal.(a)

3− 4t = 5− (2 + t)

t = −3 : 3− 4(−3) = 5− (2 + (−3)) −→ 15 = 6 not a solution

t = 0 : 3− 4 · 0 = 5− (2 + 0) −→ 3 = 3 a solution.

The two sides agree when t = 0, and disagree when t = −3. So t = 0 is a solution and t = −3is not a solution.

1.3 EQUATIONS IN ONE VARIABLE 15

(b)

3x2 + 5 = 8

x = −1 : 3(−1)2 + 5 = 8 −→ 8 = 8 a solution

x = 0 : 3 · 02 + 5 = 8 −→ 5 = 8 not a solution

x = 1 : 3 · 12 + 5 = 8 −→ 8 = 8 a solution.

Both x = −1 and x = 1 satisfy the equation and are solutions, but x = 0 is not a solution.

In Example 3 we saw that an equation can have more than one solution. We cannot be sure wehave all the solutions to an equation by trial and error. If an equation is simple enough, we can oftenreason out the solutions using arithmetic.

Example 4 Find all the solutions to the following equations.(a) x+ 5 = 17 (b) 20 = 4a (c) s/3 = 22 (d) g2 = 49

Solution (a) There is only one number that gives 17 when you add 5, so the only solution is x = 12.(b) Here the product of 4 and a is 20, so a = 5 is the only solution.(c) Here 22 must be one third of s, so s must be 3 · 22 = 66.(d) The only numbers whose squares are 49 are 7 and −7, so g = 7 and g = −7 are the solutions.

Equations and IdentitiesThe equal sign is used for more than one purpose in algebra. We have seen that the equal sign inan equation may or may not represent a true equality for particular values of the variables. On theother hand, when we have two equivalent expressions, such as x+ x and 2x, we often use an equalsign to indicate the expressions are equal for all values of x. In order to distinguish this use of theequal sign from its use in a general equation, we refer to a statement like

x+ x = 2x

as an identity. An identity is really a special equation, one that is true for all values of the variables.


In Exercises 1–6, say whether the value of the variable is asolution to the equation.

1. t+ 3 = t2 + 9, t = 3 2. x+3 = x2−9, x = −3

3. x+ 3 = x2 − 9, x = 4 4.a+ 3

a− 3= 1, a = 0

5. 3 + a

3− a = 1, a = 0 6. 4(r−3) = 4r−3, r = 1

7. Which of the following numbers is a solution to the fol-lowing equation?

−3x2 + 3x+ 8

2= 3x(x+ 1) + 1

(a) 1 (b) 0 (c) −1 (d) 2

Write an equation for each situation presented in Exercises 8–12, letting x stand for the unknown number.

8. Twice a number is 24.

9. A number increased by 2 is twice the number.

10. Six more than a number is the negative of the number.

11. Six less than a number is 30.

12. A number is doubled and then added to itself. The resultis 99.

In Exercises 13–17, say in words the statement represented bythe equation.

13. 2x = 16 14. x = x2

15. s+ 10 = 2s 16. 0.5t = 250

17. y − 4 = 3y


Solve each equation in Exercises 18–22.

18. 5x = 20 19. t+ 5 = 20

20. w/5 = 20 21. y − 5 = 20

22. 20 = 5− x

Say which of the equations in Exercises 23–26 are identities.

23. x2 + 2 = 3x 24. 2x2 + 3x2 = 5x2

25. 2u2 + 3u3 = 5u5

26. t+ 1/(t2 + 1) = (t+ 1)/(t2 + 1)

27. Write a short explanation describing the difference be-tween an expression and an equation.

Problems

A ball thrown vertically upward at a speed of v ft/sec rises adistance d feet in t seconds, given by d = 6 + vt − 16t2.In Problems 28–31, write an equation whose solution is thegiven value. Do not solve the equation.

28. The time it takes a ball thrown at a speed of 88 ft/sec torise 20 feet.

29. The time it takes a ball thrown at a speed of 40 ft/sec torise 15 feet.

30. The speed with which the ball must be thrown to rise 20feet in 2 seconds.

31. The speed with which the ball must be thrown to rise 90feet in 5 seconds.

In Problems 32–34, determine whether the sentence describesan identity.

32. Eight more than a number n is the same as two less thansix times the number.

33. Twice the combined income of Carlos and Jesse equalsthe sum of double Carlos’ income and double Jesse’s in-come.

34. In a store, N bottles of one brand of bottled water, at$1.19 a bottle, plus twice that many bottles of anotherbrand at $1.09 a bottle, cost $6.74.

1.4 SOLVING EQUATIONS IN ONE VARIABLE

How can we find the solutions to an equation? In Example 4 on page 15 the equations were simpleenough to solve by direct reasoning about numbers. To solve a more complicated equation, we tryto transform the equation into a simpler one that has the same solutions.

BalancingA solution to an equation makes the two sides equal. We can think of an equation like a scale onwhich things are weighed. When the two sides are equal, the scale balances, and when they aredifferent it is unbalanced. If we change the weight on one side of the scale, we must change theother side in exactly the same way to be sure that the scale remains in the same state as before,balanced or unbalanced. Similarly, when we transform an equation into a simpler one, we mustperform the same operation on both sides of the equal sign. In that way we can be sure that the newequation has the same solutions—the same values that make the scale balance—as the original one.The simplest form of equation that we can aim for is one of the form

x = Number,

which directly tells us the solution.

Example 1 Solve the equation 16.8x = 84.

Solution We divide both sides of the equation by 16.8:

16.8x = 8416.8x

16.8=

84

16.8divide both sides by 16.8

x = 5.

We have transformed our original equation into the simpler equation x = 5, whose solution isobviously 5. You can check that 5 is also the solution to 16.8x = 84 by multiplying: 16.8 · 5 = 84.

1.4 SOLVING EQUATIONS IN ONE VARIABLE 17

In the previous example, we solved the equation by dividing both sides by 16.8. We chosethis operation to isolate the x on one side. Being able to anticipate the effect of an operation is animportant skill in solving equations.

Example 2 What operation transforms the first equation into the second equation? Check to see that the solu-tions of the second equation are also solutions of the first equation.

(a) 3t+ 58.5 = 94.53t = 36

(b) a2/1.6 = 40a2 = 64

Solution (a) We subtract 58.5 from both sides of the first equation to get

3t+ 58.5− 58.5 = 94.5− 58.5

3t = 36.

