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In This Chapter

^ Writing word problems as algebraic equations

^ Making comparisons of numbers using mathematical operations

^ Finding answers requiring two numbers and choosing which of two numbers is the

Ord problems in algebra are problems that seem to take on a life of their own. You start out with a simple sentence in English and end up with an equation that, you hope, will answer the question that’s been posed.

In this chapter, you see how to use word clues to convert from English sentences to algebraic statements that can be solved and checked. The problems in this chapter aren’t really Themed — they’re all over the place — but they show how to put algebra to its best use.

Basic number problems are like the riddles or brainteasers that your fifth-grade teacher gave you to get you interested in doing numbers in your head. Your teacher may have said: "I’m thinking of a number. If I double it and subtract five, then I get a number that’s two more than the original number." The first member of your class to guess the number got a prize, like a new, spiffy-doodle pencil. In this section, you find number problems like this. (Before starting, you may want to refer back to Chapter 5 for a review of operations and their corresponding words.)

Answer

Changing from Words to math expressions

The word And Is equivalent to addition; Difference Refers to subtraction. It’s the variations on and the subtle use of these words that affect how the mathematical expression is written. Look at the following interpretations.

The subtleties in the wording are important in the interpretation. The implied grouping symbols need to be written when translated into mathematics. The English language can be misinterpreted (in world relations, too, of course), so it’s usually a good idea to use as many descriptive words as necessary to be sure that your mathematical meaning is understood.

Solving equations involving one number

Many algebra word problems involve a single number and some operations performed on it. The problems of this sort are good for practice in writing algebraic equations, because you only need to worry about one unknown, or variable. For simplicity, I’ll use X For the unknown.

The translations from word problems to algebraic problems have two standard rules:

The variable always represents a number.

The sentence is translated directly (in order) letting the verb be replaced by an equal sign.

The sum of three times X And Y

Three times the sum of X And Y

The product of X And four more than Y

Four more than the product of X And Y

Five less than the sum of X And Y

Five minus the sum of X And Y

The sum of four and the square of X

The square of the sum of X And 4

Twice the square of the sum of six and Y

The square of the sum of six and twice Y

3(x + y) X(y + 4) Xy+ 4

3X + Y

(x + y) – 5 5 – (x + y) 4 + X

(x + 4)2 2(6 + y)2 (6 + 2y)2

The Problem: The sum of a number and 12 is 18. What is the number?

^VLA* Let the number be represented by X. The word Sum Implies addition, so add the X And 12. Put an equal sign after the sum and before the 18. The equation you want reads: X + 12 = 18. Solving this linear equation, you subtract 12 from each side to get X = 6. Now, check your answer. Replace A number In the problem with 6, and it now reads: The sum of 6 and 12 is 18. You got it!

The Problem: The sum of a number and 12 is twice the number. What is that number?

^VLA* This problem is different from the one earlier only in the result, or the value on the right side of the equation sign. You write Twice the number Using 2x. So this equation is written: X + 12 = 2x. Subtracting X From each side, you get that 12 = X. So, checking, "The sum of 12 and 12 is twice 12." Well, sure! It just sounded more complicated in math speak.

The Problem: The sum of a number and three more than twice the number is 36. What is the number?

Vjj. VLA/V The operation that’s suggested here is addition. You want the sum of a

Number and Some stuff. You now tackle the expression Three more than twice the number. The words More than Suggests addition, too. So you write Three more than twice a number As 3 + 2X. Now, back to the original sentence, you write X + (3 + 2X) To represent the sum of a number and three more than twice the number. The parentheses aren’t really necessary. I just put them there to emphasize the two different numbers in the sum. Writing the complete equation to replace the complete sentence: X + (3 + 2X) = 36. To solve this equation, you first combine the terms on the left and get 3X + 3 = 36. Now subtract 3 from each side, giving you 3x = 33. Dividing by 3, you get X = 11. Going back to the original problem, you check to see if the sum of 11 and three more than twice 11 is 36. Three more than twice 11 is 25. And the sum of 11 and 25 is, indeed, 36.

