1. Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 16 By Herbert I. Gross and Richard A. Medeiros next

2. Problem #1 © 2007 Herbert I. Gross next For what value of x does 2(x + 3) + 5x = 7x + 6 ? Answer: Every value of x next

3. Answer: Every value of x Solution for #1: The left-hand side of the equation 2(x + 3) + 5x = 7x + 6 is not in the “mx + b” form. However, we may paraphrase by the methods we used in previous Lessons… that is, by the distributive property we may rewrite the expression… next © 2007 Herbert I. Gross 2(x + 3) + 5x next as… 2x + 6 + 5x

4. Solution for # 1: Then by the associative and commutative properties for addition, we may rewrite the expression 2x + 6 + 5x as… next © 2007 Herbert I. Gross (2x + 5x) + 6 next 7x + 6 or… We can now replace the expression 2(x + 3) + 5x by the expression 7x + 6 on the left-hand side of the equation 2(x + 3) + 5x = 7x + 6 to obtain the identity… next 7x + 6 = 7x + 6

5. Solution for #1: Since the equation 7x + 6 = 7x + 6 is equivalent to the equation 2(x + 3) + 5x = 7x + 6, it means that this equation is also an identity. next © 2007 Herbert I. Gross

6. next Notes on #1 The fact that the equation… 2(x + 3) + 5 x = 7x + 6 is an identity means that the left-hand side of the equation above is simply another way of saying the same thing the right-hand side says. © 2007 Herbert I. Gross If you don’t quite have a feeling yet for what an identity means, try substituting a few values of x into the equation and observe what happens. next

7. Notes on #1 For example, if we elect to replace x by 9 in the equation 2(x + 3) + 5x = 7x + 6, we get the true statements… © 2007 Herbert I. Gross next 2( x + 3 ) + 5( x ) = 7( x ) + 6 2(12) + 45 = 63 + 6 24 + 45 = 63 + 6 69 = 69 99 9 next

8. Notes on #1 If we had instead to replace x by, say, - 6 in the equation 2(x + 3) + 5x = 7x + 6, we would have obtained the true statements… © 2007 Herbert I. Gross next 2( x + 3 ) + 5( x ) = 7( x ) + 6 2( - 3) + - 30 = - 42 + 6 - 6 + - 30 = - 42 + 6 - 36 = - 36 -6-6 -6-6 -6-6 next

9. Notes on #1 We happened to choose 9 and - 6 as replacements for x in the equation 2(x + 3) + 5x = 7x + 6. © 2007 Herbert I. Gross The point is that no matter what values we replaced x by in our equation, the resulting equation would be a true statement. next

10. Notes on #1 In “program” format, the expression 2(x + 3) + 5x may be rewritten as… © 2007 Herbert I. Gross next Program # 1 Step 1Start with x.xStep 2Add 3.x + 3Step 3Multiply by 2.2(x + 3)Step 4Store the result.2(x + 3)Step 5Input x.xStep 6Multiply by 5.5xStep 7Add step 6 to step 4.2(x + 3) + 5xStep 8The output of step 7 is the answer. 2(x + 3) + 5x

11. next Notes on #1 What we have shown in this exercise is that Program #1 is equivalent to the simpler program… © 2007 Herbert I. Gross next Program # 2 Step 1Start with x.xStep 2Multiply by 7.7xStep 3Add 6.7x + 6Step 4The output of step 3 is the answer. 7x + 6

12. next Notes on #1 In summary… We may always replace the more complicated Program #1 by the simpler Program #2. © 2007 Herbert I. Gross next That is, for each input, the output we get in Program #1 is equal to what the output would be for the same input in Program #2.