The solution to the last equation is 12. Substituting t = 12 into the first equation gives 3(12) +58.5 = 36 + 58.5 = 94.5 so 12 is also the solution of the first equation.

(b) We multiply both sides of the first equation by 1.6 to get

1.6(a2/1.6) = 40(1.6)

a2 = 64.

There are two solutions to the second equation, 8 and−8. Substituting these values into the firstequation gives 82/1.6 = 64/1.6 = 40 and (−8)2/1.6 = 64/1.6 = 40, so 8 and −8 are bothsolutions to the first equation as well.

We summarize our observations as follows.

Principle for Solving EquationsWe can do anything to an equation that does not change whether the two sides are equal ornot. This includes• adding or subtracting the same number to both sides• multiplying or dividing both sides by the same number, provided it is not zero.

Following this rule ensures that the original equation has the same solutions as the new equa-tion, even though the expressions on each side might change.

Example 3 Without solving, explain why the equations in each pair have the same solution.(a) 2.4(v − 2.1)2 = 15

(v − 2.1)2 = 6.25(b) y3 + 4y = y3 + 2y + 7

4y = 2y + 7

Solution (a) We divide both sides of the first equation by 2.4 to obtain the second equation.(b) We subtract y3 from both sides of the first equation to get the second.


Example 4 What operation transforms the first equation into the second equation? Explain why this operationis not a valid step for solving the equation.

x2 = 3x

x = 3.

Solution We divide both sides of x2 = 3x by x to obtain x = 3. This step is not valid because we must notdivide by zero, and the variable x might be zero. In fact, both x = 3 and x = 0 are solutions to thefirst equation, but only 3 is a solution of the second equation. When we divided by x, we lost one ofthe solutions of the original equation.

Deciding Which Operation to UseIf an equation includes only one expression involving the variable, we try to analyze which opera-tions were used to build that expression, and then undo the operations by performing the oppositeoperations.

Example 5 Which operation should we use to solve each equation?(a) x+ 5 = 20 (b) 5x = 20 (c) x5 = 20

Solution (a) In this equation, 5 was added to x, so we should subtract 5 from both sides of the equation.(b) Because x is multiplied by 5, we should divide both sides of the equation by 5.(c) Here xwas raised to the fifth power, so we should take the fifth root of both sides of the equation.

Example 6 Solve each equation.

(a) 3√z = 8 (b)

1

z= 2.5

Solution (a) To undo the operation of taking the cube root we cube both sides of the equation.3√z = 8

( 3√z)3 = 83 cube both sides

z = 512.

We check to see that 512 is a solution: 3√

512 = 8.(b) To undo the operation of taking the reciprocal we take the reciprocal again.

1

z= 2.5

1

1/z=

1

2.5take reciprocal of both sides

z = 0.4.

We check to see that 0.4 is a solution: 1/0.4 = 2.5.

Transforming Equations versus Transforming ExpressionsLet’s compare what we have learned about expressions and equations. To transform an expressioninto an equivalent expression, we can use any operation that does not change the value of the expres-sion. These operations include reordering, regrouping, and using the distributive law. To transforman equation, we can use any operation that does not change the equality of the two sides. However,

1.4 SOLVING EQUATIONS IN ONE VARIABLE 19

when we solve an equation, the expressions on either side of the equal sign may change at each step.Consider the equation 2x + 5 = 13, whose solution is 4. If we subtract 5 from both sides of theequation, we get

2x+ 5 = 13

2x = 8 subtracting 5 from both sides,

whose solution is also 4. But the expression 2x is not equivalent to the expression 2x+ 5. Althoughwe can subtract 5 from both sides of an equation, we cannot subtract 5 from an expression withoutchanging the value of the expression. Thus, the list of operations we can use to solve equationsincludes some things we cannot do to create an equivalent expression.

Example 7 (a) Doesx

2+

3

4= 2x have the same solution as 2x+ 3 = 8x?

(b) Isx

2+

3

4equivalent to 2x+ 3?

Solution (a) If we multiply both sides of the first equation by 4, we get

x

2+

3

4= 2x

4

(x

2+

3

4

)= 4(2x) multiplying both sides by 4

4(x

2

)+ 4

(3

4

)= 4(2x) applying the distributive law

2x+ 3 = 8x.

Therefore, the two equations have the same solution. You can verify that x = 1/2 is the solutionto both equations.

(b) The expression x/2 + 3/4 is not equivalent to 2x + 3. This can be seen by substituting x = 0into each expression. The first expression becomes

x

2+

3

4=

0

2+

3

4=

3

4,

but the second expression is2x+ 3 = 2(0) + 3 = 3.

Since 3 is not equal to 3/4, the expressions are not equivalent. We cannot multiply this expres-sion by 4 without changing its value.


In Exercises 1–8, say whether the second equation is the re-sult of a valid operation on the first, and if so say what theoperation is.

1. 3 + 5x = 1− 2x3 + 7x = 1

2. 3 + 2x = 53 = 2x+ 5

3. 1− 2x2 + x = 1x− 2x2 = 0

4. x

3− 3

4= 0

4x− 9 = 0

5. 9x− 3x2 = 5x9− 3x = 5

6. 3

4− x

3= 2

9− 3x = 2

7. 5x2 − 20x = 90x2 − 4x = 18

8. x+ 2

5= 1− 3x

x+ 2 =1− 3x

5


In Exercises 9–16, state the operation on both sides of theequation that isolates the variable on one side, and give thesolution of the equation.

9. 0.1 + t = −0.1 10. −10 = 3 + r

11. −t+ 8 = 0 12. 5y = 19

13. −x = −4 14. 7x = 6x− 6

15. −x5

= 4 16. 0.5x = 3

Problems

In Problems 17–24, say which of the following operations onboth sides transforms the equation into one whose solution iseasiest to see.

(a) Add 5 (b) Add x (c) Collect like terms(d) Multiply by 3 (e) Divide by 2

17. x− 5 = 6 18. 2x = 2

19. x

3− 1

3= 0 20. 5− x = 0

21. 2x− 7− x = 3 22. 2− 2x = 2− x

23. 2x

3+x

3= 2 24. 5− 2x = 0

In Problems 25–28, the solution to the equation depends onthe constant a. Assuming a is positive, what is the effect ofincreasing a on the value of the solution? Does the solutionincrease, decrease, or remain unchanged? Give a reason for

your answer that can be understood without solving the equa-tion.