The Problem: When five is subtracted from twice a number, the result is two more than the number. What is that number?

»tVLA/V This equation has operations on each side. Letting the unknown number be X, You write Five subtracted from twice the number As 2x – 5. Notice the order of the subtraction. You can’t reverse subtraction and get the same result. Now, for the other side of the equation, you write Two more than the number As 2 + X. You could also have written X + 2, because addition is commutative and can be written in either order. Putting the two parts together, you get the equation 2X - 5 = 2 + X. Add 5 to each side and subtract X From each side to get X = 7. Checking this with the original problem, you see that when 5 is subtracted from twice 7, you subtract 5 from 14 and get 9. Two more than 7 is also equal to 9. It checks. Do you recognize this problem? It’s the example I accused your fifth-grade teacher of giving you (flip back to the first page of this chapter).

How many goats?

Avery and Reid both are raising goats for a 4-H project. Avery says to Reid, "If you let me have one of your goats, then I’ll have twice as many goats as what you have left." Reid then counters

With, "If you let me have one of Your Goats, then we’ll have the same number of goats." How many goats does each have right now?

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Comparing Two Numbers in a Problem

Number problems can contain situations involving one, two, or even more different integers, whole numbers, or fractions. The more numbers you have to solve for, the more interesting the problem becomes. Usually, when more than one number is involved in one of these problems, there’s some sort of relationship between the numbers — some mathematical comparison. Chapter 12 is completely devoted to Consecutive integers, So you’ll find other types of problems here. Other problems requiring that you find two or more solutions are solved with systems of equations. (You’ll find systems in Chapter 17.)

Looking at the bigger, the smaller, and the multiple

Children compare their bicycles or cookies or even their fathers by saying that theirs is bigger or faster or smarter or whatever comes to mind. Mathematical expressions take over where childish comparisons leave off, using operations to combine and compare different numbers.

The Problem: I’m thinking of two numbers. One of the numbers is four less than the other, and their sum is 100. What are the numbers?

^VLA* Select a variable to represent one of the numbers, such as X. Now write an

Expression, using the X, To describe the other number. Even though it may be tempting to let the other number be represented by a new variable, Y, You stay with just one unknown or variable. Because the other number is four less than the first, you can write the other number as X - 4. The sum of the two numbers is 100, so write the sum of X And X - 4 as being equal to 100. Your equation is X + (x – 4) = 100. Remove the parentheses and simplify the expression on the left. 2x – 4 = 100. Now add 4 to each side, and you have 2x = 104. Divide by 2, and X = 52. The other number is four less than 52. So X - 4 becomes 52 – 4 = 48. The two numbers are 52 and 48.

What if you had decided, in the preceding problem, that the XShould be the smaller number, so the larger number is written as X+ 4? Note that XIs four less than X + 4. How will that affect the answer? Try adding the numbers together and solving for X. You get X+ (x+ 4) = 100, 2X+ 4 = 100, 2X= 96, and X= 48. The smaller number is 48, and the number four bigger than that is 52. You get the same two numbers! As long as you write the relationship between the numbers correctly, it really doesn’t matter which of the two you solve for first.

The Problem: Separate the number 20 into two parts so that five times the smaller part plus eight is equal to the larger part. What are the two numbers?

I$\M\l One way to do this problem is to hope that the numbers are whole numbers and make a list or chart, guessing what they may be. The number 20 is small enough that a chart is reasonable. A more efficient way, though, is to use an algebraic equation. You’d definitely choose the equation route if the two numbers added up to 200 or 2,000 instead of a nice, civilized 20.

Let X Be one of the parts. The sum of X And some other number is supposed to be 20. So you can write the other number as 20 – X. (If the sum of two numbers is 20, and one of the numbers is 7, then the other number is 20 – 7 = 13.)

