# Linear Systems

Beloved, I pray that you may prosper in all 'good' things and be in good health, even as thy soul prospers. - 3rd John 1:2

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

I greet you this day,

First: read the notes. Second: view the videos. Third: solve the questions/solved examples. Fourth: check your solutions with my thoroughly-explained solutions. Fifth: check your answers with the calculators as applicable.

I wrote the codes for these calculators using JavaScript, a client-side scripting language, and AJAX, a Javascript library. Please use the latest Internet browsers. The calculators should work.
The Wolfram Alpha widgets (many thanks to the developers) was also used for some of the calculators.
Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. If you are my student, please do not contact me here. Contact me via the school's system. Thank you for visiting.

Samuel Dominic Chukwuemeka (SamDom For Peace) B.Eng., A.A.T, M.Ed., M.S

## Overview of Linear Systems

#### Objectives

Students will:

(1.) Define linear systems.

(2.) Discuss linear systems.

(3.) Classify linear systems.

(4.) Translate linear systems from "English" to "Math".

(5.) Solve linear systems.

(6.) Check the solution of linear systems.

(7.) Solve linear systems using the TI-84/84-Plus Calculator.

#### Skills Measured/Acquired

(1.) Use of prior knowledge

(2.) Critical Thinking

(3.) Interdisciplinary connections/applications

(4.) Technology

(5.) Active participation through direct questioning

(6.) Student collaboration in Final Project

#### Vocabulary Words

Bring it to Biology: system (cell --> tissues --> organ --> "system" --> organism)

Bring it to English: right, left, vary, equality, consistent, inconsistent, dependent, independent, relationship

Bring it to Math: coefficient, right hand side (RHS), left hand side (LHS), fraction, decimal, numerator, denominator, least common denominator (LCD), equation, linear, linear system, constant, variable, dependent variable, independent variable, input, output, consistent system, inconsistent system, dependent system, independent system, solution, graph, slope, parallel lines, graphical method, substitution method, elimination method, elimination-by-addition method, Cramer's rule, method of determinants, matrices, Gaussian elimination method, Gauss-Jordan method, matrix inverse method, triangular method, overdetermined system, underdetermined system.

#### Why Study Linear Systems?

Linear Systems are used in the:

(2.) Manufacturing sector for production and work scheduling.

Discuss how linear systems are used in these sectors. Ask students to list other applications. Ask the "how" and the "why" of any application.

How is linear system used in that application? Why is linear system used in that application?

Are there any alternatives that could have been used in that application?

#### What is a Linear System?

A linear system is a system of linear equations.

We pronounce it in terms of "the number of equations by the number of variables"

$2$ equations that has $2$ variables is a $2 * 2$ (pronounced as 2 by 2) linear system. See examples below.

$3$ equations that has $3$ variables is a $3 * 3$ (pronounced as 3 by 3) linear system. See examples below.

So, what is a "$5$ equations that has $5$ variables" called?

What if we have $2$ equations that has $3$ variables (more variables than equations)? What should we call it?

What if we have $3$ equations that has $2$ variables (more equations than variables)? What should we call it?

Ask students to write an example of the last two cases.

A complete linear system has the same number of variables as equations.

An incomplete linear system does not have the same number of variables as equations.

An incomplete linear system can either be an underdetermined system or an overdetermined system.

A linear system in which the number of variables is greater than the number of equations is an underdetermined linear system.

A linear system in which the number of equations is greater than the number of variables is an overdetermined linear system.