25. x− a = 0 26. ax = 1

27. ax = a 28. x

a= 1

29. A town’s total allocation for firemen’s wage and bene-fits in a new budget is $600,000. If wages are calculatedat $40,000 per fireman and benefits at $20,000 per fire-man, write an equation whose solution is the maximumnumber of firemen the town can employ, and solve theequation.

30. (a) Does x/3 + 1/2 = 4x have the same solution as2x+ 3 = 24x?

(b) Is x/3 + 1/2 equivalent to 2x+ 3?

31. (a) Does 8x − 4 = 12 have the same solution as2x− 1 = 3?

(b) Is 8x− 4 equivalent to 2x− 1?

1.5 TABLES AND GRAPHS

We can analyze an algebraic expression by making a table of its values for various different valuesof the variable, and plotting the result.

Example 1 Make a table of values of a2 for a = −2,−1, 0, 1 and 2, and plot the results. Explain how youranswer illustrates that a2 = (−a)2 for any number a.

Solution See Table 1.3 and Figure 1.3. Notice the pattern in the right column of the table: the values go from4 to 1 to 0, then back to 1 and 4 again. This is because (−2)2 = 22 = 4, and (−1)2 = 12 = 1. Wecan see the same thing in the symmetrical arrangement of the points in Figure 1.3 about the verticalaxis. Since both a2 and (−a)2 have the same value, the point above a on the horizontal axis has thesame value as the point above −a.

Table 1.3 Values of a2

a a2

−2 4

−1 1

0 0

1 1

2 4

−2 −1 1 2

1

2

3

4

a

Figure 1.3: Graph of values of a2

1.5 TABLES AND GRAPHS 21

We can also use graphs to visualize solutions to equations.

Example 2 A town’s population t years after it was incorporated is given by 30,000+2000t. Plot the populationat five year intervals over a 20 year period, and indicate the solution t = 10 to the equation

30,000 + 2000t = 50,000

on your plot and interpret the solution in practical terms.

Solution The initial population in year t = 0 is given by

Population = 30,000 + 2000(0) = 30,000.

In year t = 5 the population is given by

Population = 30,000 + 2000(5) = 30,000 + 10,000 = 40,000.

In year t = 10 the population is given by

Population = 30,000 + 2000(10) = 30,000 + 20,000 = 50,000.

Similar calculations for year t = 15 and year t = 20 give the values in Table 1.4. The dotted lineshows the trend in the data. The practical interpretation of the solution t = 10 is that the populationreaches 50,000 in 10 years.

Table 1.4 Population over 20 years

t, years Population

0 30,000

5 40,000

10 50,000

15 60,000

20 70,000 5 10 15 20

10,00020,00030,00040,00050,00060,00070,000

t, time (years)

Population

� Population is 50,000when t = 10

Figure 1.4: Town’s population over 20 years

In Example 1 we plotted a few values of a2. If we plot more values, we get Figure 1.5. Noticehow the points appear to be joined by a smooth curve. If we could plot all the values we would getthe curve in Figure 1.6. This is the graph of the expression.

−2 −1 1 2

1

2

3

4

a

Figure 1.5: More valuesof a2

−2 −1 1 2

1

2

3

4

a

Figure 1.6: Graph of a2


Example 3 Bernardo plans to travel 400 miles over spring break to visit his family. He can choose to fly, drive,take the train or make the journey as a bicycle road trip. If his average speed is r miles per hour,then the time taken, in hours, is given by 400/r. Make a table of values and graph them.

Solution If r = 200, we have

Time taken =400

200= 2.

This means that it will take Bernardo 2 hours traveling at 200 miles per hour. If r = 80, we have

Time taken =400

80= 5,

which means it will take 5 hours traveling at 80 miles per hour. Table 1.5 shows other values, andFigure 1.7 shows the plot of these values, along with the graph of 400/r.

Table 1.5 Valuesof 400/r

r 400/r

25 16

40 10

80 5

100 4

200 2

100 200 300 400

10

20

Speed, r (mph)

Time taken, in hours

Figure 1.7: Graph of 400/r

Example 4 Write an equation whose solution is the speed Bernardo would have to maintain to make the tripwith 25 hours of cycling. Use the graph in Figure 1.7 to estimate a solution to this equation, thencheck your answer by evaluating.

Solution We want the time taken to be 25 hours. Since the time taken at speed r is 400/r, we want to solvethe equation

25 = 400/r.

We can locate the point on the graph corresponding to a time of 25 hours, as shown in Figure 1.8.This point corresponds to about r = 16, so we estimate that r = 16. Substituting this into theright-hand side of the equation we get 400/16 = 25, so it is indeed a solution. To travel 400 milesin 25 hours of cycling, Bernardo should ride at an average speed of 16 miles per hour.

16

25 � Time is 25 hours when speed is 16 mph

Speed, r (mph)

Time taken, in hours

Figure 1.8: Finding the solution to 25 = 400/rgraphically



In Exercises 1–6, complete the table, and say which, if any, ofthe expressions in the left-hand column are equivalent to eachother. Justify your answer algebraically.

1.x −11 −7 0 7 11

2x

3x

2x+ 3x

5x

2.t −11 −7 0 7 11

2t

−3t

2t− 3t

−t

3.m −1 0 1

m4

2m2

2m2 + 2m2

4m4

4m2

4.I −2 −1 0 1 2

−I−(−I)

5.a −2 −1 0 1 2

−(1/2)(a+ 1) + 1

−(1/2)a+ (1/2)

6.x −2 −1 0 1 2

x+ 3

−(x+ 3)

−x−x+ 3

−x− 3

7. Figure 1.9 shows the number of pigeons living in a townin various years t.

(a) In what year is the population 500?(b) The number of pigeons is 300 + 100t. From the

graph, when does 300 + 100t equal 600?

1 2 3 4

100

200

300

400

500

600

700

t

pigeons

Figure 1.9

8. Figure 1.10 shows the resale value of a car given by$10,000(0.5)t, where t is in years.

(a) In which years in the figure is the car’s value greaterthan $1250?

(b) What value of t is a solution to 5000 =10,000(0.5)t?

