Equations--FullProgram.mp4

Algebra Nspirations

Variables, Equations and Functions: The Foundation of Algebra

From the mathematics of old Babylon and Egypt, to the

mathematics of ancient Greece, China, India, the Islamic

world, and all the way up to the Renaissance and the

present, equations have played a central role in the

development of algebra.

But algebraic equations haven't always been written in

present day symbolic form.

Hello.

My name is Monica Neagoy.

And I will guide your exploration into the world of

variables and equations with the help of the TI-Nspire.

In this program, I'll describe the fundamentals of the

equations and introduce the two most basic types of

equations you'll encounter in a beginning algebra class,

linear and quadratic.

In the first investigation, we'll do a quick historical

overview of equations, examine the building blocks of

equations, and look at the multiple uses of

variables in equations.

In the second investigation, we'll dive into solving

equations, beginning with linear equations and then

quadratic equations.

Two aspects of the history of algebra are of interest here,

the evolution of its notation and the development of its

subject matter.

Namely, what is algebra?

Developed by the Babylonians, rhetorical algebra was

dominant until the 1500s.

Rhetorical algebra was predominantly oral, as writing

materials were limited in most cultures.

Algebraic symbols that are

commonplace today were unknown.

Whenever mathematicians did transcribe algebraic equations

onto clay tablets, for instance, they

wrote out every word.

For example, the equation 3x equals 1/2 x plus 10 would

have looked something like this.

"Three times a certain quantity has the same value as

one half the same quantity increased by ten."

The second stage is called syncopated algebra.

Here we begin to see abbreviations for unknown

quantities and for frequently used operations.

But these varied from country to country, as each country

created its own.

And syncopated algebra was not consistent, as it did not

follow clearly stated rules.

The third and final stage, called symbolic algebra, began

with the Renaissance mathematicians.

From the mid-1500s to about 1700, algebra symbols for

operations, variables, relations, grouping,

exponents, and so on were born, evolved, and matured.

So algebra's symbolic notation used around the world today

has only been used for roughly 300 years.

Regarding algebra's subject matter, it's fair to say that

up until the 18th century, algebra was

about solving equations.

This algebra, called elementary algebra, is a

generalization of arithmetic and is the focus of most high

school algebra courses.

During the 19th century, a new kind of algebra was born.

It's called modern algebra or abstract algebra.

While this advanced algebra is beyond our scope, you can

still remember two important points.

The variables don't always represent generalized numbers,

but other objects or structures.

And therefore, the properties of operations among those

objects don't always hold.

For example, A times B equals B times A is not true if A and

B are matrices.

Even if modern algebra is beyond our scope, it's

important to know that algebra has different meanings

depending on what mathematical objects are being studied.

We're now ready to examine the building blocks of equations.

There are three main components of the basic a

algebraic equation.

Quantities that change the values are called variables

and are usually denoted by letters at the end of the

alphabet, such as x, y, and z.

Quantities that are fixed are called constants and appears

real numbers, namely natural numbers, integers, rational

numbers, or decimals, or irrational numbers.

The main symbol in an equation is the equal sign, indicating

that the left and right sides must be in perfect balance.

Other symbols include operation signs, fraction

lines, parentheses, exponents, et cetera.

Parameters are another component

of generalized equations.

And they are denoted by letters at the beginning of

the alphabet, such as a, b, and c.

The following example illustrates the

meaning of a parameter.

These three equations are different, but

have a common form.

On the left, a constant multiplies a variable.

And then another constant is added on.

On the right, there's a 0.

It can be generalized by the equation, ax plus b equals 0,

where a and b, letters from the beginning of the alphabet,

are

Constants, and x, a letter from the end of the alphabet,

is a variable.

a and b are called parameters, representing constants that

vary from one equation to the next.

If you're asked to solve for the variable, x, in this case,

it's called the unknown.

And ax plus b equals 0 is actually the general form of a

linear equation in one unknown, which

we'll explore later.

Equations can be tricky in algebra, since not all of them

begged to be solved.

Let's take a closer look at the multiple uses of variables

in equations.

There are formulas, some of which may be familiar to you,

like the perimeter of a rectangle or

the area of a triangle.

Formulas provide useful information, but do not need

to be solved per se.

Here, variables are labels.

There are special identities that all algebra students

should know.

These are called the square of the sum of two numbers and the

square of the difference of two numbers.

The variables here stand for all real numbers and not

unknowns to solve for.