13. Problem #2 © 2007 Herbert I. Gross next For what value of x does 2(x + 3) + 5x = 7x + 9 ? next Answer: No value of x

14. Solution for #2: We have already seen in our solution for Problem #1 that the left side of the equation 2(x + 3) + 5 x = 7x + 9 is equivalent to the simpler expression 7x + 6. next © 2007 Herbert I. Gross 2(x + 3) + 5x next Hence, we may replace… 7x + 9 by the simpler expression…

15. Solution for #2: Hence… 7x + 6 = 7x + 9. next © 2007 Herbert I. Gross 7x + 6 = 7x + 9 next If we now subtract 7x from both sides of equation 7x + 6 = 7x + 9, we obtain the false statement… 6 = 9 – 7x – 7x

16. Solution for #2: No matter what value we choose for x, the equation 6 = 9 is a false statement; and since the equation is equivalent to the algebraic equations 2(x + 3) + 5 x = 7x + 9 and 7x + 6 = 7x + 9 the fact that the equation 6 = 9 is always false means that neither the equation 2(x + 3) + 5 x = 7x + 9 nor the equation 7x + 6 = 7x + 9 can be true for any value of x. next © 2007 Herbert I. Gross

17. next Notes on #2 Looking at the equation 7x + 6 = 7x + 9, we might have immediately noticed that adding 9 to 7x will always give an answer that is 3 more than if we had added 6 to 7x. © 2007 Herbert I. Gross This observation blends nicely with the result we got when we subtracted 7x from both sides of the equation. next

18. Notes on #2 Namely: since 9 is 3 more than 6, the equation 6 = 9 tells us that not only is there no value of x that is a solution for the equation 2(x + 3) + 5 x = 7x + 9, but also us that for any value of x, the right-hand side of the equation is always 3 more than the left-hand side. © 2007 Herbert I. Gross

19. next Notes on #2 Do not confuse the false statement 6 = 9 with the algebraic equation x + 6 = 9. © 2007 Herbert I. Gross While there is no value of x for which 6 = 9 (that is, for 0x + 6 = 9); the equation x + 6 = 9 is satisfied by x = 3. next

20. Problem #3 © 2007 Herbert I. Gross next For what value of x does 2(x + 3) + 5x = 6x + 9 ? next Answer: x = 3

21. Solution for #3: We already know from our solutions for Problems #1 and #2 that the left-hand side of the equation 2(x + 3) + 5x = 6x + 9 can be replaced by 7x + 6. next © 2007 Herbert I. Gross 7x + 6 = 6x + 9 next Making this same substitution, we now obtain the equation…

22. Solution for #3: To solve the equation 7x + 6 = 6x + 9, we may begin by subtracting 6x from both sides to obtain the algebraic equation… next © 2007 Herbert I. Gross next If we now subtract 6 from both sides of the equation x + 6 = 9, we obtain as our solution… x + 6 = 9 x = 3

23. next Notes on #3 As a reminder, notice that we still have to check whether x = 3 is a solution. © 2007 Herbert I. Gross next That is, so far our solution has only proved that if x ≠ 3, then 2(x + 3) + 5x ≠ 6x + 9.

24. Notes on #3 To check that x = 3 is indeed a solution, we can replace x by 3 in 2(x + 3) + 5x = 6x + 9, to obtain the true statements… © 2007 Herbert I. Gross next 2( x + 3 ) + 5( x ) = 6( x ) + 9 2(6) + 15 = 18 + 9 15 + 12 = 18 + 9 27 = 27 33 3 next

25. Summary of Problems #1, #2, and #3 Every linear equation can be written in the form… © 2007 Herbert I. Gross next mx + b = nx + c Notice that the left-hand sides of the equations in Problems #1, #2, and #3 were the same, and that they were equivalent to the simpler expression 7x + 6. That is, the left-hand side of each equation could be written in the form… mx + b where m = 7 and b = 6 next

26. Summary of Problems #1, #2, and #3 In other words, each of the three equations could be written in the form… © 2007 Herbert I. Gross next 7x + 6 = nx + c Hence, we could be guaranteed that there would be one and only one solution of the equation unless the multiplier of x on the right-hand side of the above equation was also 7. next In fact, this happened in Problems #1 and #2 but not in Problem #3.