1 2 3 4

6251250

2500

5000

10000

t

value ($)

Figure 1.10

9. Table 1.6 shows values of x and the expression 3x + 2.For which values of x in the table is

(a) 3x+ 2 < 8(b) 3x+ 2 greater than 8(c) 3x+ 2 equal to 8

Table 1.6

x 0 1 2 3 4

3x+ 2 2 5 8 11 14


10. Table 1.7 shows values of z and the expression 4 − 2z.For which values of z in the table is

(a) 4− 2z less than 2 (b) 4− 2z > 2

(c) 4− 2z = 2

Table 1.7

z 0 1 2 3 4

4− 2z 4 2 0 −2 −4

Problems

11. The balance in dollars in a bank account after t years isgiven by the expression 4622(1.04)t rounded to the near-est dollar. For which values of t in Table 1.8 is

(a) 4622(1.04)t < 5199?(b) The balance greater than $5199?(c) Balance = $5199?

Table 1.8

t 0 1 2 3 4

4622(1.04)t 4622 4807 4999 5199 5407

12. The height (in meters) of a diver s seconds after begin-ning his dive is given by the expression 10 + 2s− 9.8s2.For which values of s in Table 1.9 is

(a) 10 + 2s− 9.8s2 < 9.89?(b) His height greater than 9.89 meters?(c) Height = 9.89 meters?

Table 1.9

s 0 0.25 0.5 0.75 1

10 + 2s− 9.8s2 10 9.89 8.55 5.99 2.2

13. Table 1.10 shows values of v and the expressions 12−3vand −3 + 2v. For which values of v in the table is

(a) 12− 3v less than −3 + 2v?(b) 12− 3v > −3 + 2v?(c) 12− 3v = −3 + 2v?

Table 1.10

v 0 1 2 3 4 5 6

12− 3v 12 9 6 3 0 −3 −6

−3 + 2v −3 −1 1 3 5 7 9

14. Table 1.11 shows values of a and the expressions 2 + a2

and 10− 2a. For which values of a in the table is

(a) 2 + a2 < 10− 2a?(b) 2 + a2 greater than 10− 2a?(c) 2 + a2 = 10− 2a?

Table 1.11

a 0 1 2 3 4 5 6

2 + a2 2 3 6 11 18 27 38

10− 2a 10 8 6 4 2 0 −2

15. If a company sells p software packages, its profit is given

by $10,000− 100,000

p, as shown in Figure 1.11.

(a) From the graph, estimate the number of packagessold when profits are $8000.

(b) Check your answer to part (a) by substituting it intothe equation

10,000− 100,000

p= 8000.

20 40 60 80

2000

4000

6000

8000

10,000

p

profit

Figure 1.11

16. The tuition for a semester at a small public university tyears from now is given by $3000 + 100t, as shown inFigure 1.12.

(a) From the graph, estimate how many years it will takefor tuition to reach $3700.


3000 + 100t = 3700.

1 3 5 7 9

3000

3500

4000

t years

tuition ($)

Figure 1.12


17. (a) Construct a table showing the values of the expres-sion 1 + 5x for x = 0, 1, 2, 3, 4.

(b) For what value of x does 1 + 5x = 16?

18. (a) Construct a table showing the values of the expres-sion 3− a2 for a = 0, 1, 2, 3, 4.

(b) For what value of a does 3− a2 = −6?

19. Tuition cost T (in dollars) for part-time students atStonewall College is given by T = 300 + 200C, whereC represents the number of credits taken.

(a) Make a table showing costs for taking from one totwelve credits. For each value of C, give both the tu-ition cost, T , and the cost per credit, T/C. Round tothe nearest dollar.

(b) Which of these values of C has the smallest cost percredit?

20. The population of a town, t years after it was founded, isgiven by 5000 + 350t.

(a) Write an equation whose solution is the number ofyears it takes for the population to reach 12,000.

(b) Make a plot of the population for t =14, 16, 18, 20, 22, 24 and indicate the solutiont = 20 to the equation in part (a).

21. The number of stamps in a person’s passport, t years afterthe person gets a new job which involves overseas travel,is given by 8 + 4t.

(a) Write an equation whose solution is the number ofyears it takes for the passport to have 24 stamps.

(b) Make a plot of the number of stamps for t =1, 2, 3, 4, 5, 6, and indicate the solution t = 4 to theequation in part (a).

22. The balance in a checking account set up to pay rent,m months after its establishment, is given by $4800 −400m.

(a) Write an equation whose solution is the number ofmonths it takes for the account balance to reach$2000.

(b) Make a plot of the balance for m = 1, 3, 5, 7, 9, 11,and indicate the solution m = 7 to the equation inpart (a).

23. The number of gallons of gas in a car’s tank, dmiles afterstopping for gas, is given by 15− d/20.

(a) Write an equation whose solution is the number ofmiles it takes for the amount of gas in the tank toreach 10 gallons.

(b) Make a plot of the balance for d =40, 60, 80, 100, 120, 140, and indicate the solutionm = 100 to the equation in part (a).

24. The populations of Towns A and B are in Table 1.12.

(a) For which years in the table is the population ofTown A

(i) Greater than the population of Town B?(ii) Less than the population of Town B?

(iii) Equal to the population of Town B?(b) The population, in year y, of Town A is 600 +

100(y − 2000) and of Town B is 200 + 300(y −2000). Use the table to find a value of y for which600 + 100(y − 2000) = 200 + 300(y − 2000).

(c) Check your answer to part (b) by using it to evaluate600 + 100(y − 2000) and 200 + 300(y − 2000).

Table 1.12

Year 2000 2001 2002 2003 2004

Pop A 600 700 800 900 1000

Pop B 200 500 800 1100 1400

25. The balances, t years since 2000, in two checking ac-counts, R and S, are in Table 1.13.

(a) For which years in the table is the balance in ac-count R

(i) Greater than the balance in account S?(ii) Less than the balance in account S?

(iii) Equal to the balance in account S?(b) The balance, in year t, of account R is $6312 −

1528t and of account S is $1000(1.2)t. Use the ta-ble to find a value of t for which 6312 − 1528t =1000(1.2)t.

(c) Check your answer to part (b) by evaluation.

Table 1.13

t 0 1 2 3 4

Bal R ($) 6312 4784 3256 1728 200

Bal S ($) 1000 1200 1440 1728 2074

Table 1.14 shows values of three unspecified expressions in xfor various different values of x. Give as many solutions of theequations in Problems 26–28 as you can find from the table.

