Key Stone Problems… Key Stone Problems… next © 2007 Herbert I. Gross Set 12

Key Stone Problems…Key Stone Problems…

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© 2007 Herbert I. Gross

Set 12

You will soon be assigned problems to test whether you have internalized the material

in Lesson 12 of our algebra course. The Keystone Illustrations below are

prototypes of the problems you’ll be doing. Work out the problems on your own.

Afterwards, study the detailed solutions we've provided. In particular, notice that several different ways are presented that could be used to solve each problem.

Instructions for the Keystone Problems

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As a teacher/trainer, it is important for you to understand and be able to respond

in different ways to the different ways individual students learn. The more ways

you are ready to explain a problem, the better the chances are that the students

will come to understand.

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In this lesson we discussed the fact that if the relationship between x and y was

linear, it meant that the rate of change of y with respect to x was constant; and that if the relationship was expressed in the

form y = mx + b, then m denoted the rate of change of y with respect to x.

Preface -- Review and Preview

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However, there are times when although the relationship is linear, it is not

expressed in the form mx + b. In such cases the “rules of the game” for algebra

allow us to rewrite the relationship, transforming it into y = mx + b.

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In the next lesson, we will discuss this in more detail, but for now we want to illustrate the above remarks with

a specific example.

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1(a) What is the rate of change of y with respect to x if…

y = 5[4x – 2(6 – x)]?

1(b) For what value of x does…

5[4x – 2(6 – x)] = 420?

nextKeystone Problems for Lesson 12

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Solution for Problem 1a

To begin with, we have been identifying linear with the form y = mx + b and so

the expression…

…doesn't “look” as though it represents a linear relationship.

y = 5[4x – 2(6 – x)]

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Solution for Problem 1a

However, using our rules for algebra, the expression 5[4x – 2(6 – x)] can be

rewritten as…

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5[4x – 2(6 – x)]

5[4x + -2(6 + -x)]

5[4x + -2(6) + -2(-x)]

5[4x + -12 + 2x]

5[6x + -12]

5[6x] + 5[-12]

30x + -60 = 30x – 60

In summary: the relationship expressed by y = 5[4x – 2(6 – x)] is equivalent to the

relationship…

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The expression 30x + -60 has the form y = mx + b with m = 30 and b = -60.

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y = 30x + -60

And since m represents the rate of change of y with respect to x, we see

that the rate of change of y with respect to x is an increase of 30 “y’s per x”.

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The answer to part (a) may seem easier to visualize in terms of an application of the relationship. Suppose you are buying x items at a cost of $30 each and that you

have a credit balance of $60. Then the total cash that you hand over (y) for buying x

items is given by the linear relationship…

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Note 1a

y = 30x + -60

(which is the same as y = 30x – 60).

In this case, every time x increases by 1 (that is, you buy 1 more item) y increases by $30 (that is, the cost per item is $30). The fact that b = -60 reflects the fact that you are owed $60.

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Note 1a

In other words, the first 2 items you buy are pre-paid (taken care of by your credit balance).

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But if you are still not quite comfortable with the algebraic approach, you could make a chart and see what happens for various values of x.

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x 4x 6 – x 2(6 – x) 4x – 2(6 – x) 5[4x – 2(6 – x)] (= y)

0 0 6 12 -12 -60

1 4 5 10 -6 -30

2 8 4 8 0 0

3 12 3 6 6 30

4 16 2 4 12 60

5 20 1 2 18 90

6 24 0 0 24 120

nextnextnextnextnextnext In the form y = 5[4x – 2(6 – x)], it is not apparent that the rate of change of y with respect to x is 30. However, to verify that this is the case

we could have made a chart like this...

A Note on Signed Numbersnext


*Notice the role of signed numbers. By the “add the opposite rule”; 28 – -2 = 30.

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x 4x 6 – x 2(6 – x) 4x – 2(6 – x) 5[4x – 2(6 – x)] (= y)

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7 28 -1 -2 30 150

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This is an example of why the arithmetic of signed numbers is very important in the

study of algebra.

If we were to continue the chart, the next row would be…

*

Notice that if the relationship betweenx and y is linear, we can write the relationship either in the form y = mx + b, or in the formx = ny + c.

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For example in the present situation, using symmetry, we may replace y = 30x – 60 by...

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30x – 60 = y

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We may then add 60 to both sides of the above equation to obtain…

30x = y + 60

* It’s traditional for the output, in this case x, to be written on the left hand side of the equation.

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Then we may divide both sides of this equation by 30 (that is, multiply both

sides by 1/30) to obtain…

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30x = y + 60

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or

x = 1/30 y + 2

The relationship between 30 in y = 30x – 60 and 1/30 in x = 1/30 y + 2 is that the rate,30y’s per 1x, is the same as the rate

1x per 30y’s.

1/30 1/30 ( ) ( )

In more concrete terms; a rate of 30 ounces for $1 is the same rate as

$1 for 30 ounces.

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So if we knew the value of x and wanted to find the corresponding value of y, we would tend to use y = 30x – 60.

On the other hand, we would tend to use x = 1/30 y + 2 if we knew the value of y and

wanted to find the corresponding value of x.

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Solution for Problem 1b

We want to solve the equation…

In our solution for part (a) we showed that the left hand side of the equation can be

replaced by 30x – 60. In other wordsour original equation can be replaced by

the following…

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5[4x – 2(6 – x)] = 420

5[4x – 2(6 – x)] = 42030x – 60

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Solution for Problem 1b

To solve our equation, we begin by adding 60 to both sides to obtain…

…and we then divide both sides of the equation by 30 to obtain our answer…

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x = 16

30x = 480

30x – 60 = 420

To check that our answer is correct, we

replace x by 16 in 5[4x – 2(6 – x)] = 420 and see if we get a true statement. To this end…

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Note 1b

5[4(16) – 2(6 – x)]

5[64 – 2 (–10)]

5[64 + 20]

5[84]

420

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16

When a linear relationship is written in the y = mx + b form,

it is easy to answer questions that might otherwise be messy to answer.

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The point here is that even when our equation is not in the mx + b form; just knowing that the relationship

between x and y is linear still makes it easy to predict what each row of

the chart should look like.

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For example, in part (a) of this exercise we obtained the chart…

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x 6 – x 5[4x– 2(6 – x)] (=y)4x 2(6 – x) 4x – 2(6 – x)

0 6 -600 12 -12

1 5 -304 10 -6

2 4 08 8 0

3 3 3012 3 6

4 2 6016 4 12

5 1 9020 2 18

6 0 12024 0 24

As we can see from the chart: every time x increases by 1, y will increase by 30.

The chart showed that when x = 6, y = 120. To get to the row in which y will equal 420, we see that y has to increase from 120 to 420, which is an increase of 300. That is: going from one row to the next, y increases by 30.Hence, it will take 300 ÷ 30 or 10 more rows for this to happen. 10 rows after the 6th row is the 16th row.

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x 6 – x 5[4x – 2(6 – x)] (=y)4x 2(6 – x) 4x – 2(6 – x)

16 -10 42064 -20 64 – -20 = 84

As a check, we see that…

Linear relationships are the basic building block of much mathematical analysis.For this reason, in later lessons we will be looking at linear relationships in greater detail.

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The main point to remember: that for y to be linear in x, the rate of change of y with respect to x must be constant; and in this case the rate of change of y with respect

to x is m when the relationship is expressed in the form y = mx + b.

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Key Stone Problems… Key Stone Problems… next © 2007 Herbert I. Gross Set 12