Algebra Nspirations

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.