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The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard.

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Presentation on theme: "The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard."— Presentation transcript:

1 The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard A. Medeiros next Lesson 15

2 Solving Linear Equations Solving Linear Equations © 2007 Herbert I. Gross next Part 1 (equations of the form mx + b = nx + c)

3 next ▲ There are often times when we want to equate two (or more) linear relationships. © 2007 Herbert I. Gross For example, suppose we want to order candy from a candy company's catalog. One company’s catalog advertises candy at a price of $3 per pound, but charges a service fee of $20; and a second company's catalog advertises the same candy at a price of $5 per pound, but charges a service fee of only $4. next

4 We want to compare the prices to see which catalog offers us the better deal. As we shall soon see, which is the better deal depends on how many pounds of candy we buy. © 2007 Herbert I. Gross

5 next © 2007 Herbert I. Gross For example Suppose you want to buy 2 pounds of candy. ▲ At $3 per pound the first company charges you $6 for the candy; but with the $20 service charge, your total cost for the 2 pounds of candy is $26. ▲ At $5 per pound the second company charges you $10 for the 2 pounds of candy; but with the $4 service charge, your total cost is only $14. So by using the second company you save $12. next

6 © 2007 Herbert I. Gross Suppose however, that you want to buy 12 pounds of candy. ▲ The first company charges you $36 for the candy; and with the $20 service charge, your total cost for the 12 pounds of candy is $56. ▲ The second company charges you $60 for the 12 pounds of candy; and even though the service charge is only $4, your total cost for the 12 pounds of candy is $64. So this time, by using the first company, you save $8. next

7 © 2007 Herbert I. Gross Therefore, we see that up to a certain number of pounds it is cheaper to pay more per pound but less for the service charge. After that, it is cheaper to pay more for the service charge but less per pound. next Definition The number of pounds at which the cost is the same regardless of which catalog you order from is called the equilibrium point.

8 next © 2007 Herbert I. Gross By way of illustration, in this example the equilibrium point is 8 pounds. next Namely, if we order 8 pounds using the first catalog we pay $24 for the candy and $20 for the service charge, which is a total cost of $44. If we order 8 pounds using the second catalog, we pay $40 for the candy and $4 for the service charge, which is also a total cost of $44.

9 © 2007 Herbert I. Gross next What the equilibrium point tells us here is that if you buy less than 8 pounds of candy, it is cheaper to order from the second catalog; but if you buy more than 8 pounds of candy, it is cheaper to order from the first catalog. poundsCost/ poundService chargeTotal costStore 7 pounds$3$20$41Catalog 1 8 pounds$3$20$44Catalog 1 9 pounds$3$20$47Catalog 1 poundsCost/ poundService chargeTotal costStore 7 pounds$5$4$39Catalog 2 8 pounds$5$4$44Catalog 2 9 pounds$5$4$49Catalog 2 next

10 © 2007 Herbert I. Gross To see how this discussion is related to our study of linear relationships, notice that using the first catalog, the price (in dollars) for x pounds of candy is… next 3x + 20 …and that using the second catalog, the price (in dollars) is… 5x + 4 So the equilibrium point is the solution of the equation… 3x + 20 = 5x + 4

11 © 2007 Herbert I. Gross In expressions such as 3x + 20 and 5x + 4, we refer to x as being a variable because we are allowed to choose its value in many different (varied) ways. Variables vs. Unknowns However, in satisfying the equation 3x + 20 = 5x + 4 x is no longer a variable. So in solving the above equation for x, we refer to x as being the unknown.

12 next © 2007 Herbert I. Gross The equation above looks similar to the type of linear equation we were solving in our previous lessons. The only difference is that in our previous examples, the unknown always appeared on only one side of the equation: while in this equation, the unknown appears on both sides. Now, recall that we introduced our study of algebra by discussing the power of paraphrasing. That is, we often use the “rules of the game” to replace problems we don't know how to solve by equivalent problems that we already know how to solve. 3x + 20 = 5x + 4

13 next © 2007 Herbert I. Gross So our strategy in the present situation will be to use the “rules of the game” to replace the equation by an equivalent equation in which the unknown appears only on one side of the equation. The basic principle for doing this lies in the fact that in any equation if we add or subtract the same amount from each side of the equation, we obtain an equivalent equation. 3x + 20 = 5x + 4

14 next © 2007 Herbert I. Gross For example, all that keeps the above equation from being a previously solved type of equation is that the unknown, x, appears on both sides of the equation. Our strategy, therefore, will be to use the rules of our game to “get rid of ” either the 3x term on the left-hand side of the equation, or of the 5x term on the right-hand side of the equation. 3x + 20 = 5x + 4

15 next © 2007 Herbert I. Gross Suppose we decide to “get rid of ” the term 3x that appears on the left-hand side of the equation. We notice that the 3x is being added to 20. We already know that, to “undo” adding 3x, we can subtract 3x, to keep the equation “balanced”. If we subtract 3x from one side, we must also subtract 3x from the other side. That is… 3x + 20 = 5x + 4 – 3x – 3x 20 = 2x + 4 next

16 © 2007 Herbert I. Gross This accomplishes our mission! Namely, the equation 20 = 2x + 4 is equivalent to the equation 3x + 20 = 5x + 4, but in the first equation “x" appears only on one side (the right) of the equation. We already know how to solve 20 = 2x + 4. Namely, we first subtract 4 from both sides, obtaining 16 = 2x; we then just divide both sides by 2, obtaining x = 8. A simple check lets us verify that when x = 8, both sides of 3x + 20 = 5x + 4 are indeed equal to 44.