You also have special properties of operations,

which you'll also learn, such as the associative property of

addition or the commutative property of multiplication.

Here again, variables generalized real numbers.

Finally, you have equations to be solved

with one or more unknown.

There are lots of different kinds.

This is a linear equation in x and this, a

quadratic equation in w.

You'll learn how to solve for the unknown.

But first, try this.

Take the equation we just saw, 0.75x plus 10 equals 50.

This is a linear equation in one unknown, x.

It's linear because the exponent of x is 1.

x is actually x to the first power.

If you're thinking linear has something to do with line,

you're right.

You'll see the connection shortly.

But let's begin with the algebraic solution.

Subtracting 10 from both sides does not change the balance of

this equation.

It yields 0.75x plus 0 equals 40, or 0.75x equals 40.

Next, dividing both sides by 0.75, again without changing

the equation in any way, we obtain 1x equals 40 over 0.75,

or x equals 53.3 repeating.

So 53.3 repeating is the unique value of x that makes

our equation true.

We also say that x equals 53.3 repeating

satisfies our equation.

This exercise illustrates what school algebra has consisted

of for hundreds of years.

The search for fixed value that fit

statically defined equations.

I call this the static approach to algebra, which

mainly explores the symbolic world.

Thanks to 21st century educational technology, you

can easily explore the multiple worlds of algebra.

Words that describe a real world problem accompanied in

this video series by live footage.

Tables that lists numerical values for variables.

Graphs that show how variables actually vary.

And of course, symbols that are used to write equations.

Working in these different worlds gives more meaning to

be often dry, symbolic equations because we explore

their multiple representations and the relationships both

within and among these representations.

I call this the dynamic approach to algebra.

And it's time to revisit our equation to illustrate it.

Let's start by creating a real world connection.

I bet you're familiar with walk-a-thons and bike-a-thons.

Perhaps, you've participated in one.

They're a nice way to mobilize community support to raise

money for a good cause, such as research on AIDS or cancer.

Participants raise money by collecting pledges for walking

or biking a predetermined distance.

Let's imagine that your school organized a bike-a-thon to

help send kids to camp who can't afford to pay their way.

Suppose you found a generous sponsor to pledge $0.75 for

each kilometer you bike plus a $10 donation.

How many kilometers must you personally ride to raise $50?

Let x be the number of kilometers

you'll need to ride.

The algebraic expression for the amount of money pledged is

$0.75 times each kilometer, x, plus the $10 donation.

If you wish to raise $50, set this expression equal to 50

and solve for x.

Your solution x equals 53 and 1/3 third means you would have

to bike 53 and 1/3 kilometers to raise $50.

I hope you're in good shape, because that's about 33 miles.

Let's generalize now.

Instead of $50, let's call y the amount of money you'd

raise for biking x kilometers.

Our new equation relating x and y is 0.75x plus 10 equals

y, or y equals 0.75x plus 10.

Since y depends on x, we see that y is a function of x.

And we write y equals f of x.

Next, let's use the TI-Nspire to explore the tabular and

graphical forms of the function.

Turn on the TI-Nspire.

Press the home key to open a new document.

If a previous document is open, a prompt asks if you

wish to save it.

Click to choose Yes or press Tab then click to choose No.

Create a Graphs and Geometry page.

The blinking cursor is on the function entry

line by F1 of x.

Type in 0.75x plus 10 for the money pledged.

Press Enter to graph.

Since you hope to bike more than 20 kilometers and you

want positive values, let's change the window.

Use the navigation pad or Nav Pad to move the arrow to a

clear area of quadrant one.

Press and hold the click button down until the pointer

changes to a grasping hand.

Use the Nav Pad again to drag the origin to the bottom left

corner of the monitor.

Press Escape to exit grab and drag mode.

Next, press Menu.

Under Window, select Zoom-Out.

Move the center box to the center of your monitor and

then click twice.

Press Escape.

Press Menu again.

And under Trace, select Trace setting.

Press the down arrow and select Enter Value.

Type 10 then Tab to OK and click.

Go back to Trace again.

And this time, select Graph Trace.

Your cursor is on the point 0,10, meaning if you walk 0

kilometers, you'll raise $10, your

sponsor's upfront donation.

Press the right arrow several times.

As x increases, so does y.

This yields a rising graph.

These points show the money you'll raise in the y value

for consecutive 10 kilometers increments of the x value.

Next, press Menu.

Under View, select Add Function Table.

The graph and table are now side by side.

