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A Portland Community College mathematics telecourse.

A Course in Arithmetic Review

Produced at Portland Community College.

Let's start this lesson

by adding to our list of technical vocabulary.

The word 'operation,' as used in arithmetic, simply means:

those symbols

which tell us what we're going to do

to either a number or pairs of numbers.

If you are operating on two numbers

by the plus, division, minus, and times, [ + ÷ - x ]

in math that's referred to as a 'binary' operation;

whereas, if your operation only needs one number to work upon,

than that's referred to as a 'unary' operation.

Quite frequently in advanced formulas in mathematics

when you do any kind of operation you don't do this or that.

You do a whole string of them at once,

so the question arises of all of these operations,

which do I do first?

That is, we need some kind of order for doing operations.

If it would nice if we simply agreed to go from front to back,

but that's not the way the system evolved.

So these are the order of doing operations

in all of your future mathematical work.

This is the order that computers evaluate arithmetic expressions.

It is also the order in which

calculators which have algebraic logic, scientific calculators,

will work in this order.

So first you do what's inside parentheses.

Once you're there,

you do the exponents first as you come to them.

And you go back to the beginning

and do the multiplications and/or divisions as you come to them.

Make special note:

you don't do the multiplications first then the division.

You do whichever of these two come first from beginning to end,

then back to the beginning

and do the additions and/or subtractions as you come to them

front to back.

If you come to a subtraction first, do it first.

Addition first, do it first.

So let's apply that to the problem we just had.

First we're going to get inside the parentheses.

And so we get inside the first parentheses that we see

reading left to right. [ ( 24 ÷ 2 · 3 ) ]

Now that we're inside, [ ( 24 ÷ 2 · 3 ) ]

we look for exponents.

There aren't any.

Then you look for times [x] and/or divisions [÷]

and do them as you come to them.

So there is a division. [ 24 ÷ 2 ]

It's between 24 and 2, [ 24 ÷ 2 ]

so I'll do this first, and that gives me 12. [ 24 ÷ 2 = 12 ]

Then look for more. [ ( 24 ÷ 2 · 3 ) ]

See 12 now has really replaced [ ( 12 · 3 ) ]

that entire bit. [ 24 ÷ 2 = 12 ]

Now looking for more. [ ( 12 · 3 ) ] There it is.

It's between 12, which was a result of that entry, so we do that.

12 times 3 is 36, [ 12 · 3 = 36 ]

so this whole bit here [ ( 24 ÷ 2 · 3 ) ]

has become 36. [ ( 24 ÷ 2 · 3 ) = 36 ]

Now at this point to keep from becoming confused,

let's just copy the rest of the problem. 5(36) + (2 · 3² - 4 + 1)

Now that we've reduced one parentheses, our first one,

to a single number, [36] we start looking for more parentheses.

And that would be this one. [ (2 · 3² - 4 + 1) ]

Now that we're inside the parentheses, back to the beginning,

look for exponents as you come to them.

That's not. [ 2 ] That is [ 3² ]

It's above the three [ 3² ]

so it says: take two 3s [ 3 · 3 ] and multiply them together.

3 times 3 is 9, [ 3 · 3 = 9 ]

so that [ 3² ] becomes replaced by 9.

Now continue to look for more exponents

to the end of the parentheses.

There aren't any.

Then back to the beginning.

Next you look for multiplications and/or divisions

as you come to them. [ (2 · 9 - 4 + 1) ]

There's a multiplication. [ 2 · 9 ]

It's between the 9 and the 2, so 9 times 2 is 18. [ 9 · 2 = 18 ]

So all of this [ (2 · 3²) ] has been replaced by 18.

Then we look for more multiplications.

There aren't any.

Back to the beginning of that parentheses.

And our last step,

look for additions and/or subtractions as you come to them.

18 minus, [ ( 18 - 4 + 1 ) ] I come to the subtraction first.

It's between those two.

So I do it.

So 18 minus 14 is [ 18 - 14 =

18 minus 4 is 14. [ 18 - 4 = 14 ]

Okay keep looking for more.

There is an addition.

It's between the 1 and the 14 I just did, [ ( 14 + 1 ) ]

so one plus 14 is 15. [ 1 + 14 = 15 ]

Now I'll just rewrite my problem [ 5(36) + 15 ]

to keep from getting confused in the growing mess.

Look for more parentheses.

There aren't any.

Back to the beginning.

Back to the beginning. There were no more parentheses,

so now we look for exponents.

There aren't any.

Then look for multiplications and/or divisions

as you come to them.

There is one. [ 5(36) + 15 ]

It's not written, but it is there. [ 5 · (36) ]

It says 5 times 36, which is 180. [ 5 · 36 = 180 ]

And now of course, there's just

the 15 being added left, [ 180 + 15 ]

so doing that we have 195. [ 180 + 15 = 195 ]

This process of reducing

an arithmetic expression [ 5( 24 ÷ 2 · 3) + (2 · 3² - 4 + 1) ]

to a single number [195] we call 'evaluating.'

