Key Stone Problems… Key Stone Problems… next Set 11 © 2007 Herbert I. Gross

Key Stone Problems…Key Stone Problems…

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Set 11© 2007 Herbert I. Gross

You will soon be assigned problems to test whether you have internalized the material

in Lesson 11 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

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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 beauty of logical thought is that we can often solve complicated problems by replacing them by a series of equivalent

simpler problems. In particular the goal of this set of exercises is to show how using

the rules of the game we can replace a rather complex expression such as…

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2 {5 [15 – 4 (3 – c)] – 6}by the simpler, but equivalent, expression

40 c + 18that is much easier to work with.

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We will develop this idea through a step-by-step simplification of the left side

of the expression 2 {5 [15 – 4 (3 – c)] – 6}.

The two exercises in this set will illustrate our approach. More specifically, the steps

that we will use to solve the equation in Question 2 are precisely those which make

up the different parts to Question 1.

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1. Simplify each of the following expressions…

a. -4(3 + -c)

b. 15 + -4(3 + -c)

c. 5[15 + -4(3 + -c)]

d. 5[15 + -4(3 + -c)] + -6

e. 2{5[15 + -4(3 + -c)] + -6}

2. Use the result of Problem 1e to solve the equation…

2{5[15 - 4(3 - c)] - 6} = 94

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Solution for Problem 1a

This exercise asks us to simplify the expression… -4(3 + -c).

By the distributive property…

From the arithmetic for signed numbers we know that…

If we now replace each term on the right hand side of the first equation by its value

in the second equation, we obtain…

-4(3 + -c) = -4(3) + -4(-c)

-4(3) = -12 and -4(-c) = 4c

-4(3 + -c) = -12 + 4c

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In problems such as this, we use the word “simplify” rather reluctantly. The reason is that the meaning of simplify is subjective.

That is, what seems ”simpler” to one person might not seem simpler to another. So in the spirit of our game of algebra we

will use “simplify” as another word for “paraphrase”. In this particular discussion it will mean to paraphrase each expression

into a sum consisting of two terms.

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Note 1a

In terms of the above note, someone might prefer to simplify our answer by using the

commutative property of addition andrewriting -12 + 4c in the form 4c + -12.

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And someone else might prefer replacing 4c + -12 by the “simpler”

expression 4c – 12.

Note 1a

With respect to the rules of the game, our preference is to write subtraction in the form

of addition because our rules are stated in terms of addition. However, once you feel

that you’ve internalized the connection between addition and subtraction, you may

use expressions such as…

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Note 1anext

4c + -12 and 4c –12 interchangeably.

If we were being insistent on playing the game strictly by the rules, we would have to prove such statements as: “The product oftwo negative numbers is always a positive

number”. Such an approach could be tedious and too abstract for our present

purposes. Instead, we will assume that the approach we used in Lessons 3, 4, and 5 for developing the arithmetic of signed numbers is sufficiently acceptable to all of our players.

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Note 1a

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Solution for Problem 1b

This exercise asks us to simplify the expression… 15 + -4(3 + -c)

In Problem 1a, we showed that…

We now replace -4(3 + c) by -12 + 4c to obtain…

15 + -4(3 + -c)

-4(3 + -c) = -12 + 4c

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-12 + 4c

( )

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Solution for Problem 1b

and since 15 + -12 = 3, we may rewrite the above equality as...

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15 + -4(3 + -c) = 3 + 4c

15 + -4(3 + -c) = 15 + (-12 + 4c)

By the associative property of addition we know that…

15 + -4(3 + -c) = 15 + -12 + 4c ( )

Thus…

One of the most common mistakes made by beginning students is to begin by

adding the 15 and -4 in the expression

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to obtain the incorrect expression…

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Caution

15 + -4(3 + -c)

11(3 + -c)

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Caution

15 + -4(3) = 3

For example, if we replace c by 0 in the expression 15 + -4(3 + -c) we obtain…

On the other hand, if we replace c by 0 in 11(3 + -c), we obtain…

11(3) = 33

The fact that the same input gives different outputs shows that the two expressions

can’t be equal.

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Caution

(15 + -4)(3 + -c)

That is, by our agreeing to the PEMDAS convention we do all multiplications

before we do any additions. Hence, if we had wanted the 15 and the -4 to be added

first, we would have had to write the expression as…

In words the expression 15 + -4(3 + -c) tells us…

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Note 1b

*** Start with any number c.

*** Replace it by its opposite… -c.

*** Add 3… 3 + -c.

*** Multiply by -4… -4(3 + -c).

*** Add 15… 15 + -4(3 + -c).

What we showed in this exercise is that the expression 15 + -4( 3 + -c) is equivalent to

the expression 3 + 4c.