Table 1.14

x −1 0 1 2

Expression 1 1 2 −1 0

Expression 2 1 0 −1 0

Expression 3 0 2 −1 −1

26. Expression 1 = Expression 2




1.6 UNITS

When we take a measurement, we always do so with respect to some fundamental unit. Differenttypes of measurements we might make, and the units we might use, include:• Length, measured in inches, feet, yards, miles, or (if we are using the metric system) millime-

ters, centimeters, meters, and kilometers.• Weight and mass,1 measured in ounces, pounds, grams, and kilograms.• Time, measured in seconds, minutes, days, years, and centuries.• Currency, measured in cents, US dollars ($US), British pounds (£), and Japanese yen (Y=).• Volume and liquid measure, measured in gallons, quarts, ounces, and liters.• Temperature, measured in degrees Fahrenheit (◦F), degrees Celsius (◦C), and Kelvins (K).

Anything we can measure—from velocity to electrical consumption to angle to typing speed tocaloric content to pulse rate to computer memory to cranial capacity—has its own special units ofmeasurement.

Adding and Subtracting Different UnitsWhen adding fractions, we must first convert to a common denominator. Similarly, when addingmeasurements in different units, we must first convert to a common unit.

Example 1 A person has two pieces of wood, one measuring 7 feet and the other measuring 42 inches. Find thetotal length of the two pieces, in both inches and feet.

Solution To find the total length of both pieces of wood, we should not add 7 and 42 directly, because thesenumbers are in different units. Instead, we must convert to a common unit. Since there are 12 inchesin 1 foot, we see that the 7-foot board measures 7× 12 = 84 inches, and so the total length of bothboards is 84 + 42 = 126 inches. Alternatively, we see that the 42-inch board measures 42/12 = 3.5feet. Thus, the total length of both boards is 7 + 3.5 = 10.5 feet.

In Example 1, our two answers, 126 inches and 10.5 feet, represent the same length in differentunits. Since one foot contains 12 inches, we have

Number of inches = 12× Number of feet126 = 12 · 10.5.

Example 2 Eleanor returns to the US after traveling abroad. She has $50 in US currency and £40 in Britishcurrency. How much money does she have in total?

Solution She can’t add $50 to £40 directly. This would give 90—but 90 what? Pounds or dollars? Neither iscorrect. Instead of adding $50 to £40 directly, Eleanor must first convert to a single currency.

To make the conversion, we can look up the current exchange rate on the web. The poundrecently2 traded for close to $1.68. This means that every £1 is worth $1.68, and so we can write

Value indollars = 1.68× value in

pounds .

1Technically, these are two different types of measurements. For instance, an astronaut has a different weight in spacethan she does on Earth, but her mass does not change.

2www.xe.com, accessed June 17, 2003

1.6 UNITS 27

Eleanor has £40, and so the dollar value of this amount is given by

Value of £40in dollars = 1.68× 40 = $67.20.

This means the total value of her currency is given by $50 + $67.20 = $117.20.Another approach is to convert dollars to pounds instead of pounds to dollars. To do this, we

can rewrite our conversion formula as follows:

Value inpounds =

1

1.68× value in

dollars dividing by 1.68

= 0.595× valuein dollars.

Since this person has $50, this means that

Value of $50in pounds = 0.595× 50 = £29.75.

Adding this to her £40 gives a total value of £69.75. Notice that 69.75 is quite a bit less than ourprevious answer of 117.20, but this is because we are working in pounds, which are worth morethan dollars. Because dollars are worth so much less than pounds, it takes more of them to work outto the same monetary value. Note also that £69.75 = $117.20.

Multiplying Different UnitsWhen multiplying measurements in different units, there is no need to convert to a common unit—infact, very often we can’t do this. From a practical point of view, if we want to multiply two units,the only requirement is that the result should be meaningful in the context of the problem we aretrying to solve.

Length, Area, and Volume

When we calculate areas and volumes, we often need to multiply length measurements together.A fundamental unit of length is the centimeter (cm), which is this long: . To measure

a length we put centimeters end to end along it and count how many there are. For example, thelength of a standard sheet of notebook paper is almost 28 cm. A fundamental unit for area is thesquare centimeter, (cm2), the area of a square that measures 1 cm by 1 cm:

To measure an area, we tile it with these squares and count how many there are. For example, itwould take just over 603 squares to cover an ordinary sheet of notebook paper.

For some simple shapes we can calculate the area from two length measurements. For example,the area of a rectangle is given by

Area of rectangle = length× width.


Example 3 Find the area of an ordinary sheet of notebook paper using the fact that:(a) Its length is 11 in and its width is 8.5 in(b) Its length is 27.94 cm and its width is 21.59 cm.

Solution (a) We have

Area = length× width= 8.5 in× 11 in = 93.5 in2.

(b) We haveArea = 27.94 cm× 21.59 cm = 603.2 cm2.

The volume of a box is given by

Volume of box = Length×Width× Height.

Example 4 Find the volume of a box that measures 2 ft by 4 ft by 1/2 ft.

Solution We have

Volume = Length×Width× Depth= 2 ft× 4 ft× 1/2 ft = 4 ft3.

Here, the product of three length measurements has physical meaning: It represents a volume.

The units in Example 4 are in cubic feet, or ft3. A volume of 1 ft3 equals the space inside a boxmeasuring 1 ft by 1 ft by 1 ft.

Example 5 A second box measures 2.5 feet by 2 yards by 6 inches. Is it larger or smaller in volume than thebox in Example 4?

Solution We have volume = 2.5 ft× 2 yards× 6 in. Here, it will be useful to convert all our units to feet sothat we can compare the volume of this box to the volume of the box in Example 4. Since there are12 inches in a foot, we see that 6 inches is 1/2 ft. Likewise, since there are 3 feet in 1 yard, we seethat there are 6 feet in 2 yards. We have

Volume = 2.5 ft× 6 ft× 0.5 ft = 7.5 ft3.

As you can see, this box is a good deal larger than the box in Example 4, which measured 4 ft3 involume.

Other Units

So far, our examples have been from geometry and physics. There are many non-geometrical sit-uations where numbers and units can be meaningfully multiplied together. For example, the termman-month is commonly used to measure human labor or activity (although the term person-monthwould be less biased). This unit stands for one person working for one month.