These order of operations must be memorized as soon as possible,

at least by the end of your current lesson.

So first you always do

what's inside parentheses as you come to it.

Exponents next as you come to them.

Multiplications and/or divisions as you come to them.

Don't do these [ +, - ] until you finish these. [ x ÷ ]

And back to the beginning.

Additions and/or subtractions as you come to them.

A very important lesson.

Learn it very soon.

Up to now in our preparation for algebra

we've dealt with equations where the variables is on one side

and there's only one operation associated

between that variable and a constant.

Now that we're in the order of operation

which deals with multiple operations at once,

it's a good time to handle equations

which begin to have

more than one operation associated with the variable.

Still fairly simple algebraic situations.

So let's look at those which have simply two operations.

This one has a constant on this side [ = 32 ]

On this side, the operation addition [ 3x + 5 = ]

and multiplication. [ 3 · x ]

Over here a constant on this side. [ = 2 ]

Over here [ M/7 - 3 = ] a subtraction and a division.

This strategy here will assist you

in peeling these numbers away from the variable, so to speak.

Let's apply those to these two situations.

Now notice our rule says to eliminate first

the plus or the minus. [ + or - ]

This problem has a plus. [ 3x + 5 ]

And recall that we undo adding

by subtracting. [ 3x + 5 - 5 = 32 - 5 ]

In this case, this subtraction has undone that addition [ 5 - 5 ]

leaving only the 3 times the x. [ 3 · x = ]

And of course 32 subtract 5 is 27. [ 32 - 5 = 27 ]

Then our rule says to eliminate either the times or division.

In this case, I have a times. [ 3 · x = ]

And recall you undo multiplication by division.

So this division [ 3 · x/3 ] has undone this multiplication

leaving the variable isolated [ x = ]

and of course 27 divide by 3 is 9. [ 27 ÷ 3 = 9 ]

[ x = 9 ] Then our strategy suggests that we check.

And check simply means we are claiming that

if x is 9 in my original equation, this equality will be true.

So let's try it.

This expression here says take 3 times x [ 3 · x ]

which now we're claiming is 9. [ 3 · 9 ]

So 3 times 9 and then add 5. [ 3 · 9 + 5 ]

But now we're in an order of operation situation.

So by order of operation we do this one first, [ 3 · 9 ]

not because it came first,

but because multiplication is done before addition

wherever it was.

So 3 times 9 is 27. [ 3 · 9 = 27 ]

And that added to 5 is 32. [ 27 + 5 = 32 ]

And that checks. [ 32 = 32 ]

Let's try that same strategy to the other problem.

Okay now again, our strategy suggests that we eliminate

either the additions or subtractions.

In this case it's a subtraction

and we undo subtraction by adding. [ M/7 - 3 + 3 = 2 + 3 ]

But addition of 3 undoes subtraction by 3 [ - 3 + 3 = 0 ]

and gets me to where I was before the subtraction. [ M/7 = ]

And of course this is 5. [ 2 + 3 = 5 ]

Then our strategy suggests that we eliminate

either the multiplications or divisions next.

We have a division here [ M/7 ]

and multiplication [ 7 · M/7 = 5 · 7 ] undoes division.

So this multiplication undoes this division, [ 7 · M/7 ]

isolating the variable, [ M = ]

and 5 times 7 in this case is 35. [ 5 x 7 = 35 ]

[ M = 35 ] Then check.

This says M, 35, we think, [ M/7 - 3 = 2 ]

divided by 7, subtract 3, we hope is 2. [ 35/7 - 3 = 2 ]

Well by order of operation I have to do division before subtraction

wherever it would have occurred,

so 35 divided by 7 is 5 [ 35/7 = 5 ]

and 5 minus 3 is 2. [ 5 - 3 = 2 ]

That also checks. [ 2 = 2 ]

So on this particular assignment

you will get a fair number of problems like these two.

They are fairly simple,

but they occur very frequently in formulas

and at the beginning of algebra.

So see you're really getting a running start at algebra,

which should make algebra a lot easier, when you get there,

for having taken this course

than those students who will not have taken this course.

That's one of the main reasons for this course.

For those of you who will not take algebra,

nevertheless whether it's in business math, consumer math,

you will still have many opportunities to use formulas,

and this will also give you a head start on that.

Let's work several more simple equations

before we return to a last review of order of operations.

As you'll discover in algebra or business math,

there's nothing saying that the variable

had to be on the left side of the equation or in the left member.

It could just as well be,

as it is here, in the right member. [ 37 = 4R - 7 ]

And as you will soon discover in algebra

in fact the variable usually is involved

in both members of the equation.

But nevertheless, the objective of solving an equation

as far as symbol pushing is concerned,

is to somehow isolate one variable,

1 times R, in this case, [ = 1 · R ] on one side

and the constants on the other side.