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In words the expression 3 + 4c tells us…

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Note 1b

*** Start with c… c.

*** Multiply 4… 4c.

*** Add 3… 3 + 4c.

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Note 1b In this sense, what we mean by “simplified” is that the relatively

cumbersome set of instructions…

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Start with any number c. Replace it by its opposite. Add 3. Multiply by -4. Add 15

and then write the answer.…can be replaced by the less cumbersome

but equivalent set of instructions…

Start with any number c. Multiply it by 4. Add 3 and then write the answer.

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Solution for Problem 1c

This exercise asks us to simplify the expression… 5[15 + -4(3 + -c)]

From the previous exercise we already know that…

So by replacing 15 + -4(3 + -c) by 3 + 4c, 5[15 + -4(3 + -c)] becomes…

5[3 + 4c]

15 + -4(3 + -c) = 3 + 4c

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3 + 4c

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By the distributive property, we know that…

5[3 + 4c] = 5(3) + 5(4c) = 15 + 5[4c]

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By the associative property of multiplication, we know that…

5[4c] = [5(4)]c = 20c

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If we now replace 5[4c] by 20c, we obtain…

5[3 + 4c] = 15 + 20c

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In summary, we have thus far shown that… (1) 5[15 + -4(3 + -c)] = 5[3 +4c]

and(2) 5[3 + 4c] = 15 + 20c

5[15 + -4(3 + -c)] = 15 + 20c

Thus, by transitivity…

5[15 + -4(3 + -c)]

15 + 20c

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It’s quite possible that in the above equation you looked at 5[4c] and immediately concluded that it was equal to

20c without even thinking consciously about the associative property of multiplication. Hopefully, such observations will become quite common as the course proceeds.

For the time being, we’ll try to be as complete as we feel is necessary in demonstrating how certain results follow inescapably from others just by using the accepted rules of the game.

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Note 1c

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Solution for Problem 1d

This exercise asks us to simplify the expression… 5[15 + -4(3 + -c)] + -6

In exercise 1c we showed that…

Therefore by substitution5[15 + -4(3 + -c)] + -6 becomes…

5[15 + -4(3 + -c)] + -6

(15 + 20c)

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5[15 + -4(3 + -c)] = 15 + 20c

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By the commutative property of addition we know that…

So we may replace 15 + 20c by 20c + 15 to obtain…

Therefore, 5[15 + -4(3 + -c)] + -6 may be replaced by …

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5[15 + -4(3 + -c)] + -6

(20c + 15)5[15 + -4(3 + -c)] = 15 + 20c

20c + 1515 + 20c = 20c + 15

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Then by the associative property of addition…

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20c + 15 + -6

In summary…5[15 + -4(3 + -c)] + -6 = (20c + 15) + -6

and(20c + 15) + -6 = 20 c + 9

5[15 + -4(3 + -c)] + -6 = 20c + 9

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( ) = 20c + 9

So again by transitivity…

5[15 + -4(3 + -c)] + -6

20 c + 9

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Solution for Problem 1e

This exercise asks us to simplify the expression… 2{5[15 + -4(3 + -c)] + -6}

From the previous exercise we know …

Hence, we replace 5[15 + -4(3 + -c)] + -6 in the first expression by its equivalent value in

the above expression, to obtain…

2{5[15 + -4(3 + -c)] + -6} = 2{20c + 9}

5[15 + -4(3 + -c)] + -6 = 20c + 9

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By the distributive property we know that…

2{20c + 9} = 2{20c} + 2{9} = 40 c + 18

And if we now replace 2{5[15 + -4(3 + -c)] + -6} In the previous expression with the value in the above expression we obtain...

2{5[15 + -4(3 + -c)] + -6} = 40c + 18

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In any game there is usually more than one correct strategy for obtaining a given objective. In the game of

mathematics, this may be restated as...

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EnrichmentNote

Therefore, do not try to memorize our proofs. Rather, concentrate on how we elected to use a particular strategy to achieve each of

the given objectives.

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“In a well-defined problem there is only one correct answer, but hardly ever only one correct way to derive the answer”.

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Solution for Problem 2

In this exercise we are asked to use the result of Problem 1e to solve the equation…

2 {5[15 – 4(3 – c)] – 6} = 94

We begin our solution by observing that by the “add the opposite” rule, the equation…

2{5[15 + -4(3 + -c)] + -6} = 94

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2 {5[15 – 4(3 – c)] – 6} = 94

…can be rewritten in the equivalent form…

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The left side of our equation is precisely the expression we simplified in Exercise 1e.