#### Examples of Linear Systems

2 x 2 Linear System 3 x 3 Linear System Underdetermined Linear System Overdetermined Linear System
$2x - 5y = 1$ $$4x + 3y = 15$$ x and y are "lined up"
$3x = - y + 5$ $$-10 = - 6x - 2y$$ x and y are not "lined up"
So, we need to "line them up"
$3x + y = 5$ $$6x + 2y = 10$$
$-\dfrac{1}{4} - \dfrac{2}{3} x = \dfrac{1}{4} y$ $$\dfrac{2}{5} y - \dfrac{23}{60} = \dfrac{1}{2}x$$ We have coefficients as fractions
We need to convert them to integers
1st Equation: LCD of 4, 3, 4 = 12
Multiply each term by 12
$12 * -\dfrac{1}{4} - 12 * \dfrac{2}{3} x = 12 * \dfrac{1}{4} y$ $$-3 - 8x = 3y$$
Line them up
$-3 = 3y + 8x$ $$8x + 3y = -3$$ 2nd Equation: LCD of 5, 60, 2 = 60
Multiply each term by 60
$60 * \dfrac{2}{5} y - 60 * \dfrac{23}{60} = 60 * \dfrac{1}{2}x$ $$24y - 23 = 30x$$
Line them up
$-23 = 30x - 24y$ $$30x - 24y = -23$$ So, we have:
$8x + 3y = -3$ $$30x - 24y = -23$$
$y = 0.3x$ $$0.07x + 0.12y = 3.18$$ We have coefficients as decimals
We need to convert them to integers
1st Equation: Multiple of 10 to multiply = 10
Multiply each term by 10
$10 * y = 10(0.3x)$ $$10y = 3x$$
Line them up
$3x = 10y$ $$3x - 10y = 0$$ 2nd Equation: Multiple of 10 to multiply = 100
Multiply each term by 100
$100(0.07x) + 100(0.12y) = 100(3.18)$ $$7x + 12y = 318$$
So, we have:
$3x - 10y = 0$ $$7x + 12y = 318$$
$\dfrac{1}{x} + \dfrac{1}{y} = 7$ $$\dfrac{6}{x} - \dfrac{2}{y} = 10$$
There are fractions.
However, the variables are written as denominators.
In this case, use appropriate substitutions.
Use the substitution
Let $c = \dfrac{1}{x}\: and\: d = \dfrac{1}{y}$

$\dfrac{1}{x} + \dfrac{1}{y} = 7$ $$\dfrac{6}{x} - \dfrac{2}{y} = 10$$ becomes:
$c + d = 7$ $$6c - 2d = 10$$

#### Observations from the Examples

If the variables are not "lined up", we need to "line them up".

If any variable(s) is/are missing, we can put it/them by assigning zeros as the coefficients.

If the coefficients of any of the variables is a fraction, it is better to multiply all the terms by the LCD of the denominators so you can deal with integer coefficients. (Or do you like fractions? ☺)

If the coefficients of any of the variables is a decimal, it is better to multiple all the terms by the appropriate multiples of ten so you can deal with integer coefficients. (Or do you like decimals? ☺)

If there are fractions but the variables are written as denominators, use appropriate substitution techniques to write them well.

Some students may ask why we need to line the variables in order. Please be prepared to answer.

## Classification of Linear Systems

#### Linear Systems are classified as:

Consistent OR

Inconsistent

A linear system is consistent if it has at least one solution.

"At least one" means "one or more than one"

Therefore, a consistent linear system has one or more solutions.

A linear system is inconsistent if it has no solution.

#### Consistent Linear Systems are further classified as:

Independent OR

Dependent

A consistent linear system is independent if it has only one solution.

A consistent linear system is dependent if it has many solutions.

Imagine when someone is independent. You pay a lot of taxes in the 'Uncle Sam'!

Imagine when someone has several dependents. You get several tax credits when filing taxes!

#### So, we have these three main types of linear systems:

(1.) Consistent-Independent Linear System

(2.) Consistent-Dependent Linear System

(3.) Inconsistent Linear System

#### Compare and Contrast the Linear Systems

Types of Linear Systems /
Properties
Consistent-Independent Consistent-Dependent Inconsistent
(1.) "Line up" and Check for Relationship
Remember: Variables are on the LHS; Constants are on the RHS
For 2 x 2 systems
There are no "similar" relationships between coefficients of same variables.
$2x- 5y = 1$ $$4x + 3y = 15$$ Compare 2 and 4 ($2 * 2 = 4$)
Compare -5 and 3 (But $-5 * 2 \ne 3$)
There is a relationship between 2 and 4. (Relationship for the x-variable)
That relationship is that you multiply Equation (1.) by 2 to get Equation (2.)
It works for the x-variable.
However, it does not work for the y-variable (between $-5\: and\: 3$).
In addition, it does not work for the constants (between $1\: and\: 15$).