So our strategy suggested that first,

when we have combinations

of multiplications and/or divisions [ 4R ]

with subtractions and/or additions, [ 4R - 7 ]

that we undo the subtraction or addition first.

So we undo subtraction by adding 7. [ = 4R - 7 + 7 ]

Please note, it's not the 7 where our focus is

but the operation [ - ] we're trying to undo. [ + ]

The 7 is a secondary consideration.

But once we've decided it's addition that we want to do,

and addition of 7, then the Algebra Law,

called the Addition Law of Equality,

instructs we must add 7

to both sides of the equation. [ 37 + 7 = 4R - 7 + 7 ]

So this addition [ + 7 ] undoes this subtraction [ - 7 ]

leaving us with 4 times R [ 4 · R ]

in the right member of the equation. [ = 4 · R ]

and this becomes 44. [ 37 + 7 = 44 ]

[ 44 = 4 · R ] Then we undo multiplication.

See here again our concentration is upon the operation [ · ]

We undo multiplication by division, [ = 4 · R/4 ]

and we'll divide both sides. [ 44/ = 4 · R/ ]

Now secondarily

it's the multiplication of 4 that we wish to undo,

so we divide by 4. [ 44/4 = 4 · R/4 ]

So division by 4 undoes multiplication of 4

leaving us with one times R [ = R ] on this side of the equation.

Then completing the arithmetic of this side, [ 44/4 = 11 ]

we have 11. [ 11 = R ]

So we're claiming that 11 is the value for R. [ 11 = R ]

Or reading backwards, [ 11 = R ]

which you can do in some cases in mathematics, R is 11.

So we go back to the original equation

and then use the order of operation

which says I must do multiplication before subtraction.

4 times R, 4 times 11 is 44. [ 4 · 11 = 44 ]

Subtract 7 is 37. [ 44 - 7 = 37 ]

[ 37 = 37 ] So this checks.

Let's do one more.

Through this example we wish to again illustrate

that the variable may be on the right or the left side.

But at the same time to now remind ourselves

that there's no reason why these numbers need to be simple.

They could be very, very involved.

Most text books will start with simple numbers,

not because that's the way the world works,

but because that gives us an easy starting point

in learning this game of algebra.

When we begin to apply these to reality

these numbers become quite messy.

Decimals, fractions, very large numbers, very small numbers.

But in any case, where we have two operations

operating on this side with the variable

in our simple cases we're giving you in these lessons,

again you'll undo the addition or subtraction first.

So we undo adding [ A/15 + 102 ]

by subtracting 102 from both sides [131 - 102 = A/15 + 102 - 102]

And this subtraction of 102 [ 102 - 102 ] undoes adding 102,

leaving us with my unknown A,

being divided by 15. [ = A/15 ]

And then in this case, [ 131 - 102 = ]

following through on the subtraction, we get 29. [131 - 102 = 29]

So we're a step closer to isolating our variable.

So now we wish to undo division [A/15 ]

and you undo division by multiplication, [ ___· A/15 ]

and specifically [ ___· A/15 ] we're dividing by 15,

so we wish to multiply by 15, [ 15 · A/15 ]

but multiply both sides [ 15 · 20 = A/15 · 15 ]

So this multiplication by 15 undoes division by 15,

isolating the A, [ = A ]

which is the objective of our symbol pushing game of algebra.

Then over here simply completing

the indicated multiplication, [ 15 · 29 = ]

we would get 435. [ 15 · 29 = 435 ]

Then to check that this is correct, [ 435 = A ]

wherever there's an 'A' in my original equation,

we replace the 'A' by this number [ 435 ]

So doing that A, [ 131 = A/15 + 102 ]

which we claim is 435, [ A = 435 ]

being divided by 15 plus 102, [ 435/15 + 102 ]

Now we're back to our order of operations.

In order to evaluate this, [ 435/15 + 102 ]

that is to replace this expression by a single number,

which we're hoping will be 131,

which would tell us

that our procedure in solving this equation was correct.

So the order of operation says

I have to do a division before addition.

So 435 divided by 15 is 29. [ 435/15 = 29 ]

Then 29 plus 102 is 131, [ 29 + 102 = 131 ]

and our problem checks. [ 131 = 131 ]

Now this you'll find is a rather handy way

to begin to introduce yourself towards algebra.

Throughout this course we'll very, very

slowly and progressively make these equations more involved.

But in any one lesson it will be done so slowly,

you really won't feel it

until, by the end of this course,

you'll be working some relatively good algebra

before you even start the course.

And when you get into a course in algebra or business math,

believe me, you will appreciate that head start.

Another matter we'll try to help you tackle

while reviewing your basic arithmetic skills

is the matter of, what we used to call

in elementary and high school, 'story problems'

that is problems involving words as well as numbers.

You will find in time that the real problem with story problems

is not math; it's the words.

So what we want to do is begin to play

very slowly and very comfortably with this matter

of how do we go from words to math.