17 next © 2007 Herbert I. Gross Traditionally, the unknown appears on the left-hand side of the equation. Note that we can get the same result by using the arithmetic of signed numbers. Namely, if we had subtracted 5x from both sides of the equation… 3x + 20 = 5x + 4 We would still have obtained the answer… x = 8 next Positive vs. Negative

18 next © 2007 Herbert I. Gross First subtract 5x from both sides of the equation… 3x + 20 = 5x + 4 – 5x – 5x - 2x + 20 = +4 + - 20 = + - 20 - 2x = - 16 Then we just divide both sides of the equation by - 2, and have our answer… - 2 - 2 x = 8 next Then we add - 20 to both sides of the equation…

19 © 2007 Herbert I. Gross In terms of the equilibrium point: it means that if the first catalog advertises a price of $3 per pound plus a $20 service charge, and the second catalog advertises a price of $5 per pound plus a $4 service charge; a purchase of 8 pounds costs $44, no matter which of the two catalogs you use. More generally: you get a better deal using the first catalog if you order more than 8 pounds. However, if you are ordering less than 8 pounds, it is cheaper to order from the second catalog. next

20 © 2007 Herbert I. Gross A non-algebraic way to see why 8 pounds is the equilibrium point is, to notice that the difference in the service charge is $20 – $4 or $16; but each pound you buy from the first catalog is $2 cheaper than from the second catalog. Hence, as soon as you buy 8 pounds from the first catalog, you have saved the $16 difference in the handling charge; and every additional pound you buy from the first catalog saves you an additional $2 than if you had ordered it from the second catalog. next

21 © 2007 Herbert I. Gross In terms of a chart… # of pounds Cost using Catalog 1 (A) A – B 6$38$4 7$41$2 8$44$0 9$47$-2$-2 10$50 $-4 $-4 Cost using Catalog 2 (B) $34 $39 $44 $49 $54 (The negative entries in the A – B column indicate that B is greater than A.) ---------------------------------- next

22 © 2007 Herbert I. Gross While it may be helpful to visualize the equilibrium point specifically in terms of buying candy, we should remember that this type of equation can occur in many practical ways. For example, in economics we are often interested in “supply and demand” problems. For instance, when there is a huge demand for an item that is in short supply, the price of the item is usually quite high. However, if the manufacturer produces too many items they become a glut on the market and the price falls. Therefore, the manufacturer looks for the equilibrium point to balance the demand and supply.

23 next © 2007 Herbert I. Gross Therefore, rather than depend on one interpretation, it might be better to view 3x + 20 = 5x + 4 more generally in terms of our “program” format. That is, we may view the expression 3x + 20 as being Program #1, and we may view the expression 5x + 4 as Program #2; where… Program #1 1Input xx 2Multiply by 33x 3Add 203x + 20 4Output is y3x + 20 Program #2 1Input xx 2Multiply by 55x 3Add 45x + 4 4Output is y5x + 4 next

24 © 2007 Herbert I. Gross In terms of Programs #1 and #2, the fact that the equilibrium point of the equation 3x + 20 = 5x + 4 is x = 8 means that the only time both programs will give the same output is when the input (x) is 8. Moreover, whenever the input is less than 8, the output in Program #1 will be greater than the output in Program #2; but whenever the input is greater than 8, the output in Program #2 will be greater than the output in Program #1. next Summary

25 next © 2007 Herbert I. Gross The method for solving the equation 3x + 20 = 5x + 4 can be used to solve the general linear equation… To see what this method is, let’s write the linear equation in general terms and go through the various steps we used in solving 3x + 20 = 5x + 4. next mx + b = nx + c

26 © 2007 Herbert I. Gross For example, we could begin by subtracting nx from both sides of the equation mx + b = nx + c to obtain the new equation… next mx + b = nx + c – nx – nx mx – nx + b = + c or… (m – n)x + b = + c

27 © 2007 Herbert I. Gross We may then subtract b from both sides of the equation to obtain… next (m – n)x + b = + c – b – b (m – n)x = c – b Finally, we may divide both sides of the equation by m – n to obtain… m – n next

28 © 2007 Herbert I. Gross The above equation expresses the general solution for the equation mx + b = nx + c in terms of the constants, m, b, n, and c. x = c – b m – n

29 next © 2007 Herbert I. Gross In our example, 3x + 20 = 5x + 4; x = c – b m – n x = c – b m – n 3 5 20 4 and c = 4m = 3,n = 5,b = 20, = - 16 - 2 = 8 next

30 © 2007 Herbert I. Gross The only time we can't get from the equation (m – n)x = c – b to the equation x = (c – b) / (m – n) is when m – n = 0 because we are not allowed to divide by 0. However, the only way in which m – n can equal 0 is if m = n. What happens when m = n is discussed in Lesson 16. Important Observation

31 next © 2007 Herbert I. Gross In this Lesson, we have only looked at cases in which m ≠ n. Thus, the only “tricky“ thing has been that one or both sides of the linear equation may not already have been in the standard “mx + b" form. In such cases, we use the techniques discussed in our previous lessons to paraphrase the linear equation into a form in which both sides of the equation have the “mx + b" form.


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