For 3 x 3 systems
There are no similar relationships between the coefficients of any of the variables and the constant in one equation, and the coefficients of the corresponding variables and the constant in another equation.
Compare any one variable and the constant in an equation
Compare that same variable and the constant in another equation
If there are no similar relationships, the system is consistent-independent. OR you may say:
There is a similar relationship between the coefficient of a variable in one equation and the coefficient of the corresponding variable in another equation.
However, that same relationship does not apply to the constants for both equations.
$4c + 3d + e = 16$ $$c - 3d + 2e = -29$$ $11c - 2d + 3e = -19$
Let us compare the variable, c and the constant
Equations 1 and 2: Compare $(4 \:and\: 1)$ and $(16 \:and\: -29)$
$4 * \dfrac{1}{4} = 1$ $$16 * \dfrac{1}{4} \ne -29$$ No similar relationships.
Equations 1 and 3: Compare $(4 \:and\: 11)$ and $(16 \:and\: -19)$
$4 * \dfrac{11}{4} = 11$ $$16 * \dfrac{11}{4} \ne -19$$ No similar relationships.
Equations 2 and 3: Compare $(1 \:and\: 11)$ and $(16 \:and\: -19)$
$1 * 11 = 11$ $$16 * 11 \ne -19$$ No similar relationships.
If you do so for each of the variables $d$, or $e$ and the constants, there are no similar relationships.
For 2 x 2 systems
There is a "similar" relationship between coefficients of same variables.
There is also the "same relationship" between the constants.
You are saying the same thing, but in various ways.
$2x - 5y = 1$ $$4x - 10y = 2$$ $-6x + 15y = -3$ $$\dfrac{2}{7}x - \dfrac{5}{7}y = \dfrac{1}{7}$$ Do you realize the four equations are the same?
Multiply Equation (1.) by 2 to get Equation (2.)
Multiply Equation (1.) by $-3$ to get Equation (3.)
Multiply Equation (1.) by $\dfrac{1}{7}$ to get Equation (4.)

Multiply Equation (2.) by $-\dfrac{3}{2}$ to get Equation (3.)

What will you multiply by Equation (4.) to get Equation (3.)?

For 3 x 3 systems
There is "at least a" similar relationship between the coefficient of any one variable and the constant in one equation and the coefficient of the corresponding variable and constant in another equation.
Compare any one variable and the constant in an equation
Compare that same variable and the constant in another equation
If there is a similar relationship, the system is consistent-dependent.
$c - 2d - e = -4$ $$-c + 6d - 3e = 12$$ $2c - 11d + 5e = -22$
Let us compare the variable, d and the constant
Equations 1 and 2: Compare $(-2 \:and\: 6)$ and $(-4 \:and\: 12)$
$-2 * -3 = 6$ $$-4 * -3 = 12$$
There is a relationship between the coefficients of $d$ and the constants in Equations 1 and 2.
There is no need for further comparison. "At least one" means "one or more".
The system is a consistent-dependent system.
For 2 x 2 systems
There is a "similar" relationship between coefficients of same variables.
However, that same relationship does not apply to the constants.
You are being inconsistent. You say something today. Tomorrow, you say a different thing.
You say the same thing on the LHS, but a different thing on the RHS.
Can you give a typical example of people that are inconsistent? Some Politicians!!!
$2x - 5y = 1$ $$4x - 10y = 7$$ There is a relationship between the coefficients of the x-variable.
There is the same relationship between that of the y-variable.
But, it does not apply to the constants.
Multiply Equation (1.) by 2 to give Equation (2.) works for the LHS.
But, it does not work for the RHS.
$2 * 2 = 4$ $$-5 * 2 = -10$$ $$But, 1 * 2 \ne 7$$
For 3 x 3 systems
There is "at least" a similar relationship between the "coefficients of all three variables" in one equation and the coefficients of the corresponding variables in another equation.
However, that relationship does not apply to the constants.
Compare all three variables and the constant in an equation
Compare those same variables and the constant in another equation
There is a similar relationship between all three variables, but not for the constants.
$2c - 5d - 5e = -7$ $$6c - 15d - 15e = -5$$ $2c - 2d - 5e = 4$
Let us compare the variables: c, d, and e and the constant
Equations 1 and 2: Compare $(2 \:and\: 6)$, $(-5 \:and\: -15)$, $(-5 \:and\: -15)$, and $(-7 \:and\: -5)$
$2 * 3 = 6$ $$-5 * 3 = -15$$ $-5 * 3 = -15$ $$-7 * 3 \ne -5$$
There is a relationship between the coefficients of $c$, $d$, and $e$ in Equations 1 and 2.
However, that same relationship does not apply to the constants in Equations 1 and 2.
The system is inconsistent.
Nature of Solution A unique solution (only one solution) Many solutions No solution
Nature of Graph
Please use a graphing utility to demonstrate this property.
You may sketch as well.
You may use the previous examples.
Graphs intersect at a point.
The intersection gives the solution of the system.
Graphs are a single line. Graphs are parallel lines.
Nature of Slope
Recall: $y = mx + b$
The slope-intercept equation of a straight line
m = slope
b = y-intercept
Different slopes
$2x - 5y = 1$ $$4x + 3y = 15$$ Let us re-write them in the slope-intercept form
Equation (1.)
$2x - 1 = 5y$ $$5y = 2x - 1$$ $$y = \dfrac{2}{5}x - \dfrac{2}{5}$$ Slope of Equation (1.) = $\dfrac{2}{5}$
Equation (2.)
$3y = -4x + 15$ $$y = -\dfrac{4}{3}x + \dfrac{15}{3}$$ $$y = -\dfrac{4}{3}x + 5$$ Slope of Equation (2.) = $-\dfrac{4}{3}$
Same slope
$2x - 5y = 1$ $$4x - 10y = 2$$ The equations can be "reduced" to just a single equation.
Because the equations are the same, the slopes are the same.
Same slope
$2x - 5y = 1$ $$4x - 10y = 7$$ Let us re-write them in the slope-intercept form
Equation (1.)
$2x - 1 = 5y$ $$5y = 2x - 1$$ $$y = \dfrac{2}{5}x - \dfrac{1}{5}$$ Slope of Equation (1.) = $\dfrac{2}{5}$
Equation (2.)
$10y = 4x - 7$ $$y = \dfrac{4}{10}x - \dfrac{7}{10}$$ $$y = \dfrac{2}{5}x - \dfrac{7}{10}$$ Slope of Equation (2.) = $\dfrac{2}{5}$
Nature of y-intercept
Same or Different y-intercepts
The y-intercept of Equation (1.) = $-\dfrac{1}{5}$
The y-intercept of Equation (2.) = $5$
This is a case of different y-intercepts.
Ask students to write an independent-consistent linear system that has same y-intercepts.
Same y-intercept
Because the equations are the same, the y-intercepts are the same.
Different y-intercepts
The y-intercept of Equation (1.) = $-\dfrac{1}{5}$
The y-intercept of Equation (2.) = $$-\dfrac{7}{10}$$