So let's take a few problems

and illustrate some of the pointers

we'll begin to give you in this lesson

and expand upon in future lessons.

And let's begin to build together

a small sort of words-to-math dictionary,

which we can sort of store away and eventually learn.

Let's begin with a specific problem.

In looking at this, many people when they see the word 'quotient'

they think that means 'to divide' and the answer's no, not quite.

And when they see the word 'sum,' they think it means 'to add.'

And the answer is, again, no, not quite.

And they see the word 'difference,'

and they believe it means 'to subtract,'

and the answer again is no, not quite.

And it's that 'no, not quite,' that bothers people

from going to math out of English

is that we have to be more precise

about what exactly these words mean,

and what a we mean by 'precisely' is when we say

that this [ quotient ] means to divide,

that's not quite the whole story.

What we really mean by the word 'quotient' is

the results of having divided.

This says a little bit more than just 'to divide.'

It says the results of having divided.

And this phrase [ the results ] is what's going to be our clue.

Likewise, the word 'product' does not mean to multiply

but the 'results of having multiplied.'

And likewise, the word 'sum' means the 'results of having added,'

not just 'to add,' but the 'results of having added.'

And 'difference' means in this same sense

the 'results of having subtracted.'

So now let's focus on this word 'results.'

And now the clinker.

The word 'results' in arithmetic and math

frequently, not always, means parentheses. ( )

So any time you say 'the results of' you're saying parentheses ( )

Frequently, when you have used the parentheses,

you'll find that you can drop them once again,

but if you're not sure

then any time you will have used the word 'results'

or 'results of,'

put parentheses on your paper at that moment.

Let's apply that to a problem using words.

This one:

What is the quotient of the sum of 21 and 49

and the difference of 42 and 7?

Our problem in reading this is

we're seeing too much at once and we're getting confused.

So as soon as we see enough

to go from the English words to a math mark,

let's do so before we even finish reading.

So we start by reading "What is the quotient of" right there,

that's enough to say 'the results of having divided.'

So immediately I know I'm going to divide

and that tells me I need to divide two quantities,

and the 'results' is the total thing I get having divided.

So now I'm looking for these two pieces.

So I keep reading expecting to see the first piece,

and I get the word the 'sum.'

So that tells me, my first piece of my division process

is going to be the results of having added.

So now I look for the two pieces being added,

which is 21 and 49. (( 21 + 49 ) ÷ )

Now I look for the other piece involved in the division.

So 'and,' that tells me I'm getting something else,

the difference. (( 21 + 49 ) ÷ ( - ))

And the difference is the results of having subtracted.

What? 42 and 7. (( 21 + 49 ) ÷ ( 42 - 7 ))

So the results of dividing, (( 21 + 49 ) ÷ ( 42 - 7 ))

the results of adding 21 and 49, that sum, [ ( 21 + 49 ) ]

and the results of the difference

or subtracting 42 and 7. [ ( 42 - 7 ) ]

Then at this point

it's simply a matter of the order of operation,

get inside of the parentheses. (( 21 + 49 ) ÷ ( 42 - 7 ))

Once we're here, in this case, there's going to be

only one operation, division, so we didn't need these,

and continuing to get inside of parentheses,

the first one first, there's only addition to be done.

And we would do that and this whole quantity parentheses

would be replaced by the number 70. [ ( 21 + 49 ) = 70 ]

And we don't do this [ ÷ ]

until we've done what's inside the parentheses first,

which is 42 subtract 7 which is 35. [ ( 42 – 7 ) = 35 ]

And now we only have division left

between 70 and 35, [ 70 ÷ 35 ]

so 70 divide by 35 is 2. [ 70 ÷ 35 = 2 ]

Note this, though.

Once we got rid of the words,

that is we translated correctly from words to symbols,

the symbols, through the order of operation,

told us exactly what to do and when.

That's first how powerful the order of operations are,

and secondly, how powerful it is

to know precisely what any word is saying

rather than being just too loose and mushy with words

as unfortunately all of us tend to be

with those things that we're not quite familiar with, yet.

One more before our lesson ends.

What is the difference?

That's the results of having subtracted.

Of the product?

Product means the results of having multiplied

8 and 9. [ ( 8 · 9 ) ]

And the quotient?

The quotient is the results of having divided

325 and 13. [ ( 325 ÷ 13 ) ]

[ (8 · 9) - (325 ÷ 13) ] Now the order of operation takes over

doing what's in parentheses first. [ ( 8 · 9 )= 72 ]

What's in parentheses first. [ ( 325 ÷ 13 ) ]

This might take a little bit more paper,

but doing it we get 25. [ ( 325 ÷ 13 ) = 25 ]

And then finally subtracting these two [ 72 - 25 ]

because it's telling us to, and we get 47. [ 72 - 25 = 47 ]

That's rather fast,

but we will review this more and more in future lessons.

Particularly, you'll see several more in the next lesson,

so, until then this is your host, Bob Finnell.