Now you have the two parts of 20, X, And 20 – X. Back to the problem. The problem says to multiply 5 times the smaller part. You choose the smaller part to be X. Yes, you Can Do that! Writing Five times the smaller part plus eight, You get 5X + 8. This sum is equal to the larger part (the other number), so your equation becomes 5X + 8 = 20 – X. To solve the equation, add X To each side and subtract 8 from each side to get 6x = 12. Now divide each side, and you get that X = 2. If the number 2 is one Part Of 20, then the other is 20 – 2 = 18. Does this check with the original problem? Is 5 x 2 + 8 equal to 18? Doing the math, 5(2) + 8 = 10 + 8 = 18. By golly, it works!

The Problem: One number is three less than two times another number, and their sum is 21. What are the numbers?

Choose X To be one of the numbers, and then write the other number using mathematical operations to express how they compare. If the second number is Three less than two times x, You write that as 2x – 3. Notice that the 3 is subtracted from the 2X. So now you have the two numbers, X And 2X - 3. The sum of the two numbers is 21. The equation expressing this is X + (2x – 3) = 21. Combine the terms on the left to get 3X - 3 = 21. Add 3 to each side, and the equation becomes 3X = 24. Divide by 3, and X = 8. The number 8 and another number are supposed to add up to 21. That other number has to be 13. Is the number 13 equal to three less than twice 8? Yes, 2(8) – 3 = 16 – 3 = 13.

Varying the problems with variation

The word Variation Has a special definition in mathematics. When two numbers Vary directly, It means that one of the numbers is a Direct Multiple of the other. The numbers 7 and 21 vary directly, because 21 is three times 7. When two numbers Vary inversely, It means that one number is some multiple of the reciprocal of the other.

If Y Varies directly with X, Then Y = Kx, Where K Is some constant.

If Y Varies inversely with X, Then Y = X, where k is some constant.

The usual situation with variation problems is that you’ll be told that two numbers vary either directly or inversely with a specific result. Then you have to figure out what that particular variation is and how it applies to another number.

Varying directly

When two values vary directly, then one of the numbers is a multiple of the other. You need to determine what the multiplier (the value of the constant, K) Is. Formulas and applications from the sciences often use direct variation.

The Problem: If Y Varies directly with X, And Y Is equal to 20 when X Is equal to 5, then what is the value of Y When X Is 2?

First solve the direct variation equation Y = Kx For the value of K, The constant that is particular to this problem. Substituting the values of Y And X That are given, you get 20 = K5. Dividing each side of the equation by 5, you find that K = 4. Now rewrite the relationship equation as Y = 4X. When X Is 2, you solve for Y By putting 2 into the formula. Y = 4 x -2 = 8. When X Is 2, Y Is 8.

The Problem: If the square of Y Varies directly with the cube of X, And if Y Is equal to 3 when XIs equal to 2, then what is the value of X When Y Is 24?

The variation equation is written Y2 = Kx3, And you solve for K By solving the equation (3)2 = K(2)3. You get that 9 = 8K. Dividing each side of the equation

By 8, you see that K = 9. Now use the relationship Y2 = 9 X3, replacing the Y

With 24, to solve for X. (24) = – g X3, 576 = 9X3. Multiply each side of the

Equation by 8, and the equation becomes 512 = x3. Taking the cube root of 9

Each side, you get that X = 8.

The Problem: The volume of a sphere varies directly with the cube of its radius. If a sphere with a radius of 3 yards has a volume of about 113.10 cubic yards, then what is the volume of a sphere that has a radius of 5 yards?

The variation equation is V = Kr3, Where V Represents the volume of the sphere and R Represents the radius. Using the values given, rewrite the equation as 113.10 = k(3)3 or 113.10 = 27k. Dividing each side by 27, K Is about 4.19. Now the relationship between the volume of a sphere and its radius is written V = 4.19r3. Replacing the radius, r, with 5, you get V = 4.19(5)3 = 4.19(125) = 523.75 cubic yards.

The standard formula for the volume of a sphere is V = 3 Nr3. Letting n be

Approximately 3.1416, then 4 N = 4(3.1416) or about 4.19.

33

Varying inversely

When one number varies Inversely With another, then the reciprocal of the one is some multiple of the other number.

