key stone problem… key stone problem… next set 16 © 2007 herbert i. gross

Key Stone Problem…Key Stone Problem…

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Set 16© 2007 Herbert I. Gross

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

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© 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.

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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

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For what value of x does…

Keystone Problem for Lesson 16

2(x + 3) + 5x = 7x + 6

Problem #1

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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…

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(2x + 6) + 5x

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Then by the associative and commutative properties for addition, we may rewrite the

expression (2x + 6) + 5x as…

(2x + 5x) + 6

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or…

7x + 6

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Solution for Problem #1next

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.

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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.

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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.

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For example, if we elect to replace x by 9 in our equation…

Substituting

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2(x + 3) + 5x = 7x + 6

…which is a true statement.

(= 69)

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24 + 45 = 63 + 6

2(12) + 5(9) = 7(9) + 6

2(9 + 3) + 5(9) = 7(9) + 6 2(x + 3) + 5x = 7x + 6

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Or if we had replaced x by, say, -6 in our equation, we would have obtained…

Substituting

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2(x + 3) + 5x = 7x + 6

…which is a true statement.

(= -36)

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-6 + -30 = -42 + 6

2(-3) + 5(-6) = 7(-6) + 6

2(-6 + 3) + 5(-6) = 7(-6) + 6 2(x + 3) + 5x = 7x + 6

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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

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2(x + 3) + 5x = 7x + 6

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The above expression may be represented by Program 1…

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2(x + 3) + 5x

Program 11. Input x.2. Add 3. 3. Multiply by 2. 4. Store the result. 5. Input x.6. Multiply by 5.

7. Add the answer in step 6to the number stored in step 4.

8. The output is…

xx + 3

2(x + 3)2(x + 3)

x5x

2(x + 3) + 5x

2(x +3) + 5x

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What we have shown in this problem is that Program 1 isequivalent to Program 2, where…

Notenext

2(x + 3) + 5x

Program 21. Input x.2. Multiply by 7.3. Add 6.4. 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).

77x

7x + 67x + 6

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For example, suppose we are asked to solve the equation…

Notenext

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.

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Notenext

2(x + 3) + 5x = 7x + 6

30 + 60 = 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) = 90 2(x + 3) + 5x = 90

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2(x + 3) + 5x = 7x + 9

Problem #2

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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 …

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7x + 6 = 7x + 9If we now subtract 7x from both sides of this equation, we obtain the false statement…

6 = 9

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Since we saw that the numerical equation 6 = 9 is equivalent to the algebraic equation

2(x + 3) + 5x = 7x + 9that 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 + 9can be true, for any value of x.

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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.

Note2(x + 3) + 5x = 7x + 9

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For example, if we elect to replace x by 6 in our equation we get…

Substituting

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2(x + 3) + 5x = 7x + 9

48 = 51

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18 + 30 = 42 + 9

2(9) + 5(6) = 7(6) + 9

2(6 + 3) + 5(6) = 7(6) + 9 2(x + 3) + 5x = 7x + 9

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This is a false statement: the right

side of the equal sign exceeds the left side

by 3.

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2(x + 3) + 5x = 6x + 9

Problem #3

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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...

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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

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Then all we have to do is to subtract 6 from both sides of x + 6 = 9, and we obtain…

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x = 3To verify that x = 3 is the solution to

Problem #3, we replace x by 3 in Problem #3’s equation, to obtain…

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12 + 15 = 18 + 92(6) + 5(3)= 6(3) + 9

2(3 + 3) + 5(3) = 6(3) + 92(x + 3) + 5x = 6x + 9

…which is a true

statement.

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Caution

Do not confuse the numerical false statement 6 = 9

with the algebraic equationx + 6 = 9

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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

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Summary

Every linear equation can be written in the form…

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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…

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7x + 6 = nx + c

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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.

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In terms of (looking back to) the review at the end of Lesson 16…

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2(x+3) + 5x = 7x + 6

Problem #1 is an example of Case 3

…is an identity.

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That is, the equation…

It is true for all values of x.

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2(x+3) + 5x = 7x + 9


…is inconsistent.

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It is false for every value of x.

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2(x+3) + 5x = 6x + 9


…has

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a unique solution:x = 3.

key stone problem… key stone problem… next set 16 © 2007 herbert i. gross

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