A Course in Arithmetic Review

Produced at Portland Community College.

Let's start this lesson

by adding to our list of technical vocabulary.

The word 'operation,' as used in arithmetic, simply means:

those symbols

which tell us what we're going to do

to either a number or pairs of numbers.

If you are operating on two numbers

by the plus, division, minus, and times, [ + ÷ - x ]

in math that's referred to as a 'binary' operation;

whereas, if your operation only needs one number to work upon,

than that's referred to as a 'unary' operation.

Quite frequently in advanced formulas in mathematics

when you do any kind of operation you don't do this or that.

You do a whole string of them at once,

so the question arises of all of these operations,

which do I do first?

That is, we need some kind of order for doing operations.

If it would nice if we simply agreed to go from front to back,

but that's not the way the system evolved.

So these are the order of doing operations

in all of your future mathematical work.

This is the order that computers evaluate arithmetic expressions.

It is also the order in which

calculators which have algebraic logic, scientific calculators,

will work in this order.

So first you do what's inside parentheses.

Once you're there,

you do the exponents first as you come to them.

And you go back to the beginning

and do the multiplications and/or divisions as you come to them.

Make special note:

you don't do the multiplications first then the division.

You do whichever of these two come first from beginning to end,

then back to the beginning

and do the additions and/or subtractions as you come to them

front to back.

If you come to a subtraction first, do it first.

Addition first, do it first.

So let's apply that to the problem we just had.

First we're going to get inside the parentheses.

And so we get inside the first parentheses that we see

reading left to right. [ ( 24 ÷ 2 · 3 ) ]

Now that we're inside, [ ( 24 ÷ 2 · 3 ) ]

we look for exponents.

There aren't any.

Then you look for times [x] and/or divisions [÷]

and do them as you come to them.

So there is a division. [ 24 ÷ 2 ]

It's between 24 and 2, [ 24 ÷ 2 ]

so I'll do this first, and that gives me 12. [ 24 ÷ 2 = 12 ]

Then look for more. [ ( 24 ÷ 2 · 3 ) ]

See 12 now has really replaced [ ( 12 · 3 ) ]

that entire bit. [ 24 ÷ 2 = 12 ]

Now looking for more. [ ( 12 · 3 ) ] There it is.

It's between 12, which was a result of that entry, so we do that.

12 times 3 is 36, [ 12 · 3 = 36 ]

so this whole bit here [ ( 24 ÷ 2 · 3 ) ]

has become 36. [ ( 24 ÷ 2 · 3 ) = 36 ]

Now at this point to keep from becoming confused,

let's just copy the rest of the problem. 5(36) + (2 · 3² - 4 + 1)

Now that we've reduced one parentheses, our first one,

to a single number, [36] we start looking for more parentheses.

And that would be this one. [ (2 · 3² - 4 + 1) ]

Now that we're inside the parentheses, back to the beginning,

look for exponents as you come to them.

That's not. [ 2 ] That is [ 3² ]

It's above the three [ 3² ]

so it says: take two 3s [ 3 · 3 ] and multiply them together.

3 times 3 is 9, [ 3 · 3 = 9 ]

so that [ 3² ] becomes replaced by 9.

Now continue to look for more exponents

to the end of the parentheses.

There aren't any.

Then back to the beginning.

Next you look for multiplications and/or divisions

as you come to them. [ (2 · 9 - 4 + 1) ]

There's a multiplication. [ 2 · 9 ]

It's between the 9 and the 2, so 9 times 2 is 18. [ 9 · 2 = 18 ]

So all of this [ (2 · 3²) ] has been replaced by 18.

Then we look for more multiplications.

There aren't any.

Back to the beginning of that parentheses.

And our last step,

look for additions and/or subtractions as you come to them.

18 minus, [ ( 18 - 4 + 1 ) ] I come to the subtraction first.

It's between those two.

So I do it.

So 18 minus 14 is [ 18 - 14 =

18 minus 4 is 14. [ 18 - 4 = 14 ]

Okay keep looking for more.

There is an addition.

It's between the 1 and the 14 I just did, [ ( 14 + 1 ) ]

so one plus 14 is 15. [ 1 + 14 = 15 ]

Now I'll just rewrite my problem [ 5(36) + 15 ]

to keep from getting confused in the growing mess.

Look for more parentheses.

There aren't any.

Back to the beginning.

Back to the beginning. There were no more parentheses,

so now we look for exponents.

There aren't any.

Then look for multiplications and/or divisions

as you come to them.

There is one. [ 5(36) + 15 ]

It's not written, but it is there. [ 5 · (36) ]

It says 5 times 36, which is 180. [ 5 · 36 = 180 ]

And now of course, there's just

the 15 being added left, [ 180 + 15 ]

so doing that we have 195. [ 180 + 15 = 195 ]

This process of reducing

an arithmetic expression [ 5( 24 ÷ 2 · 3) + (2 · 3² - 4 + 1) ]

to a single number [195] we call 'evaluating.'

