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

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. next © 2007 Herbert I. Gross

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 next © 2007 Herbert I. Gross next

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. © 2007 Herbert I. Gross 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. next

© 2007 Herbert I. Gross 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? next Keystone Problems for Lesson 12

next © 2007 Herbert I. Gross 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)] next

© 2007 Herbert I. Gross Solution for Problem 1a However, using our rules for algebra, the expression 5[4x – 2(6 – x)] can be rewritten as… next 5[4x – 2(6 – x)] 5[4x + - 2(6 + - x)] 5[4x + - 2(6) + - 2( - x)] 5[4x x] 5[6x ] 5[6x] + 5[ - 12] 30x = 30x – 60

In summary: the relationship expressed by y = 5[4x – 2(6 – x)] is equivalent to the relationship… next © 2007 Herbert I. Gross The expression 30x has the form y = mx + b with m = 30 and b = next y = 30x 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”. next

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… © 2007 Herbert I. Gross Note 1a y = 30x (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. next © 2007 Herbert I. Gross Note 1a In other words, the first 2 items you buy are pre-paid (taken care of by your credit balance). next 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.

© 2007 Herbert I. Gross next x4x6 – x2(6 – x)4x – 2(6 – x) 5[4x – 2(6 – x)] (= y) next 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 Numbers © 2007 Herbert I. Gross * Notice the role of signed numbers. By the “add the opposite rule”; 28 – - 2 = 30. next x4x6 – x2(6 – x)4x – 2(6 – x) 5[4x – 2(6 – x)] (= y) next next 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 between x and y is linear, we can write the relationship either in the form y = mx + b, or in the form x = ny + c. © 2007 Herbert I. Gross For example in the present situation, using symmetry, we may replace y = 30x – 60 by... next 30x – 60 = y next 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. next *

Then we may divide both sides of this equation by 30 (that is, multiply both sides by 1/30) to obtain… © 2007 Herbert I. Gross next 30x = y + 60 next 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 () ( )

In more concrete terms; a rate of 30 ounces for $1 is the same rate as $1 for 30 ounces. next © 2007 Herbert I. Gross 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. next

© 2007 Herbert I. Gross 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 words our original equation can be replaced by the following… next 5[4x – 2(6 – x)] = x – 60

next © 2007 Herbert I. Gross 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… next x = 16 30x = x – 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… next Note 1b 5[4(16) – 2(6 – x)] 5[64 – 2 ( – 10)] 5[ ] 5[84] 420 next 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. next © 2007 Herbert I. Gross 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. next

For example, in part (a) of this exercise we obtained the chart… © 2007 Herbert I. Gross next x6 – x5[4x– 2(6 – x)] (=y)4x2(6 – x)4x – 2(6 – x) 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. next © 2007 Herbert I. Gross next x6 – x5[4x – 2(6 – x)] (=y)4x2(6 – x)4x – 2(6 – x) – - 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. next © 2007 Herbert I. Gross 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. next