Carnival game

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The Problem: If Y Varies inversely with the square root of X, And if Y Is 9 when X Is 4, then what is the value of Y When X Is 36?

VLAiV Start with the inverse variation equation, Y = -h=. Replace the Y With 9 and

" ~ kk

The X With 4 and solve for K. 9 = YJ, 9 = "2 Multiplying each side by 2, you

Get that K = 18. Now replace the K With 18 and the X With 36 in the new inverse

Variation equation and solve for Y. y = \^ = 18 = 3.

The Problem: The rate of vibration of a taut string varies inversely with the length of the string. If a particular string is 50 inches long, it vibrates 250 times per second. How long is a string that vibrates 100 times per second?

.tVLA/v Using the inversely varying equation V = KL With V Representing vibrations and L Representing length, you solve for K 250 = 5°, 5 = 250 • 50 = 12,500. The value of K Holds for this particular string at different lengths. Now, letting K = 12,500 and V = 100, solve for L. 100 = 12,j500, L = 12,5q° = 125. The length of the string is 125 inches.

Squaring Off Using Quadratic Equations

A quadratic equation has a squared term in it. For instance, x2 + 2x – 3 = 0 is a quadratic equation. An interesting feature of these equations is that many quadratic equations have two completely different solutions. In the equation X2 + 2X - 3 = 0, X Is equal to either +1 or -3. Either number works. Number problems using quadratic equations in their solutions may or may not use both answers. After solving the problem, you go back and look at the original question to see if both answers make sense.

Painting the number 9

Stefanie was hired to paint the numbers on the How many times did Stefanie have to paint the doors of the offices in a new academic building. number 9? There are 100 offices, numbered 1 through 100.

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Doubling your pleasure, doubling your fun

Number problems that need two answers or end up with two solutions have to be checked carefully. The equations used when solving these number problems frequently involve expressions relating one of the numbers to the other number using mathematical operations.

The Problem: The sum of a number and eight times its reciprocal is 6. What are the two numbers?

8. X.

First write expressions for the number and its reciprocal. Letting X Be the number, then its reciprocal is x. Multiplying that reciprocal by 8, 8 ^X Now write an equation that represents the sentence exactly. The word Sum Indicates that the number and its reciprocal are added. The word Is Indicates

8

Where the equal sign goes. The equation you write is X + X = 6. Multiply each term in the equation by X, And the quadratic equation X2 + 8 = 6X Emerges. Subtract 6X From each side to set the equation equal to 0.

When solving a quadratic equation of the form Ax2 + Bx + C = 0, you either factor the expression and set the factors equal to 0 to solve for X, Or you use the quadratic formula. (You’ll find the quadratic formula in the Cheat Sheet.)

The quadratic equation X2 - 6X + 8 = 0 factors into (x - 2)(x - 4) = 0. Setting the factor X - 2 = 0, you get that X = 2. Setting the factor X - 4 = 0, you get the answer X = 4. Each answer needs to be checked. Is the sum of 2 and eight

Times its reciprocal equal to 6? Checking this out, 2 + 8 ^ 2 J = 2 + 2 =

2 + 4 = 6. It’s true, so 2 works. You’ll find that the 4 works, also.

4 + 814

4 + f:

4 + 2 = 6.

The Problem: The sum of two numbers is 7 and the sum of the squares of the two numbers is 29. What are the numbers?

^VLA* You first write the two numbers in terms of the same unknown or variable. If the first number is X, Then the other number is 7 – x. How did I pull the 7 – X Out of my hat? Think about two numbers having a sum of 7. If one of them is 5, then the other is 7 – 5, or 2. If one of them is 3, then the other number is 7 – 3, or 4. Sometimes, when you do easy problems in your head, it’s hard to figure out how to write what you’re doing in math speak. So, if the two numbers are X And 7 – X, Then you have to square each of them, add them together, and set the sum equal to 29. The equation to use is X2 + (7 – X)2 = 29. To solve this equation, you square the binomial, combine like terms, subtract 29 from each side, factor the quadratic equation, and then set each of the factors equal to 0.