These order of operations must be memorized as soon as possible,

at least by the end of your current lesson.

So first you always do

what's inside parentheses as you come to it.

Exponents next as you come to them.

Multiplications and/or divisions as you come to them.

Don't do these [ +, - ] until you finish these. [ x ÷ ]

And back to the beginning.

Additions and/or subtractions as you come to them.

A very important lesson.

Learn it very soon.

Up to now in our preparation for algebra

we've dealt with equations where the variables is on one side

and there's only one operation associated

between that variable and a constant.

Now that we're in the order of operation

which deals with multiple operations at once,

it's a good time to handle equations

which begin to have

more than one operation associated with the variable.

Still fairly simple algebraic situations.

So let's look at those which have simply two operations.

This one has a constant on this side [ = 32 ]

On this side, the operation addition [ 3x + 5 = ]

and multiplication. [ 3 · x ]

Over here a constant on this side. [ = 2 ]

Over here [ M/7 - 3 = ] a subtraction and a division.

This strategy here will assist you

in peeling these numbers away from the variable, so to speak.

Let's apply those to these two situations.

Now notice our rule says to eliminate first

the plus or the minus. [ + or - ]

This problem has a plus. [ 3x + 5 ]

And recall that we undo adding

by subtracting. [ 3x + 5 - 5 = 32 - 5 ]

In this case, this subtraction has undone that addition [ 5 - 5 ]

leaving only the 3 times the x. [ 3 · x = ]

And of course 32 subtract 5 is 27. [ 32 - 5 = 27 ]

Then our rule says to eliminate either the times or division.

In this case, I have a times. [ 3 · x = ]

And recall you undo multiplication by division.

So this division [ 3 · x/3 ] has undone this multiplication

leaving the variable isolated [ x = ]

and of course 27 divide by 3 is 9. [ 27 ÷ 3 = 9 ]

[ x = 9 ] Then our strategy suggests that we check.

And check simply means we are claiming that

if x is 9 in my original equation, this equality will be true.

So let's try it.

This expression here says take 3 times x [ 3 · x ]

which now we're claiming is 9. [ 3 · 9 ]

So 3 times 9 and then add 5. [ 3 · 9 + 5 ]

But now we're in an order of operation situation.

So by order of operation we do this one first, [ 3 · 9 ]

not because it came first,

but because multiplication is done before addition

wherever it was.

So 3 times 9 is 27. [ 3 · 9 = 27 ]

And that added to 5 is 32. [ 27 + 5 = 32 ]

And that checks. [ 32 = 32 ]

Let's try that same strategy to the other problem.

Okay now again, our strategy suggests that we eliminate

either the additions or subtractions.

In this case it's a subtraction

and we undo subtraction by adding. [ M/7 - 3 + 3 = 2 + 3 ]

But addition of 3 undoes subtraction by 3 [ - 3 + 3 = 0 ]

and gets me to where I was before the subtraction. [ M/7 = ]

And of course this is 5. [ 2 + 3 = 5 ]

Then our strategy suggests that we eliminate

either the multiplications or divisions next.

We have a division here [ M/7 ]

and multiplication [ 7 · M/7 = 5 · 7 ] undoes division.

So this multiplication undoes this division, [ 7 · M/7 ]

isolating the variable, [ M = ]

and 5 times 7 in this case is 35. [ 5 x 7 = 35 ]

[ M = 35 ] Then check.

This says M, 35, we think, [ M/7 - 3 = 2 ]

divided by 7, subtract 3, we hope is 2. [ 35/7 - 3 = 2 ]

Well by order of operation I have to do division before subtraction

wherever it would have occurred,

so 35 divided by 7 is 5 [ 35/7 = 5 ]

and 5 minus 3 is 2. [ 5 - 3 = 2 ]

That also checks. [ 2 = 2 ]

So on this particular assignment

you will get a fair number of problems like these two.

They are fairly simple,

but they occur very frequently in formulas

and at the beginning of algebra.

So see you're really getting a running start at algebra,

which should make algebra a lot easier, when you get there,

for having taken this course

than those students who will not have taken this course.

That's one of the main reasons for this course.

For those of you who will not take algebra,

nevertheless whether it's in business math, consumer math,

you will still have many opportunities to use formulas,

and this will also give you a head start on that.

Let's work several more simple equations

before we return to a last review of order of operations.

As you'll discover in algebra or business math,

there's nothing saying that the variable

had to be on the left side of the equation or in the left member.

It could just as well be,

as it is here, in the right member. [ 37 = 4R - 7 ]

And as you will soon discover in algebra

in fact the variable usually is involved

in both members of the equation.

But nevertheless, the objective of solving an equation

as far as symbol pushing is concerned,

is to somehow isolate one variable,

1 times R, in this case, [ = 1 · R ] on one side

and the constants on the other side.