27. Summary of Problem #1 Problem #1 is an example of an identity. © 2007 Herbert I. Gross next In terms of the “ordering from a catalog” model discussed in the lesson, it corresponds to the case in which two catalogs offer the same item for both the same price and the same shipping charge. $44.99 Shipping $10.50 $44.99 Shipping $10.50

28. next Summary of Problem #2 Problem #2 is an example of “No Solution”. © 2007 Herbert I. Gross next It corresponds to the case in which two catalogs offer the same item for the same price but with a different shipping charge. $44.99 Shipping $10.50 Shipping $12.50

29. next Summary of Problem #3 Problem #3 is an example of a unique solution. © 2007 Herbert I. Gross next It corresponds to the case in which two catalogs offer the same item at different prices and different shipping charges. $39.99 $44.99 Shipping $10.50 Shipping $5.50

30. Problem #4 © 2007 Herbert I. Gross next For what value of x does 2(5 – 3x) – 4(5 – 2x) = 2x + 12 ? next Answer: No value of x

31. Solution for #4: The right-hand side of 2(5 – 3x) – 4(5 – 2x) = 2x + 12 is already in the mx + b form. next © 2007 Herbert I. Gross next So we will use our rules of arithmetic to rewrite the left-hand side of the above equation in the mx + b form also.

32. Solution for # 4: To this end we see that… 2(5 – 3x) – 4(5 – 2x) = next © 2007 Herbert I. Gross next 2(5 + - 3x) + - 4(5 + - 2x) = 2(5) + 2( - 3x) + - 4(5) + - 4( - 2x) = 10 + - 6x + - 20 + 8x = (10 + - 20) + ( - 6x + 8x) = 2x + - 10

33. Solution for #4: Therefore, in the equation 2(5 – 3x) – 4(5 – 2x) = 2x + 12 we may replace the left-hand side by 2x + - 10 to obtain the equivalent equation… next © 2007 Herbert I. Gross next 2(5 – 3x) – 4(5 – 2x) = 2x + 12 then by subtracting 2x from both sides of the equation above, - 10 = 12 2x + - 10 – 2x – 2x we get the false statement… - 10 = 12 next

34. Solution for #4: if we now add 10 to both sides of - 10 = 12, we get the equivalent false statement… next © 2007 Herbert I. Gross next Since the equations 2(5 – 3x) – 4(5 – 2x) = 2x + 12, - 10 = 12, and 0 = 22 are equivalent, it means that there is no value of x for which the equation 2(5 – 3x) – 4(5 – 2x) = 2x + 12 is a true statement. 0 = 22

35. next Note on Problem #4 Moreover, the equation 0 = 22 tells us that for any value of x, the number on the right side of the equation 2(5 – 3x) – 4(5 – 2x) = 2x + 12 is 22 greater than the number on the left side of the equation. © 2007 Herbert I. Gross

36. next Note on Problem #4 © 2007 Herbert I. Gross 2(5 – 3x) – 4(5 – 2x) = 2x + 12 10 – 20 = 12 2(5 – 0) – 4(5 – 0) = 0 + 12 - 10 = 12 …and 12 exceeds - 10 by 22. Choosing x to be 0 will simplify our computation. Doing this we obtain… next

37. Note on Problem #4 Notice that our demonstration only validated our claim in the case when x was replaced by 0. It told us nothing about what would happen for other values of x. © 2007 Herbert I. Gross In contrast, using our “rules of the game”: the algebraic solution validates our claim, independently of the number which we use to replace x. next

38. Problem #5a © 2007 Herbert I. Gross next For what value of x does 5[2(x – 3) + 7] – 8x = 2x + 5 ? next Answer: All values of x

39. Solution for #5a: Just as in our solution to the previous problem, the right-hand side of 5[2(x – 3) + 7] – 8x = 2x + 5 is already in the mx + b form. next © 2007 Herbert I. Gross next So we will use our rules of arithmetic to rewrite the left-hand side of the above equation in the mx + b form also.

