Algebra Nspirations

Solving Systems of Equations

As an algebra student, you've learned by now that many day

to day operations in business, science, government, and

elsewhere can be described by equations

in one or more variables.

The variables represent quantities that change, and

the equations model relations among the variables.

But perhaps more important in all fields of study, are

problem situations involving, not only two or more

variables, but also two or more equations.

Such multi-variable and multi-relation mathematical

models are called systems.

Solving algebraic systems of equations will expand your

knowledge of variables, equations, and functions.

In the process, you'll also become more proficient with

the TI-Nspire technology.

We'll start by solving linear systems with the

Nspire in many ways.

Then, algebraically, using two classical methods.

And finally, we'll summarize the different types of

solutions you can encounter.

Without further delay let's go straight to our first

investigation.

Have you ever found yourself in the situation of having to

decide between two competitive cellphone plans?

Or having to choose between two rental car companies for a

short out-of-town trip?

Systems of equations really come in handy for these types

of problems.

Suppose that you and a few friends or relatives need to

make an out-of-town trip for the day.

You're debating between two companies.

EZ Ride charges $0.20 per mile for the compact car of your

choice, plus a fixed daily fee of $22.

For the same car, Just4U, Just charges for mileage at a rate

of $0.30 per mile.

Which offer would you choose?

First, we need to formulate equations for your total cost

y as a function of your number of miles x for each company.

$0.20 or $0.2 for each mile translates into 0.2 times x.

We add to that 22 for the fixed daily fee of $22.

For Just4U, it's just $0.30 times the number of miles x.

Jot down these equations for our next step.

In a technological world, it's not only important to know how

to use technology, but also, to use it efficiently.

The Nspire offers several representations you can use to

solve this problem.

I'd like to walk you through four of them, and then,

depending on the particular problem, you'll decide which

method is most efficient.

The first approach uses the spreadsheet application.

Turn on the TI Nspire, press Home, Open a New Document.

If another document is open, you'll be prompted to Save it.

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

Create a New Spreadsheet Page.

Arrow up to the top of Column A and type x for miles, our

independent variable.

Arrow down one line to the Formula Line, press Menu, and

under Data, select Generate Sequence.

We're going to generate the sequence of every 10 miles

starting with zero.

Head down, and enter zero for the starting value.

Enter 30 for the maximum number of terms, which stands

for 300 miles.

Then tab down to OK and click.

You can see the first few multiples of 10

in the miles column.

For the cost of EZ Ride, arrow right and up to the top of

Column B, type y1 and press Enter.

That's the second way to move down to the formula line.

Type 2x plus 22.

Notice that the equal sign is inserted by default.

Press Enter.

Press the Down Arrow to resolve the Detected Conflict,

and select Variable Reference.

Head down and click OK when you're done.

These are the costs for the respective numbers of miles x.

Let's do the same for the second company, Just4U, arrow

right and up to the top of Column C, type y2 this time,

press Enter and type the formula point 3x.

Press Enter again, and make the same choice as before,

then tab down to OK.

You can now compare both companies.

Scroll down the first 20 lines or so, and notice Just4U is

the better deal.

But at line 23, both costs are equal to $66.

Thereafter, EZ Ride becomes more affordable.

So your choice between the two companies depends on the

length of your trip.

If your round-trip is more than 220 miles, EZ Ride is

your best choice.

If it's less than 220 miles, you'll pick Just4U.

This method is also called the Tabular or Numerical approach,

as the table of numbers gives you a good sense of

what's going on.

You can see the change in y variable or cost as a function

of the change in the x variable or

miles for each company.

You even found that at x equals 220 miles,

both costs are equal.

This specific fact helps illustrate the meaning of

solving a system of equations.

First, the notation.

To denote the system comprising our two rental car

equation, we use a left brace.

Here, the equations are in slope-intercept form,

rearranging the terms, and then multiplying the first

equation by five and the second by 10 to get integer

coefficients, you obtain the equivalent standard form.

I'll let you verify that.

Solving a system means finding the solution or solutions for

x, y that satisfy the equation simultaneously.

In other words, they must make both equations true.

From our spreadsheet, we saw that 220 comma 66 satisfies

both equations.

So as far as we could see, it is the unique solution of the

system of equations.

I said, as far as we could see, because the spreadsheet

was incomplete.

It did not show all the possible x and y values.

The graphical method we will explore next, will bring us

more certainty about the uniqueness of this solution.

But, once again, first, a quick review.

Since both equations are linear, they

graph a straight line.

But what are the possible relative positions of two

lines in the x,y plane?

The two lines can intersect at one point.

The lines can be parallel, or the two lines can be

coincident, meaning they're the same.

With that in mind, let's use the graphical method to solve

a slightly different problem.

What number of miles x will the cost of renting a compact

car from either EZ Ride or Just4U be the same?

In other words, you must solve this system.

While we may know the numerical answer, let's

explore a second technological approach onto the TI Nspire.

Press Home to add a Graphs in Geometry Page to our document.

Note the one point two at the top left of the screen

indicating the second page.