So our strategy suggested that first,

when we have combinations

of multiplications and/or divisions [ 4R ]

with subtractions and/or additions, [ 4R - 7 ]

that we undo the subtraction or addition first.

So we undo subtraction by adding 7. [ = 4R - 7 + 7 ]

Please note, it's not the 7 where our focus is

but the operation [ - ] we're trying to undo. [ + ]

The 7 is a secondary consideration.

But once we've decided it's addition that we want to do,

and addition of 7, then the Algebra Law,

called the Addition Law of Equality,

instructs we must add 7

to both sides of the equation. [ 37 + 7 = 4R - 7 + 7 ]

So this addition [ + 7 ] undoes this subtraction [ - 7 ]

leaving us with 4 times R [ 4 · R ]

in the right member of the equation. [ = 4 · R ]

and this becomes 44. [ 37 + 7 = 44 ]

[ 44 = 4 · R ] Then we undo multiplication.

See here again our concentration is upon the operation [ · ]

We undo multiplication by division, [ = 4 · R/4 ]

and we'll divide both sides. [ 44/ = 4 · R/ ]

Now secondarily

it's the multiplication of 4 that we wish to undo,

so we divide by 4. [ 44/4 = 4 · R/4 ]

So division by 4 undoes multiplication of 4

leaving us with one times R [ = R ] on this side of the equation.

Then completing the arithmetic of this side, [ 44/4 = 11 ]

we have 11. [ 11 = R ]

So we're claiming that 11 is the value for R. [ 11 = R ]

Or reading backwards, [ 11 = R ]

which you can do in some cases in mathematics, R is 11.

So we go back to the original equation

and then use the order of operation

which says I must do multiplication before subtraction.

4 times R, 4 times 11 is 44. [ 4 · 11 = 44 ]

Subtract 7 is 37. [ 44 - 7 = 37 ]

[ 37 = 37 ] So this checks.

Let's do one more.

Through this example we wish to again illustrate

that the variable may be on the right or the left side.

But at the same time to now remind ourselves

that there's no reason why these numbers need to be simple.

They could be very, very involved.

Most text books will start with simple numbers,

not because that's the way the world works,

but because that gives us an easy starting point

in learning this game of algebra.

When we begin to apply these to reality

these numbers become quite messy.

Decimals, fractions, very large numbers, very small numbers.

But in any case, where we have two operations

operating on this side with the variable

in our simple cases we're giving you in these lessons,

again you'll undo the addition or subtraction first.

So we undo adding [ A/15 + 102 ]

by subtracting 102 from both sides [131 - 102 = A/15 + 102 - 102]

And this subtraction of 102 [ 102 - 102 ] undoes adding 102,

leaving us with my unknown A,

being divided by 15. [ = A/15 ]

And then in this case, [ 131 - 102 = ]

following through on the subtraction, we get 29. [131 - 102 = 29]

So we're a step closer to isolating our variable.

So now we wish to undo division [A/15 ]

and you undo division by multiplication, [ ___· A/15 ]

and specifically [ ___· A/15 ] we're dividing by 15,

so we wish to multiply by 15, [ 15 · A/15 ]

but multiply both sides [ 15 · 20 = A/15 · 15 ]

So this multiplication by 15 undoes division by 15,

isolating the A, [ = A ]

which is the objective of our symbol pushing game of algebra.

Then over here simply completing

the indicated multiplication, [ 15 · 29 = ]

we would get 435. [ 15 · 29 = 435 ]

Then to check that this is correct, [ 435 = A ]

wherever there's an 'A' in my original equation,

we replace the 'A' by this number [ 435 ]

So doing that A, [ 131 = A/15 + 102 ]

which we claim is 435, [ A = 435 ]

being divided by 15 plus 102, [ 435/15 + 102 ]

Now we're back to our order of operations.

In order to evaluate this, [ 435/15 + 102 ]

that is to replace this expression by a single number,

which we're hoping will be 131,

which would tell us

that our procedure in solving this equation was correct.

So the order of operation says

I have to do a division before addition.

So 435 divided by 15 is 29. [ 435/15 = 29 ]

Then 29 plus 102 is 131, [ 29 + 102 = 131 ]

and our problem checks. [ 131 = 131 ]

Now this you'll find is a rather handy way

to begin to introduce yourself towards algebra.

Throughout this course we'll very, very

slowly and progressively make these equations more involved.

But in any one lesson it will be done so slowly,

you really won't feel it

until, by the end of this course,

you'll be working some relatively good algebra

before you even start the course.

And when you get into a course in algebra or business math,

believe me, you will appreciate that head start.

Another matter we'll try to help you tackle

while reviewing your basic arithmetic skills

is the matter of, what we used to call

in elementary and high school, 'story problems'

that is problems involving words as well as numbers.

You will find in time that the real problem with story problems

is not math; it's the words.

So what we want to do is begin to play

very slowly and very comfortably with this matter

of how do we go from words to math.