When X - 2 = 0, you get that X = 2. Subtracting 7 – 2 for the other number, you get 5. The square of 2 is 4, and the square of 5 is 25. The sum of 4 and 25 is 29, so the 2 works. You don’t have to check the other solution from the quadratic equation, because X - 5 = 0 gives you the number 5. Both solutions work and are just repeats of one another. But sometimes the numbers that appear as the two solutions aren’t the two numbers that answer the problem.

The Problem: The product of two integers is 48, and one of the integers is two less than the other. What are the two numbers?

48

^VLA/V If one of the numbers in the product is X, Then the other number is —.

Writing that one number is two less than the other, you can either write an equation where one of the numbers equals the other minus 2 or where one is equal to the other plus 2. It doesn’t really matter, because you’d be writing a statement making one side two more than the other. One equation you may

48

Use to solve this problem is X = – + 2. Multiply each term by X And solve the equation.

48

X = ~ + 2 X2 = 48 + 2x X2 – 2x – 48 = 0 (X – 8)(X + 6) = 0

If you let X - 8 = 0, you get that X = 8.

The other number is found by dividing 48 by 8, so you get that the other number is 6. And the number 6 is 2 less than 8. The other factor of the equation gives you that X + 6 = 0. The value of X This time is -6. The -6 doesn’t work with the 8, because the product of -6 and 8 is -48, and the two numbers are different by 14, not 2. But what about the -6? If it’s one of the numbers, then dividing 48 by -6 you get -8. The product of -6 and -8 is 48, and the -8 is two less than -6. So there are two different sets of answers, each with the two numbers required.

Guessing the money in the box

A new television game show starts out by eliminating contestants from the audience by having them guess the amount of money in four different boxes. One hundred contestants are shown the four boxes and told to guess which box has $1, which has $5, which has $1,000, and which has $10,000. Those who guess all four boxes correctly get to come on stage for round two.

When the entries are tallied, the game-show host announces that 50 people have none correct, 20 people have one correct guess, and 15 people have two correct. Those who got three correct would be given $100 each, and those who got all four correct would get to come up on stage to compete again. How many people got to come up on stage?

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Disposing of the nonansWers

Quadratic equations may yield two different answers. Sometimes both answers work in a word problem. But in some instances, only one works or neither of the answers actually answers the question. Just because you can solve a quadratic equation for a correct answer to the equation, it doesn’t mean that what you get will solve the original question. The question may be unsolvable.

The Problem: The square of a positive number is six more than five times that number. What is the number?

Tf\Ml Let the number be represented by X. The square of X Is written x2. The expres-‘ 0¥\ Sion Six more than five times the number Is written 6 + 5x. Put an equal sign §^Xt ) Between the two expressions, to represent the Is, And you have X2 = 6 + 5x. ^Imlw Subtracting 6 and 5x from each side, X2 - 5x – 6 = 0 factors into (x – 6)(x + 1) = 0. When X - 6 = 0, you get the answer 6. The number 6 squared is 36, and six more than five times 6 is 6 + 30 = 36. The other factor gives you that X + 1 = 0, or X = -1. You don’t consider this answer, because the problem asks for a positive number. This problem has just the one answer.

The Problem: One positive integer is three more than twice another positive integer, and the difference of their squares is nine. What are they?

^VLA/V The two numbers are represented by X And 3 + 2x. Their squares are X2 And

(3 + 2x)2. Their difference can be written as either X2 - (3 + 2xJ2 or (3 + 2xJ2 – x2. Either works. I’m going to use the second version, because I don’t want to have to distribute the negative sign over the three terms in the square of the binomial. Writing that the difference is equal to 9, (3 + 2X)2 - X2 = 9.

(3 + 2x)2 – X2 = 9 9 + 12x + 4×2 = 9 3×2 + 12x = 0 3x (X + 4) = 0

When you set 3X= 0, you get that X= 0. When X+ 4 = 0, you get that X= -4. Neither of these numbers works in the original problem, because you need a Positive Integer, and neither is positive. This problem has no answer. Even though the quadratic equation has solutions, neither works with the question.

Chapter 12