Type point two x plus 22 for f1 of x.

You see no graph, because the window needs to be adjusted.

We'll do so after defining f2.

Type point 3 x, then press Enter.

Press Menu, and under Window, select Window Settings.

Enter negative 10 and 300 for x, skip Auto, then negative 5

and 100 for y.

Skip Auto again, then tab down and click OK.

To find the coordinates of the point where these two graphs

intersect, press Menu and under Points and Lines select

Intersection Points.

You can see the intersection icon at the

top left of the screen.

Now, use the mouse pad to move the arrow cursor to either

line until it begins flashing.

Click or press Enter to select that line.

Notice that as you move away from it, it continues to

blink, but more slowly.

This confirms that it has been selected.

Do the same for the second graph.

Click or press Enter.

The Nspire generates the cornice of the intersection

point, of this pair of line graphs is the graphical

representation of our linear system.

And their intersection point, 220,66 is a graphical

representation of this system's solution.

Let's construct a function table to confirm our findings.

Press Menu, and under View, select Add Function Table.

To Edit, press Menu again, and under Function Table select

Edit Function Table Settings.

Insert 218 for Table Start, then tab down to OK.

Lastly, to fully view both columns, f1 and f2, under Page

Layout, pick Custom Split.

Arrow left until both columns are visible, press Enter.

The table confirms that both y values or function values are

equal for x equals 220.

For x less than 220, f1 is greater than F2.

For x greater than 220, it's the reverse.

Can you make the visual connection with the relative

position of the two graphs?

For exactly 220 miles, both rental car companies would

charge $66.

We'll assume that other optional costs, like

insurance, are the same.

This graphical method works for any system

two or more equations, two or more variables, and linear or

nonlinear graphs.

This method provides helpful information, even if you are

not too familiar with the types of equations involved.

A well known business application of systems of two

equations is the concept of equilibrium point.

Suppose the red graph represents the company supply

curve of some commodity, like sneakers, for example.

And the blue curve represents the consumers demand curve for

those particular sneakers.

Equilibrium occurs when the amount consumers wish to

purchase at price p, is the same as the amount producers

are willing to offer for sale at the same price.

So the equilibrium point is where supply meets demand, and

both parties are satisfied.

Now, it's your turn to try this.

There are at least two other ways to solve our system of

two linear equations in two variables with the Nspire

technology.

The first uses the feature called nSolve.

The second uses Matrices.

We'll work through these two methods in calculator mode, so

let's begin by introducing a calculator page to our already

open document.

Here we go.

Press Home to Add a Calculator Page.

Note, once again, that 1.3 at the top left indicates our

document's third page.

Press the Catalog key right under the Clear button.

Press the letter n to scroll down to the entries

beginning with an n.

Use the mouse pad to scroll further down to nSolve.

Click or press Enter to select it.

It gets pasted on to your new blank page.

Since we are looking for the value x that makes both the

equations equal, enter point 2x plus 22 equals point 3x.

To indicate we're solving for x, insert a comma followed by

the letter x, then press Enter.

The x value of 220 we've already found that satisfies

both equations, appears on the right.

Using the simpler of the two equations to find y, enter

point 3 times 220, then press Enter.

The y value of 66 also appears on the right.

You've just discovered the teacher's best kept secret,

namely, that you're Nspire can work out the algebraic

manipulations of equation solving.

That doesn't mean it's not important to know how to solve

them by hand.

It is, and we'll learn how to later.

But, in some circumstances, when the focus of a lesson or

a test is not on solving equations, it is good to know

a quick and efficient way to find algebraic solutions

technologically.

And now, on to a sophisticated approach to solving systems of

equations using matrices, plural for matrix.

Yes, like the title of the 1999 science

fiction action film.

For this, we'll introduce a new linear system three

equations and three variables.

Suppose your school puts on a play, and donates half the

proceeds to help send children from developing

countries to school.

You packed the auditorium, which tells you the total

number of tickets sold was 400.

The tickets cost $5 for students, $10 for adults, $7

for seniors, and the school made $2,760.

Lastly, the number of student tickets sold equal the number

of adults and senior tickets combined.

The question is, how many tickets were

sold in each category?

Our variables or unknowns are the number of tickets sold in

each category.

s for student, a for adults, and g for golden-agers since s

is already taken.

The first piece of information generates our first equation.

s plus a plus g equals 400, the total number of tickets.

The second gives 5s, plus 10a plus 7g equals $2,760, the

total amount of money.

The third tells us something about the relative number of

tickets sold.

s equals a plus g, which in standard form, can be written

as s minus a minus g equals zero.

We've built our system of three linear equations, and

now must solve for our three unknowns s, a, and g.

Keep in mind that if and when your goal is a quick and

correct solution, then the method we're about to see is

the shortest path to that goal.

It's called RREF for Row Reduction to Echelon Form on

the augmented matrix.

You probably know that a matrix is an array of numbers

arranged in [? m ?] rows and [? m ?] columns.

The augmented matrix to our system is made up of the nine

numerical coefficient augmented by the three

constants to the right of the equal sign.