Example 6 A project requires the labor of 4 people working 3 months. How many man-months does it require?How many months would it take 3, 2, or 6 people to complete the same project?

1.6 UNITS 29

Solution We have

Amount of labor = Number of people · Time required = 4 · 3 = 12 man-months.

Other possibilities are 3 people working 4 months, 2 people working 6 months, or 6 people working2 months. All of these activities require 12 man-months of labor:

Amount of labor = 3 people× 4 months= 2 people× 6 months= 6 people× 2 months= 12 man-months.

In physics, the term work has a very special meaning. One unit used to measure work is thefoot-pound, or ft-lb. This is the amount of work required to lift a 1-lb weight vertically through adistance of 1 ft. A formula for calculating work is given by

Work = distance× weight.

Notice that in the formula, we are multiplying a distance (measured in feet) by a weight (measuredin pounds) to find work (measured in foot-pounds).

Example 7 Find the amount of work required to lift a 3-lb weight a distance of 4 feet.

Solution The amount of work we do is given by 3 lbs × 4 ft = 12 ft-lbs. This is the same amount of workrequired to lift a 4-lb weight a distance of 3 ft, or a 2-lb weight a distance of 6 ft, or a 24-lb weighta distance of 0.5 ft.

Dividing Different Units: RatesWhen we calculate the speed of a moving object, we divide the distance it travels by the time ittakes.

Example 8 You travel 120 miles in 2.5 hours at a constant speed. How fast are you going?

SolutionSpeed =

DistanceTime

=120 miles2.5 hours

= 48 miles per hour.

Notice that in the last example it makes sense to divide units of distance (miles) by units oftime (hours). The resulting ratio is a unit for speed: miles per hour or mph. Other common units forspeed include feet per second, meters per second, and kilometers per hour.

A rate expresses how one quantity changes with respect to another quantity. For example, thespeed of a moving object tells you how fast its distance traveled changes with respect to time. Gasmileage is another example of a rate.

Example 9 A car travels 160 miles on 5 gallons of gas. What is its gas mileage?

Solution We compute the ratio160 miles traveled

5 gallons of gas consumed= 32 miles per gallon (mpg).

Gas mileage is a rate that measures the number of miles the car can go on a gallon of gas. In thisexample, the units of gas mileage are miles per gallon.


Rates often appear in algebraic expressions.

Example 10 A car travels at a constant speed of 60 mph. Write an expression for the distance d, in miles, the cartravels in terms of the number of hours t the trip takes. If the trip lasts 4 hours, how far does the cartravel?

Solution We knowDistance = Rate× Time.

Since the rate is a constant 60 mph, we have

Distance = 60× Timed = 60t.

If t = 4, we have d = 60(4) = 240 miles.

Example 11 Before the invention of the compact disc (CD), vinyl long-playing records (vinyl LPs) were com-monplace. By design, an LP revolves on a record player 33 1

3 times per minute. Write an expressionfor the number of revolutions in terms of the time elapsed.

Solution The rate is 33 13 revolutions per minute, so

Revolutions = Revolutions per minute×Minutes = 331

3×Minutes.

If N is the number of revolutions and t is the number of minutes, then

N = 331

3t.


In Exercises 1–6, pick units for each measurement from thefollowing list: cm, cm2, cm3.

1. Area of circle 2. Perimeter of circle

3. Volume of sphere

4. Surface area of closed cylinder

5. Surface area of open cylinder

6. Length of the equator drawn on sphere

In Exercises 7–10, give the appropriate units if(a) Length is measured in meters, time in seconds, and mass

in grams

(b) Length is measured in inches, time in hours, and mass inpounds.

7. Area 8. Volume 9. Speed 10. Work

11. If sand costs $15 a cubic yard how much does it cost acubic foot?

12. If distance, d, is in cm, speed, v, is in cm/sec, and time,t, is in sec, what are the units and the meaning of thefollowing?

(a) vt (b)d

t(c)

d

v(d)

v

t

Problems

13. If n birds eating continuously consume V in3 of seed inT hours, how much does one bird consume per hour?

14. If n birds eating continuously consume W ounces ofseed in T hours, what are the units of W/(nT )? Whatdoes W/(nT ) represent in practical terms?

REVIEW EXERCISES AND PROBLEMS FOR CHAPTER ONE 31

In Problems 15–19, use units to interpret the practical mean-ing of the expression, where the variables refer to a truckdriven at a constant speed of 70 mph for d miles, and3

m = mileage in miles per gallon

f = fuel cost per gallon in dollars

w = driver’s wages per hour in dollars

c =operating cost per mile in dollars,excluding fuel cost and driver’s wages.

15. dc 16. d/70 17. dw/70 18. df/m

19. df/m+ dw/70 + dc

20. John has to run m miles then swim s meters. Which offollowing expressions could give the time in minutes thatit takes him to do this?

(a) 60m+ 0.01s (b) 7m+ s

(c) 7m+ 0.02s (d) 60 + 7m+ 0.1s

21. Mary has one hour in which to run m miles and swim smeters. Which of the following expressions could repre-sent the number of minutes she has left after finishing therunning portion?

(a) 7m+ 0.02s (b) 60− 7m

(c) 60 + 0.2s (d) 60− 7m+ 0.2s

22. A 200 gallon container contains 100 gallons of water. Attime t = 0 minutes, water is added to the container at therate of 5 gallons per minute, but drains out at the rate of 1gallon per minute. How much water is in the container af-ter t minutes? Is the container emptying or filling? Whenis the container empty/full? What happens after that?

23. The weight of one cubic foot of gasoline is 58 pounds,while the weight of one cubic foot of motor oil is 78pounds. If a shipment of gasoline and motor oil containsx cubic feet of gasoline and y cubic feet of motor oilwrite an expression for the total weight in the shipment.If shipping costs are calculated at 6 dollars per poundwrite an expression for the total shipping cost.

24. If you mix x pounds of tea costing $5 per pound with ypounds of tea costing $3 per pound, write an expressionthat represents the cost per pound of the mixture.

25. Hamburger costsH dollars per pound, and beans costsBdollars per pound. A Mexican restaurant spends 75H +150B dollars each week on hamburger and beans. Howmuch of each ingredient does it buy each week?