So let's take a few problems

and illustrate some of the pointers

we'll begin to give you in this lesson

and expand upon in future lessons.

And let's begin to build together

a small sort of words-to-math dictionary,

which we can sort of store away and eventually learn.

Let's begin with a specific problem.

In looking at this, many people when they see the word 'quotient'

they think that means 'to divide' and the answer's no, not quite.

And when they see the word 'sum,' they think it means 'to add.'

And the answer is, again, no, not quite.

And they see the word 'difference,'

and they believe it means 'to subtract,'

and the answer again is no, not quite.

And it's that 'no, not quite,' that bothers people

from going to math out of English

is that we have to be more precise

about what exactly these words mean,

and what a we mean by 'precisely' is when we say

that this [ quotient ] means to divide,

that's not quite the whole story.

What we really mean by the word 'quotient' is

the results of having divided.

This says a little bit more than just 'to divide.'

It says the results of having divided.

And this phrase [ the results ] is what's going to be our clue.

Likewise, the word 'product' does not mean to multiply

but the 'results of having multiplied.'

And likewise, the word 'sum' means the 'results of having added,'

not just 'to add,' but the 'results of having added.'

And 'difference' means in this same sense

the 'results of having subtracted.'

So now let's focus on this word 'results.'

And now the clinker.

The word 'results' in arithmetic and math

frequently, not always, means parentheses. ( )

So any time you say 'the results of' you're saying parentheses ( )

Frequently, when you have used the parentheses,

you'll find that you can drop them once again,

but if you're not sure

then any time you will have used the word 'results'

or 'results of,'

put parentheses on your paper at that moment.

Let's apply that to a problem using words.

This one:

What is the quotient of the sum of 21 and 49

and the difference of 42 and 7?

Our problem in reading this is

we're seeing too much at once and we're getting confused.

So as soon as we see enough

to go from the English words to a math mark,

let's do so before we even finish reading.

So we start by reading "What is the quotient of" right there,

that's enough to say 'the results of having divided.'

So immediately I know I'm going to divide

and that tells me I need to divide two quantities,

and the 'results' is the total thing I get having divided.

So now I'm looking for these two pieces.

So I keep reading expecting to see the first piece,

and I get the word the 'sum.'

So that tells me, my first piece of my division process

is going to be the results of having added.

So now I look for the two pieces being added,

which is 21 and 49. (( 21 + 49 ) ÷ )

Now I look for the other piece involved in the division.

So 'and,' that tells me I'm getting something else,

the difference. (( 21 + 49 ) ÷ ( - ))

And the difference is the results of having subtracted.

What? 42 and 7. (( 21 + 49 ) ÷ ( 42 - 7 ))

So the results of dividing, (( 21 + 49 ) ÷ ( 42 - 7 ))

the results of adding 21 and 49, that sum, [ ( 21 + 49 ) ]

and the results of the difference

or subtracting 42 and 7. [ ( 42 - 7 ) ]

Then at this point

it's simply a matter of the order of operation,

get inside of the parentheses. (( 21 + 49 ) ÷ ( 42 - 7 ))

Once we're here, in this case, there's going to be

only one operation, division, so we didn't need these,

and continuing to get inside of parentheses,

the first one first, there's only addition to be done.

And we would do that and this whole quantity parentheses

would be replaced by the number 70. [ ( 21 + 49 ) = 70 ]

And we don't do this [ ÷ ]

until we've done what's inside the parentheses first,

which is 42 subtract 7 which is 35. [ ( 42 – 7 ) = 35 ]

And now we only have division left

between 70 and 35, [ 70 ÷ 35 ]

so 70 divide by 35 is 2. [ 70 ÷ 35 = 2 ]

Note this, though.

Once we got rid of the words,

that is we translated correctly from words to symbols,

the symbols, through the order of operation,

told us exactly what to do and when.

That's first how powerful the order of operations are,

and secondly, how powerful it is

to know precisely what any word is saying

rather than being just too loose and mushy with words

as unfortunately all of us tend to be

with those things that we're not quite familiar with, yet.

One more before our lesson ends.

What is the difference?

That's the results of having subtracted.

Of the product?

Product means the results of having multiplied

8 and 9. [ ( 8 · 9 ) ]

And the quotient?

The quotient is the results of having divided

325 and 13. [ ( 325 ÷ 13 ) ]

[ (8 · 9) - (325 ÷ 13) ] Now the order of operation takes over

doing what's in parentheses first. [ ( 8 · 9 )= 72 ]

What's in parentheses first. [ ( 325 ÷ 13 ) ]

This might take a little bit more paper,

but doing it we get 25. [ ( 325 ÷ 13 ) = 25 ]

And then finally subtracting these two [ 72 - 25 ]

because it's telling us to, and we get 47. [ 72 - 25 = 47 ]

That's rather fast,

but we will review this more and more in future lessons.

Particularly, you'll see several more in the next lesson,

so, until then this is your host, Bob Finnell.