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Key Stone Problem… Key Stone Problem… next Set 16 © 2007 Herbert I. Gross.

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Presentation on theme: "Key Stone Problem… Key Stone Problem… next Set 16 © 2007 Herbert I. Gross."— Presentation transcript:

1 Key Stone Problem… Key Stone Problem… next Set 16 © 2007 Herbert I. Gross

2 You will soon be assigned problems to test whether you have internalized the material in Lesson 16 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. Instruction for the Keystone Problem 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 The Keystone Problems for Lesson 16 will be in three parts in order to emphasize all three possibilities for the solution of a linear equation. Note next

5 © 2007 Herbert I. Gross For what value of x does… Keystone Problem for Lesson 16 2(x + 3) + 5x = 7x + 6 Problem #1

6 next © 2007 Herbert I. Gross Solution for Problem #1 The right-hand side of the equation 2(x + 3) + 5x = 7x + 6 is in the “mx + b” form; but the left-hand side is not. However, the same methods used in previous Lessons will let us paraphrase the left-hand side into this “mx + b” form. That is, by the distributive property, we may first rewrite 2(x + 3) + 5x as… next (2x + 6) + 5x

7 next © 2007 Herbert I. Gross Solution for Problem #1 Then by the associative and commutative properties for addition, we may rewrite the expression (2x + 6) + 5x as… (2x + 5x) + 6 next or… 7x + 6

8 next © 2007 Herbert I. Gross Solution for Problem #1 next And since the equation 2(x + 3) + 5x = 7x + 6 is an identity, it yields a true statement for every value of x. That is, we can now replace the expression 2(x + 3) + 5x by 7x + 6 and this gives us the resulting equation, 7x + 6 = 7x + 6, which is an identity.

9 © 2007 Herbert I. Gross The fact that our equation is an identity means that the left-hand side of the equation is another way of expressing the same thing as the (simpler) right-hand side. Note next 2(x + 3) + 5x = 7x + 6 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.

10 next © 2007 Herbert I. Gross For example, if we elect to replace x by 9 in our equation… Substituting next 2(x + 3) + 5x = 7x + 6 …which is a true statement. (= 69) next = (12) + 5(9) = 7(9) + 6 2(9 + 3) + 5(9) = 7(9) + 62(x + 3) + 5x = 7x + 6 next

11 © 2007 Herbert I. Gross Or if we had replaced x by, say, - 6 in our equation, we would have obtained… Substituting next 2(x + 3) + 5x = 7x + 6 …which is a true statement. (= - 36) next = ( - 3) + 5( - 6) = 7( - 6) + 6 2( ) + 5( - 6) = 7( - 6) + 62(x + 3) + 5x = 7x + 6 next

12 © 2007 Herbert I. Gross The point is that no matter what values we replaced x by in the equation, the resulting equality would be a true statement, because the expressions on each side of the equation are equivalent. Key Point next 2(x + 3) + 5x = 7x + 6

13 next © 2007 Herbert I. Gross The above expression may be represented by Program 1… Note next 2(x + 3) + 5x Program 1 1. Input x. 2. Add Multiply by Store the result. 5. Input x. 6. Multiply by Add the answer in step 6 to the number stored in step The output is… x x + 3 2(x + 3) x 5x 2(x + 3) + 5x

14 next © 2007 Herbert I. Gross What we have shown in this problem is that Program 1 is equivalent to Program 2, where… Note next 2(x + 3) + 5x Program 2 1. Input x. 2. Multiply by Add The output is … = 7x + 6 This equivalence means that we may replace the more complicated Program 1 by the simpler Program 2. That is: for any input (x), the output (answer) we get with Program 1 is equal to the output given by Program 2 for the same input (x). 7 7x 7x + 6

15 next © 2007 Herbert I. Gross For example, suppose we are asked to solve the equation… Note next 2(x + 3) + 5x = 7x + 6 We can replace the expression on the left by 7x + 6 to obtain the equivalent equation… 2(x + 3) + 5x = 90 7x + 6 = 90 Then to solve this equation, we need only subtract 6 from both sides and then divide by 7 to obtain as our solution, x = 12. next

16 © 2007 Herbert I. Gross Note next 2(x + 3) + 5x = 7x = 90 As a check: we replace x by 12 in the equation below, to obtain the true statement… 2(15) + 5(12) = 90 2(12 + 3) + 5(12) = 902(x + 3) + 5x = 90 next