REVIEW EXERCISES AND PROBLEMS FOR CHAPTER ONE

Exercises

In Exercises 1–5, say in words the statement represented bythe equation.

1. 1 + 3r = 15 2. s3 = s2

3. w + 15 = 3w 4. 0.25t = 100

5. 2a− 7 = 5a

Solve each equation in Exercises 6–10.

6. 7r = 21 7. a+ 7 = 21

8. J/8 = 4 9. b− 15 = 25

10. 5 = 1− d

For Exercises 11–14, write an expression that gives the resultof performing the indicated operations on the number x. Youdo not need to simplify your answer.

11. Add 2, double the result, then subtract 4.

12. Divide by 5, subtract 2 from the result, then divide by 3.

13. Subtract 1, square the result, then add 1.

14. Divide by 2, subtract 3 from the result, then add the re-sult to the result of first subtracting 3 from x and thendividing the result by 2.

For Exercises 15–18, which of the given values of the variableis a solution?

15. x2 + 2 = 3x for x = 0, 1, 2.

16. 2x2 + 3x3 = 5x5 for x = −1, 0, 1.

17. t+ 1/(t2 + 1) = (t+ 1)/(t2 + 1) for t = −1, 0, 1.

18. 2(u−1) + 3(u−2) = 7(u−3) for u = 1, 2, 3, and 6.5.

In Exercises 19–23, explain what operation can be used totransform the first equation into the second equation.

19. 2x+ 5 = 13 and 2x = 8

20. x− 11 = 26 and x = 37

21. x/2 = 40 and x = 80

22. (2y)/3 = 20 and y = 30

23. t/2 = (t+ 1)/4 and 2t = t+ 1

3Adapted from Cliff Sloyer, Fantastiks of Mathematiks (Rhode Island: Janson, 1986).


Establish the identities in Exercises 24–25.

24. 12(a+ b) + 1

2(a− b) = a

25. 12(a+ b)− 1

2(a− b) = b

In Exercises 26–29, which of the values (if any) are solutionsto the given equation?

26. 2(r − 6) = 5r + 12, for r = 8,−8

27. n2 − 3n = 2n+ 24, for n = 8,−3

28. n2 − 3n = 2n− 24, for n = −8, 3

29. s3 − 8 = −16, for s = −2, 2

In Exercises 30–35, identify each description as either an ex-pression or equation and write it algebraically using the vari-ables given.

30. Twice n plus three more than n.

31. Twice n plus three more than n is 21.

32. The combined salary of Jason, J , and Steve, S.

33. Twice the combined salary of Jason, J , and Steve, S, is$140,000.

34. 225 pounds is ten pounds more than Will’s weight, w.

35. 50 pounds less than triple Bob’s weight, w.

Problems

36. Find the expression for the volume of a rectangular solidin terms of its width w where the length l is twice thewidth and the height h is five more than the width.

37. A company uses two different sized trucks to deliversand. The first truck can transport x cubic yards, and thesecond y cubic yards. The first truck makes S trips toa job site, while the second makes T trips. What do thefollowing expressions represent in practical terms?

(a) S + T(b) x+ y(c) xS + yT(d) (xS + yT )/(S + T )

38. A company out-sources the manufacturing of widgets totwo companies, A and B, which supply a and b widgetsrespectively. However, 10% of a and 5% of b are defec-tive widgets. What do the following expressions repre-sent in practical terms?

(a) a+ b(b) a/(a+ b)(c) 10a+ 5b(d) 0.1a(e) 0.1a+ 0.05b(f) (0.1a+ 0.05b)/(a+ b)(g) (0.9a+ 0.95b)/(a+ b)

39. (a) Doesx

4+

1

2= 3x have the same solution as

x+ 2 = 12x?(b) Is

x

4+

1

2equivalent to x+ 2?

40. (a) Does5x

3+2 = 1 have the same solution as 5x+6 =

1?(b) Is

5x

3+ 2 equivalent to

5x+ 6

3?

41. Which of the following is not the result of a valid opera-tion on the equation 3(x− 1) + 2x = −5 + x2 − 2x?

(a) 5x− 3 = x2 − 2x− 5

(b) 4x+ 3(x− 1) = −5 + x2

(c) 3(x− 1) = −5 + x2 − 2x+ 2x

(d) 3(x− 1) + 2x− 5 = −10 + x2 − 2x

42. You plan to drive 300 miles at 55 miles per hour, stop-ping for a two-hour rest. You want to know t, the numberof hours the journey is going to take. Which of the fol-lowing equations would you use?

(A) 55t = 190 (B) 55 + 2t = 300

(C) 55(t+ 2) = 300 (D) 55(t− 2) = 300

43. (a) Plot the values of 7 + 5q for q = 0, 1, 2, 3, 4.(b) Use your graph to determine at which q-value 7+5q

is equal to 17.

44. (a) Plot the values of 2b2 − b3 for b = 0, 1, 2, 3, 4.(b) Use your graph to determine at which b-values 2b2−

b3 = 0.

45. The number of dirty socks on your roommate’s floor, tdays after the start of exams, is given by 10 + 2t.

(a) Write an equation whose solution is the number ofdays it takes for the number of socks to reach 26.

(b) Make a plot of the number of socks for t =2, 4, 6, 8, 10, 12, and indicate the solution t = 8 tothe equation in part (a).

46. For accounting purposes, the value of a machine, t yearsafter it is purchased, is given by $100,000− 10,000t.

(a) Write an equation whose solution is the numberof years it takes for the machine’s value to reach$70,000.

(b) Make a plot of the value of the machine at t =1, 2, 3, 4, 5, 6, and indicate the solution t = 3 to theequation in part (a).

REVIEW EXERCISES AND PROBLEMS FOR CHAPTER ONE 33

47. The amount of glass used in a door is 4500 + w2 cm2,where w cm is the width of a window in the door. Fig-ure 1.13 shows the graph of this expression.

(a) From the graph, estimate the width of the window ifthe total glass used is 7000 cm2.

(b) Check your answer to part (a) by substituting it intothe equation 4500 + w2 = 7000.

50 100

4500

7000

9500

12,000

14,500

w (cm)

Amount of glass (cm2)

Figure 1.13

48. The population of bacteria in some rotting vegetables af-ter t days is given by 2000(2)t, as shown in Figure 1.14.

(a) From the graph, estimate how many days it takes forthe population to reach 32,000.


