1. Key Stone Problem… Key Stone Problem… next Set 13 © 2007 Herbert I. Gross

2. You will soon be assigned problems to test whether you have internalized the material in Lesson 13 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

3. 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

4. next © 2007 Herbert I. Gross If… 3y = 2x + 12 …what is the rate of change of y with respect to x? Keystone Problem for Lesson 13

5. next © 2007 Herbert I. Gross Solution for Problem 1 Let's begin by observing that when we said that m denoted the rate of change of y with respect to x, it was assumed that the linear relationship was written in the form… Notice that the formula 3y = 2x + 12 doesn't have this form because the left side of the formula is 3y rather than y. y = mx + b next

6. However, we can paraphrase the form 3y = 2x + 12 into the form y = mx + b by dividing both sides of our equation by 3; and since dividing by 3 is the same as multiplying by 1 / 3, we see that… © 2007 Herbert I. Gross or… next 3y = 2x + 12 next 1/31/3 1/31/3 () ( ) y = 2x + 12 1/31/3 1/31/3 ()() or… y = 2 / 3 x + 4

7. next © 2007 Herbert I. Gross y = 2 / 3 x + 4 is in the “y = mx + b” form with m = 2 / 3 and b = 4. The rate of change of y with respect to x is 2 / 3 y's per x. Answer for Problem 1 next

8. Since “per”, when it occurs between two nouns, can be replaced by “ ÷ ” we may replace the fractional rate… © 2007 Herbert I. Gross next by 2/3 y ÷ x; and then multiply both sides of this quotient by 3 to obtain…2 Note 1 y’s per x 2323 or… 2y’s per 3x’s next 2y’s ÷ 3x

9. With respect to the above note, most merchants use this form of representing a fractional rate. That is, it is more likely that an advertised price will be “2 pounds for (per) $3” rather than “ 2 / 3 of a pound per dollar”. next © 2007 Herbert I. Gross

10. As a check, notice that since “2 dollars for 3 pounds” means the same thing as 2 pounds ÷ 3 dollars, we may rewrite this rate in the form… next © 2007 Herbert I. Gross Thus, we see that the cost y, in dollars, of x pounds is given by the direct proportion y = 2 / 3 x, which has the form y = mx, where m = 2 / 3. 2 dollars 3 pound = 2323 dollars pound dollars per pound 2323 = next

11. To see how the formula y = 2 / 3 x + 4 is related to this discussion, notice that if the price was 3 pounds for $2 plus a $4 charge for shipping and handling, the formula for the cost in dollars of x pounds would now be given by… © 2007 Herbert I. Gross …and if we now multiply both sides of the form y = 2 / 3 x + 4 by 3 we obtain… next y = 2 / 3 x + 4 33()() or… 3y = 2x + 12 next

12. This discussion can be applied to any fractional rate. For example, if you feel uncomfortable dealing with a rate such as 14 / 15 miles per minute, rewrite the rate in the form… © 2007 Herbert I. Gross Psychologically, probably because it involves only whole numbers, the phrase “14 miles per 15 minutes" seems less threatening to most people than the phrase “ 14 / 15 miles per minute". next 14 15 = 14 miles 15 min 14 miles per 15minutes = miles min ( )

13. To go one step further, since there are four 15 minute segments in an hour, 14 miles per 15 minutes means the same thing as 56 miles per hour. Most people have an easier time visualizing a speed of 56 miles per hour than a speed of 14 miles per 15 minutes. next © 2007 Herbert I. Gross In a similar way, if we write y = 6 / 11 x + 5, it is sometimes more convenient to interpret the fractional rate m = 6 / 11 as meaning that y increases by 6 every time x increases by 11 rather than by saying that the rate of change of y with respect to x is 6 / 11. next

14. There is a tendency to overlook the 3 in the formula… 3y = 2 x + 12 © 2007 Herbert I. Gross …and confuse it with the formula y = 2 x + 12; thus leading to the incorrect result that m = 2 and b = 12. So we have to remember that if we are going to use the result that the rate of change of y with respect to x is m, the equation has to be in the form y = mx + b. That's why we divided both sides of 3y = 2x + 12 to arrive at the form… y = 2 / 3 x + 4. next

15. However, we didn't have to convert 3y = 2x + 12 to y = 2 / 3 x + 4 other than as a matter of convenience. For example, if we replace x in our first formula by, say, 3, 6 or 9 we would see that… © 2007 Herbert I. Gross next 3y = 2(3) + 12 = 6 + 12 = 18; or y = 6 33()() 3y = 2(6) + 12 = 12 + 12 = 24; or y = 8 3y = 2(9) + 12 = 18 + 12 = 30; or y = 10 next

16. And we see from the forms above that when x increases by 3, y increases by 2. © 2007 Herbert I. Gross When x = 3; y = 6. next 3y = 2(3) + 12 = 6 + 12 = 18; or y = 6 33()() 3 next 6 3y = 2(6) + 12 = 12 + 12 = 24; or y = 8 68 3y = 2(9) + 12 = 18 + 12 = 30; or y = 10 910 When x = 6; y = 8. When x = 9; y = 10. next

17. © 2007 Herbert I. Gross In the above discussion we replaced x by multiples of 3, so that m would be a whole number when we divided both sides of the formula 3y = 2x + 12 by 3. next 3y = 2(3) + 12 = 6 + 12 = 18; or y = 6 33()() 3y = 2(6) + 12 = 12 + 12 = 24; or y = 8 3y = 2(9) + 12 = 18 + 12 = 30; or y = 10 3 6 9