Scrolling down the function column with the down arrow,

you can see all the points on the line that have an integer

x-coordinate.

While 50 is not listed in the y column, it corresponds

indeed to an x value between 53 and 54.

To end, press Control Tab to switch back to the graph.

Type 53.333 then press Enter.

Your cursor is now on the solution

point we found earlier.

Biking 53 and 1/3 kilometers will yield $49.999 dollars,

which concretely is $50.

So an equation has a symbolic, graphic, and tabular

representation and can sometimes model

a real world problem.

These help you see the relationship between x and y

dynamically.

Solving for x, given a desired y value, yields

one specific point.

Calculating y, given in desired x value, say 25, also

yields a specific point.

There are infinite ordered pairs, (x,y), that satisfy the

equation y equals 0.75x plus 10.

And each pair is a point on the functions graph.

And now, onto our quadratic equation.

Since this ax plus b equals 0 is the most basic form of a

linear equation in one unknown, ax squared plus bx

plus c equals 0 is the most basic form of a quadratic

equation in one unknown.

Linear, because the highest exponent x is 1.

Quadratic, because the highest exponent is two.

Like in the linear case, let's begin with a simple equation,

solve it algebraically, then generalize into a two variable

quadratic, and make sense of it by examining different

representations.

Take the equation negative 2x squared plus 40x equals 150.

One way of solving algebraically is to set one

side equal to 0 and factor the other.

It doesn't matter what side the 0 is on.

Subtracting negative 2x squared plus 40x from both

sides yields 0 equals 2x squared minus 40x plus 150.

Factoring the right side gives 2 quantity x squared

minus 20x plus 75.

75 is five times 15.

And their sum is conveniently 20.

So factoring further gives 2 times quantity x minus 5 times

quantity x minus 15.

For a product of three factors to equal 0, at least one of

them must be 0.

2 can't be 0, because it's 2.

x minus 5 is 0 when x is 5.

And x minus 15 is 0 when x is 15.

So 5 and 15 are the solutions of our equation, which can be

written equivalently in many ways.

The solutions are also called the zeros or the roots, simply

because they make the quadratic expression 0.

Next, the real story behind this equation

I created for you.

Suppose you recently acquired a pet and wanted to design a

rectangular pen in your backyard.

You have 40 feet of fence.

And you're using the building wall for one of the sides of

the rectangle.

What dimensions would maximize the area of

the rectangular pen?

From the diagram, we see that the fence part of the

perimeter is 2w plus l.

And this totals 40 feet.

Solving this equation for l in terms of w gives l

equals 40 minus 2w.

The area of the rectangle is width times length.

Plugging in 40 minus 2w for l gives 40w minus 2w squared, or

negative 2w squared plus 40w.

Here, the area is a function of one variable, w, the width.

If you wanted an area of 150 square feet, for example,

you'd set negative 2w squared plus 40w equals 150.

And you'd obtain the equation we just solved.

Instead, lets call y the general area and go back to

the TI-Nspire to graph our new quadratic area function.

Remember, w will be x on our handheld.

Open a New Document.

Save or Delete the old one.

Select Graphs and Geometry.

Enter negative 2x squared plus 40x for f1 of x.

Press Enter to graph.

Press Menu.

And under Window, select Window Settings.

Enter the following, negative 5, 25, negative 100, and 225.

Press Tab from one to the next and select OK.

This graph of a quadratic function is called a parabola.

Press Menu and under Trace, select graph trace.

The cursor is on the origin.

And the z indicates the function's first 0.

A 0 width obviously gives 0 area.

Factoring f of x gives 20 as our second 0.

So type 20 next, then press Enter.

That's our second 0.

A width of 20 gives a length of 0, which

also yields a 0 area.

Notice that the functions 0 occur on the

x-axis, where y is 0.

Press Menu.

And under View, select Function Table.

Scroll down and you'll see all the points we've encountered.

0,0, 5, 150, 15, 150, and 20, 0.

Scroll back up and you'll find the maximum area of 200 when

the width is 10.

In this case, the length would be 20.

And 10 times 20 is 200 square feet.

Lastly, press control Tab to switch back to the graph.

Type pen then press Enter.

The maximum area corresponds to the highest

point of the graph.

Solving a specific quadratic equation in one variable gives

finite solutions.

Graphing the corresponding quadratic function reveals

those solutions and infinitely many more.

Gradually, you'll begin to appreciate the connections

between equations and their solutions

graphing their points.

I hope you'll solve other equations algebraically and

using the TI-Nspire.

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