2000(2)t = 32,000.

1 2 3 4 5

16,000

32,000

48,000

64,000

t

population

Figure 1.14

49. A tank contains 20−2t cubic meters of water, where t isin days.

(a) Construct a table showing the number of m3 of waterat t = 0, 2, 4, 6, 8, 10.

(b) Use your table to determine when the tank(i) Contains 12 m3

(ii) Is empty

50. A vending machine contains 40 − 8h bags of chips hhours after 9 am.

(a) Construct a table showing the number of bags ofchips in the machine at h = 0, 1, 2, 3, 4, 5, 6.

(b) How many bags of chips are in the machine at 9 am?(c) At what time does the machine run out of chips?(d) Explain why the number of bags you found for h =

6 using 40 − 8h is not reasonable. How many bagsare probably in the machine at h = 6?

51. If a solution containing 3 pounds of salt per gallon flowsinto a tank at the rate of 5 gallons per minute, how manypounds of salt flow into the tank per minute?

52. If a solution containing x pounds of salt per gallon isflowing into a tank at the rate of y gallons per minute,what does the quantity xy represent?

53. Someone says that the surface area S of a can of heighth and radius r is 2πr2 + πr2h. How can you tell usingunits that this formula cannot be correct?

54. In the summer of 2002, the cost of gas was about 0.74£/liter in the UK, and about 1.40 $/gallon in the US. Ex-plain what additional information is needed to comparethese prices. How would you then compare prices?

For Problems 55–57, assume that movie tickets cost $p foradults and $q for children.

55. Write expressions for the total cost of tickets for:

(a) 2 adults and 3 kids (b) No adults and 4 kids

(c) No kids and 5 adults (d) A adults and C kids

56. Rework Problem 55, this time giving expressions for theaverage cost per ticket.

57. A family of two adults and three children has an enter-tainment budget equal to the cost of 10 adult tickets. Anadult ticket costs twice a child ticket. How much moneywill they have left after seeing two movies? Can they af-ford to see a third movie?

58. It costs a contractor $p to employ a plumber, $e to em-ploy an electrician, and $c to employ a carpenter.

(a) Write an expression for the total cost to employ 4plumbers, 3 electricians, and 9 carpenters.

(b) Write an expression for the fraction of the total costin part (a) that is due to plumbers.

(c) Suppose the contractor hires P plumbers, E electri-cians, and C carpenters. Write expressions for thetotal cost for hiring these workers and the fraction ofthis cost that is due to plumbers.


59. A contractor is managing three different job sites. It costsher $p to employ a plumber, $e to employ an electrician,and $c to employ a carpenter. The total cost to employplumbers, carpenters, and electricians at each site is givenby

Cost at site 1 = 12c+ 2p+ 4e

Cost at site 2 = 14c+ 5p+ 3e

Cost at site 3 = 17c+ p+ 5e.

Write expressions in terms of c, p, and e for:

(a) The total employment cost for all three sites.(b) The difference between the employment cost at site

1 and site 3.(c) The amount remaining in the contractor’s budget af-

ter accounting for the employment cost at all threesites, given that originally the budget is 50c+10p+20e.

60. If m is the number of males and f is the number of fe-males in a population P , which of the following formulasexpresses the fact that 47% of the population is male and53% of the population is female?

(a) P = 0.47m+ 0.53f(b) P = 0.53m+ 0.47f(c) m = 0.47P and f = 0.53P(d) P = 0.47m and P = 0.53f

61. Weight Watchers c© assigns points to various foods, andlimits the number of points you can accumulate in a day.If c is the number of calories, g the grams of fat, and fthe grams of dietary fiber, then the number of points fora piece of food is

c

50+

g

12− f

4.

Which of the following statements can you concludefrom this formula?

(a) One fiftieth of a calorie has the same number ofpoints as one twelfth of a gram of fat.

(b) You can trade one gram of fiber for three grams offat.

62. Here are the line-by-line instructions for calculating thededuction on your federal taxes for medical expenses.

1 Enter medical expenses

2 Enter adjusted gross income

3 Multiply line 2 by 7.5% (0.075)

4Subtract line 3 from line 1. If line 3 ismore than line 1, enter 0.

Write an expression for your medical deduction (line 4)in terms of your medical expenses, E, and your adjustedgross income, I .

63. Antonio and Lucia are both driving through the desertfrom Tucson to San Diego, which takes each of them 7hours of driving time. Antonio’s car starts out full with 14gallons of gas and uses 2 gallons per hour. Lucia’s SUVstarts out full with 30 gallons of gas and uses 6 gallonsper hour.

(a) Construct a table showing how much gas is in eachof their tanks at the end of each hour into the trip. As-sume neither stops for gas until just before the tankis empty, and then the tank is filled.

(b) Use your table to determine when they have thesame amount of gas.

(c) If they drive at the same speed while driving andonly stop for gas, which of them gets to San Diegofirst?

(d) Now suppose that between 1 hour and 6.5 hours out-side of Tucson, all of the gas stations are closed un-expectedly. Does Antonio arrive in San Diego? DoesLucia?

(e) The amount of gas in Antonio’s tank after t hours is14 − 2t gallons, and the amount in Lucia’s tank is30− 6t gallons. When does

(i) 14− 2t = 30− 6t?

(ii) 14− 2t = 0?

(iii) 30− 6t = 0?

A government buys x fighter planes at f dollars each, and ytons of wheat at w dollars each. It spends a total of B dollars,where B = xf + yw. In Problems 64–66, write an equationwhose solution is the given value.

64. The number of tons of wheat it can afford to buy if itspends a total of $100 million, wheat costs $300 per ton,and it must buy 5 fighter planes at $15 million each.

65. The price of fighter planes if it bought 3 of them,10,000 tons of wheat at $500 a ton, and spent a total of$50,000,000.

66. The price of a ton of wheat if it buys 20 fighter planesand 15,000 tons of wheat for a total expenditure of$90,000,000, and a fighter plane costs 100,000 times aton of wheat.

In Problems 67–69, decide if the statement is true or false.Justify your answers using algebraic expressions.

67. The sum of three consecutive integers is a multiple of 3.

68. The sum of three consecutive integers is even.

69. The sum of three consecutive integers is three times themiddle integer.

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The Basic Ideas of Algebra - COPYRIGHTED MATERIAL