In other words the fact that…

…means that we may replace the equation 2{5[15 + -4(3 + -c)] + -6} = 94

by the equivalent but simpler equation…

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40c + 18 = 94

2{5[15 + -4(3 + -c)] + -6} = 40c + 18

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To solve equation 40c + 18 = 94, we first subtract 18 from both sides to obtain the equivalent equation…

Finally, we divide both sides of the above equation by 40 to obtain…

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40c = 76

c = 76/40 = 19/10

or in decimal form…

= 1.9

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We could have omitted Exercise 1 in its entirety and just asked

you to find the value of c for which...

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Note 2

In that case one of the ways we could have used to answer the question would have

been to rewrite the equation in the form…

2 {5[15 – 4(3 – c)] – 6} = 94

2{5[15 + -4(3 + -c)] + -6} = 94

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Notice that the left side of 2{5[15 + -4(3 + -c)] + -6} = 94 is the expression that we simplified in Exercise 1e.

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Note 2

Then since grouping symbols are removed starting from the innermost set first, we

would have begun the simplifying processby starting with the removal of the

parentheses; and this is precisely what we did in Exercises 1a, 1b, 1c, 1d, and 1e.

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More visually…2{ 5 [15 + -4(3 + -c)] + -6 }-4(3 + -c)[15 + -4(3 + -c)]5 [15 + -4(3 + -c)]5[15 + -4(3 + -c)] + -6

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1a,

2 {5[15 + -4(3 + -c) + -6}

1b, 1c, 1d, 1e.

In summary, we used the rules in Lessons 9 and 10 to show that the

more complicated expression…

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Summary

…could be replaced by the equivalent, but much simpler expression…

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2{5[15 + -4(3 + -c)] + -6}

40c + 18

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At least until you become more comfortable with complicated algebraic expressions; it is a good idea to rewrite every subtraction problem in terms of

the add-the-opposite rule. That is, wherever we see a minus sign we

replace it by a plus sign and change the sign of the number immediately after the

minus sign.

Feeling Comfortable

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*** we first locate the minus signs…

*** Then we replace each minus sign by a plus sign…

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2 {5[15 – 4(3 – c)] – 6} = 94

2 {5[15 – 4(3 – c)] – 6} = 94

…and we replace each number immediately after the changed minus sign by its opposite…

2 {5[15 + 4(3 + c)] + 6} = 94- - -

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+ + +

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*** Notice that the steps we used in solving the equation…

…had nothing to do with the fact that the right side was 94. That is, we demonstrated

that 2{5[15 – 4(3 – c)] – 6} = 40c + 18without any reference to the right hand side

of the above equation.

By way of Review

2 {5[15 – 4(3 – c)] – 6} = 94

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Recall that to evaluate the expression…

2 {5[15 – 4(3 – c)] – 6}

…for any given value of c, we start with c and remove the grouping symbols by

working our way from the inside to the outside.

The above comments may seem more impressive if you look at them in terms of

“programs”.

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Program #1(1) Start with c(2) Subtract it from 3(3) Multiply the result by 4(4) Subtract this result from 15(5) Multiply the result by 5(6) Subtract 6(7) Multiply by 2(8) Write the answer

Following this procedure we seethat the expression

2 { 5 [ 15 – 4 ( 3 – c ) ] – 6 } is equivalent to Program #1, where…

c( 3 – c )4 ( 3 – c )15 – 4 ( 3 – c )5 [ 15 – 4 ( 3 – c ) ]5 [ 15 – 4 ( 3 – c ) ] – 6 2 { 5 [ 15 – 4 ( 3 – c ) ] – 6 }

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Program #2(1) Start with c(2) Multiply by 40(3) Add 18(4) Write the answer

On the other hand, the expression 40 c + 18 is equivalent to

Program #2, where…c40 + 18

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If we had not used the rules of the game, how likely is it that we could have guessed that Programs #1 and #2, as different as they look,

are equivalent? It is in this sense that the “game” is important. That is, we use our rules to demonstrate that certain things which may not seem obvious are indeed true. Because it contains fewer steps, Program #2 is preferable to Program #1. In other words, the fewer steps

we use, the less chance there is that we will make a computational mistake. This is true in other things as well. For example, the fewer

“moving parts” a machine has, the easier it is to maintain the machine.

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For example, we used Program #2 to prove that if the output of Program #1 was 94 then the input had to be 1.9. To check that our solution was correct, we would

replace c by 1.9 in Program #1 andshow that we got 94 as the output.