The matrix, thus, created is an array of numbers enclosed

in brackets.

It's a three by four matrix since it has three rows and

four columns.

Each represents what I call the skeleton of our problem.

The variables and the equal signs

are absent, but implicit.

The Nspire will perform a series of operations on the

equations, or on the rows of the augmented matrix, each

time, either multiplying through one or more rows by a

constant, adding one or more rows together, or

both actions combined.

At each step, new but equivalent

equations will be generated.

Let's direct the TI Nspire to carry out these row

operations.

Then we'll come back to interpret the results.

We'll work on the same Calculator Page.

Press Menu, and under Matrix and Vector, select Reduced

Row-Echelon Form.

We now must create our augmented matrix.

Select the largest Matrix in the middle of Row 2.

Three is the default number of rows, so tab down and enter

four for the number of columns.

Tab down again and click OK.

We'll enter the 12 numbers, pressing Tab after each.

One, one, one, and 400 for equation one.

Five, 10, seven 2,760 for equation two.

And one, negative one, negative one, zero for

equation three.

Press Enter and the Row Reduced Form for matrix

appears instantly on the right.

Notice that the main diagonal of the first few columns is

populated by one.

And the numbers off this diagonal are zero.

Fairly simple to interpret this output matrix, which is

equivalent to the one we input.

Since we know that the first column is s, the second column

a, and the third column g, this matrix yields the same

solutions of our system in the last column.

We can see that clearly, by translating this Row Reduced

Matrix back into Equation Form.

Simplifying the system, we find that 200 students

tickets, 120 adult tickets, and 80 senior

tickets were sold.

This system, therefore, has a unique solution

200, 120, 80.

Let's go back and see how we can check our solution with

the Nspire.

Enter 200, press Control bar followed by the letter s and

colon to store 200 in the variable s.

Likewise, store 120 in a, and store 80 in g.

Press Enter.

On the next line, enter any one of the three equations.

For instance, the second one, 5s plus 10a

plus 7g equals 2,760.

Then press Enter.

The word, true, appears confirming that the equation

entered is true.

Given the values we've stored for the three variables.

You can check the other two equations the same way.

Now it's time to learn, or review for some of you, how to

solve a system algebraically.

The first technique is called the Substitution Method.

Take the following system.

You can solve for one variable in terms of the other in

either equation.

So for example, we can solve equation 2 for x and obtain x

equals one minus 3 y.

Substituting one minus 3 y for x in equation one, gives two

quantity 1 minus 3 y, minus 5y equals four.

Multiplying out gives 2 minus 6 y minus 5y equals four.

Or 2 minus 11y equals four.

Subtracting two from both sides, yields negative 11y

equals two.

And finally, dividing both sides by negative 11 gives y

equals negative 2 over 11.

To find x, substitute negative 2 over 11

for y in either equation.

Using the second one yields x plus three quantity negative 2

over 11 equals one.

Or x minus 6 over 11 equals one.

Adding 6 over 11 to both sides, gives x equals

one plus 6 over 11.

Or 11 over 11, plus 6 over 11, or 17 over 11.

So our unique solution xy is 17 over 11 negative 2 over 11.

The second algebraic technique has two names.

The Elimination or Linear Combinations Method.

Let's use this method to solve the following riddle.

The sum of two natural numbers in 94.

The difference between this first and twice

the second is 25.

Find the two numbers.

Using f and s for the first and second numbers

respectively, here is our system

of two linear equations.

We could eliminate s by multiplying equation one by

two, then adding it to equation two.

Or, we could eliminate f by multiplying equation two by

negative one, then adding it to equation one.

Let's choose the second option.

Adding the multiple of equation two to the original

equation one, eliminates f and yield.

3s equals 69 or s equals 23.

To find f, like in the substitution method, we

substitute 23 for s into either equation.

Say the first.

We obtain f plus 23 equals 94.

Or f equals 94 minus 23.

Finally, f equals 71.

Our two numbers are f equals 71 and s equals 23.

And a solution to this system is the ordered

pair 71 comma 23.

You now have an idea of the two classical algebraic

methods for solving systems of equations.

In closing, I'd like to summarize the three possible

solutions to a system of two linear

equations and two variables.

This system of equations has a unique solution, and is said

to be consistent and independent.

In this case, the two lines intersect at one point.

This system of equations has an infinite number of

solutions and is said to be consistent and dependent.

The two lines are coincident, meaning they're the same

having all points in common.

Lastly, this system of equations has no solution, and

is said to be inconsistent.

The lines are parallel having no common point.

You've heard of a variety of ways to solve systems of

linear equations.

These techniques form the basis of a large part of what

is called linear algebra, an advanced mathematics course

with many real world applications you

may take later on.

Advances in technology make solving systems much easier

than in the past, enabling you to use computer assisted

methods in tandem with algebraic methods.

I suggest you explore further with the help of the Nspire

nonlinear systems that either cannot be solved

algebraically, or that you have not yet learned how to

solve algebraically.

Have fun.