40. Solution for #5a: To this end, we see that… 5[2(x – 3) + 7] – 8x = next © 2007 Herbert I. Gross next 5[2(x + - 3) + 7] + - 8x = 5[2x + - 6 + 7] + - 8x = 5[2x + ( - 6 + 7)] + - 8x = 5[2x + 1] + - 8x = 10x + 5 + - 8x = (10x + - 8x) + 5 = 2x + 5 =

41. Solution for #5a: So in the equation 5[2(x – 3) + 7] – 8x = 2x + 5, we replace 5[2(x – 3) + 7] – 8x by 2x + 5 to obtain the equivalent equation… 5[2(x – 3) + 7] – 8x = 2x + 5 next © 2007 Herbert I. Gross next which is a true statement for all values of x. 2x + 5

42. next If you tried to solve the equation 5[2(x – 3) + 7] – 8x = 2x + 5 by working on both sides of the equation at the same time, avoid the temptation to “cancel” the 5 in the expression 5[2(x – 3) + 7] – 8x with the 5 in the expression 2x + 5. © 2007 Herbert I. Gross

43. next It would have been okay to do that if the equation had been… 5 + [2(x – 3) + 7] – 8x = 2x + 5 in which case we could subtract 5 from both sides of the equation. © 2007 Herbert I. Gross Or if it had been 5[2(x – 3) + 7] – 8x = 5(2x) in which case we could divide by 5 on both sides of the equation. next However, if we divide one side by 5 but subtract 5 from the other side, we do not get an equivalent equation.

44. Problem #5b © 2007 Herbert I. Gross next For what value of x does 5[2(x – 3) + 7] – 8x = 17 ? next Answer: x = 6

45. Solution for #5b: In part (a) of this problem, we saw that 5[2(x – 3) + 7] – 8x was equivalent to 2x+ 5. next © 2007 Herbert I. Gross next Hence, solving 5[2(x – 3) + 7] – 8x = 17 is equivalent to solving the simpler equation 2x + 5 = 17 To solve this equation, we subtract 5 from both sides to obtain… 2x = 12 And if we then divide both sides of 2x = 12 by 2, we obtain… x = 6

46. © 2007 Herbert I. Gross next 5[2(x – 3) + 7] – 8x ≠ 17 Notes on #5b Actually, all we’ve shown so far is that if x ≠ 6, then… To check that x = 6 is the solution, we now replace x by 6 in the expression 5[2(x – 3) + 7] – 8x and see if we obtain 17 as the answer.

47. next © 2007 Herbert I. Gross next 5[2(x – 3) + 7] + - 8x = Notes on #5b To this end… 5[2(6 – 3) + 7] + - 8(6) = 5[2(3) + 7] + - 8(6) = 5[6 + 7] + - 8(6) = 5[13] + - 48 = 65 + - 48 = 17

48. next Notice the power of paraphrasing. The problem asked us to find that value of x for which the output will be 17. That is… © 2007 Herbert I. Gross next Program # 1 Step 1Start with x.xStep 2Subtract 3.x – 3Step 3Multiply by 2.2(x – 3)Step 4Add 7.2(x – 3) + 7Step 5Multiply by 5. 5[2(x – 3) + 7]Step 6Store the answer.5[2(x – 3) + 7]Step 7Input xxStep 8Multiply by 8.8x Step 9 Subtract the answer in step 8 from the number stored in step 6. 5[2(x – 3) + 7] – 8x Step 10Write the answer.5[2(x – 3) + 7 ] – 8x next 17 next

49. And we can also find the answer by using the much simpler Program #2, where… © 2007 Herbert I. Gross next Program # 2 Step 1Start with x.xStep 2Multiply by 2.2xStep 3Add 5.2x + 5Step 10Write the answer.2x + 5 17 next

50. As a check, notice that… © 2007 Herbert I. Gross next Program # 1 Step 1Start with x.x Step 2Subtract 3.x – 3 Step 3Multiply by 2.2(x – 3) Step 4Add 7.2(x – 3) + 7 Step 5Multiply by 5. 5[2(x – 3) + 7] Step 6Store the answer.5[2(x – 3) + 7] Step 7Input xx Step 8Multiply by 8.8x Step 9 Subtract the answer in step 8 from the number stored in step 6. 5[2(x – 3) + 7] – 8x Step 10Write the answer.5[2(x – 3) + 7 ] – 8x 6 3 6 13 65 6 48 17

51. next © 2007 Herbert I. Gross Key Point Notice that if we had tried to solve for x in Program #1 by using the undoing method, we would have been stymied when we tried to undo Step 9 because, in undoing, we wouldn’t yet know what the number stored in Step 6 was.