17 © 2007 Herbert I. Gross For what value of x does… Keystone Problem for Lesson 16 2(x + 3) + 5x = 7x + 9 Problem #2

18 next © 2007 Herbert I. Gross Solution for Problem #2 2(x + 3) + 5x = 7x + 9 We have already seen in our solution of Problem #1 that the left side of the equation, 2(x + 3) + 5x, is equivalent to the simpler expression on the right side, 7x + 6. Hence, we may replace 2(x + 3) + 5x in the equation by 7x + 6 to obtain … next 7x + 6 = 7x + 9 If we now subtract 7x from both sides of this equation, we obtain the false statement… 6 = 9 next

19 © 2007 Herbert I. Gross Solution for Problem #2 Since we saw that the numerical equation 6 = 9 is equivalent to the algebraic equation 2(x + 3) + 5x = 7x + 9 that is, to 7x + 6 = 7x + 9, the fact that equation 6 = 9 is always false (that is, cannot be true for any value of x) means that neither the equation 2(x + 3) + 5x = 7x + 9 nor the equation 7x + 6 = 7x + 9 can be true, for any value of x. next

20 © 2007 Herbert I. Gross Since 9 is 3 more than 6, the equation 6 = 9 told us that there cannot be any value of x that solves the above equation. Moreover, it also told us that, for any value of x, the right-hand side of this (false) equation will always be 3 more than the left-hand side. Note 2(x + 3) + 5x = 7x + 9

21 next © 2007 Herbert I. Gross For example, if we elect to replace x by 6 in our equation we get… Substituting next 2(x + 3) + 5x = 7x = 51 next = (9) + 5(6) = 7(6) + 9 2(6 + 3) + 5(6) = 7(6) + 92(x + 3) + 5x = 7x + 9 next This is a false statement: the right side of the equal sign exceeds the left side by 3.

22 © 2007 Herbert I. Gross For what value of x does… Keystone Problem for Lesson 16 2(x + 3) + 5x = 6x + 9 Problem #3

23 next © 2007 Herbert I. Gross Solution for Problem #3 We already know (from our solution for Problems #1 and #2) that the left-hand side of the equation, 2(x + 3) + 5x = 7x + 9, can be replaced by 7x + 6. Making this substitution, we obtain the equation... next 2(x + 3) + 5x = 6x + 97x + 6 From here, the simple step of subtracting 6x from both sides of the equation gives us the algebraic equation… x + 6 = 9 next

24 © 2007 Herbert I. Gross Solution for Problem #3 Then all we have to do is to subtract 6 from both sides of x + 6 = 9, and we obtain… next x = 3 To verify that x = 3 is the solution to Problem #3, we replace x by 3 in Problem #3’s equation, to obtain… next = (6) + 5(3)= 6(3) + 9 2(3 + 3) + 5(3) = 6(3) + 92(x + 3) + 5x = 6x + 9 …which is a true statement. next

25 © 2007 Herbert I. Gross Caution Do not confuse the numerical false statement 6 = 9 with the algebraic equation x + 6 = 9 next The statement 6 = 9 is simply a false statement (independently of the value of x); while the algebraic equation x + 6 = 9 becomes a true statement when, and only when, x = 3

26 © 2007 Herbert I. Gross Summary Every linear equation can be written in the form… next mx + b = nx + c Notice that the left-hand side in each of the equations in Problems #1, #2, and #3 was [2(x + 3) + 5x], which is equivalent to the simpler expression, 7x + 6. In other words, each equation in this presentation could be written in the form… next 7x + 6 = nx + c

27 next © 2007 Herbert I. Gross Summary Hence, we could be guaranteed that there would be one and only one solution of the equation, unless the multiplier of x on right-hand side of the equation 7x + 6 = nx + c was also 7. This happened in Problems #1 and #2, but not in Problem #3.

28 next © 2007 Herbert I. Gross In terms of (looking back to) the review at the end of Lesson 16… next 2(x+3) + 5x = 7x + 6 Problem #1 is an example of Case 3 …is an identity. next That is, the equation… It is true for all values of x.

29 © 2007 Herbert I. Gross next 2(x+3) + 5x = 7x + 9 Problem #2 is an example of Case 2 …is inconsistent. next That is, the equation… It is false for every value of x.

30 © 2007 Herbert I. Gross next 2(x+3) + 5x = 6x + 9 Problem #3 is an example of Case 1 …has next That is, the equation… a unique solution: x = 3.


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