Moreover, if we have to evaluate a program for many different values of the

input, the shorter program has the potential to save us significant amounts of time.

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Program #1(1) Start with C(2) Subtract it from 3(3) Multiply the result by 4(4) Subtract this result from 15(5) Multiply the result by 5(6) Subtract 6(7) Multiply by 2(8) Write the answer

That is…

1.91.14.4

10.653479494

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Program #2(1) Start with c(2) Multiply by 40(3) Add 18(4) Write the answer

If now we wanted to find the output in Program #1 when the input was, say, 6.5,

we could go through the same tedious process we used above, or we could replace Program #1 by the equivalent

Program #2 and see that…

6.5260278278

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The important thing isn't just that Program #2 is simpler than Program #1;

but rather that it is equivalent to Program #1. That is, if we replace c by any given value using Program #2, we will find

the same answer more simply than we would have found it by using the more

cumbersome Program #1.

Important Point

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Keep in mind that while we simplify expressions by removing the grouping

symbols from the inside to the outside; we simplify equations by removing the grouping symbols from the outside to the inside. The

point is that when a complicated expression is part of an equation, we may not have to

simplify the complicated expression first.

For example, another way to solve the equation 2{5[15 + -4(3 + -c)] + -6} = 94

without first doing the steps in Exercise 1 is by working on both sides of the equation at the

same time.

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More specifically, in solving the equation 2{5[15 + -4(3 + -c)] + -6} = 94 by the undoing method, we notice that c is inside

the parentheses; the parentheses areinside the brackets; and the brackets are inside the braces. So the last step we did

to obtain 94 was to multiply by 2.

Hence, we divide both sides of the equation by 2 to obtain…

5[15 + -4(3 + -c)] + -6 = 472{5[15 + -4(3 + -c)] + -6} = 94

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There is a tendency for beginning students to begin the solution of

2{5[15 + -4(3 + -c)] + -6} = 94 by adding 6 to both sides of the equation. However, notice that since -6 is inside the

braces, and the braces are being multiplied by 2, it behaves as 12.

More specifically, PEMDAS tells us that we multiply by 2 after we add -6.

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Caution

To undo subtracting 6, we add 6 to both sides of 5[15 – 4(3 – c)] – 6 = 47 to obtain…

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To undo multiplying the left side of 5[15 – 4(3 – c)] = 53 by 5, we divide both

sides of by 5 to obtain…

5[15 – 4(3 – c)] = 53

15 – 4(3 – c) = 10.6

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Since it's often easier to deal with undoing addition than with undoing subtraction, we may prefer to replace each subtraction by the add-the-opposite rule and rewrite equation 15 – 4(3 – c) = 10.6 in the form…

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Since the last step we did in the above equation to obtain 10.6 was to add 15, we

now subtract 15 from both sides of 15 + -4(3 + -c) = 10.6 to obtain…

15 + -4(3 + -c) = 10.6

-4(3 + -c) = -4.4

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Since the last step in obtaining -4(3 + -c) = -4.4, was to multiply by -4, we divide both sides of the equation by -4 to obtain…

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Next we subtract 3 from both sides of equation 3 + -c = 1.1 to obtain…

3 + -c = 1.1

-c = -1.9

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Since the opposite of c (that is, -c) is -1.9, c itself must be equal to 1.9…

c = 1.9

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We may solve Question 2 by either of the previous two methods.

However, the method we used in Question 1 is the only one we can use if we are asked to simplify an expression

rather than to solve an equation.

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Important Point

As a final note observe the subtlety involved in the language ofsubtraction. For example, it makes no

difference whether we say“Pick a number and add 7”

(in the language of algebra, x + 7) or

“Pick a number and add it to 7”(in the language of algebra, 7 + x)

because addition obeys the commutative rule.

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FinalNote

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However, it makes a big difference whether we say…

“Pick a number and subtract 7 (in the language of algebra, x – 7)

or“Pick a number and subtract it from 7”

(in the language of algebra 7 – x)

…because subtraction doesn't obey the commutative property.

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This is one reason we prefer to rewrite subtraction in terms of

“add the opposite”. That is, while 15 – 7 ≠ 7 – 15;

15 + -7 = -7 + 15

In fact, as we’ll discuss in more detail in the Exercise Set, x – 7 and 7 – x are

opposites of one another.

For example, 15 – 7 = 8 while 7 – 15 = -8.

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Key Stone Problems… Key Stone Problems… next Set 11 © 2007 Herbert I. Gross