#### Check for Understanding

Students must explain their answers. No guess work. Reasons may vary. They could explain using any or all of the properties discussed earlier.

 $2x - 5y = 1$ $$4x + 3y = 15$$ Answer: Consistent-Independent $15x - 20y = 21$ $$3x - 4y = 7$$ Answer: Inconsistent $2x - 5y = 10$ $$6x - 15y = 30$$ Answer: Consistent-Dependent $y = 0.3x$ $$0.07x + 0.12y = 3.18$$ Answer: Consistent-Independent $3x = -y + 5$ $$-10 = -6x - 2y$$ Answer: Consistent-Dependent $y = -\dfrac{3}{2}x + 3$ $$3x - 4y = 24$$ Answer: Consistent-Independent $15y = -7 + 10x$ $$4x - 6y = -7$$ Answer: Inconsistent $\dfrac{2}{3}x + \dfrac{1}{4}y = -\dfrac{1}{4}$ $$\dfrac{1}{2}x - \dfrac{2}{5}y = -\dfrac{23}{60}$$ Answer: Consistent-Independent $y = 4x - 3$ $$y = 7 - x$$ Answer: Consistent-Independent $y = \dfrac{1}{4}x + 3$ $$y = 5 - \dfrac{1}{4}x$$ Answer: Consistent-Independent $5x + y = -4$ $$2x - y = -3$$ Answer: Consistent-Independent $x = 4 - y$ $$5 = -y + 5x$$ Answer: Consistent-Independent $-2y = 4 - x$ $$3x = -4 - 2y$$ Answer: Consistent-Independent $-2y = -x + 8$ $$-3x = -5 - 6y$$ Answer: Inconsistent $97 - 0.2n = 0.3m$ $$0.1m + 47 = 0.5n$$ Answer: Consistent-Independent $\dfrac{1}{3}c + \dfrac{1}{4}d - \dfrac{4}{15} = 0$ $$-d + \dfrac{4}{5} + \dfrac{1}{6}c = 0$$ Answer: Consistent-Independent $4e - 16 = -4f$ $$4 = e + f$$ Answer: Consistent-Dependent $2v = w$ $$w - 6 = -4v$$ Answer: Consistent-Independent $k - 2 - 5j = 0$ $$10j - 6 -2k = 0$$ Answer: Inconsistent $-6q + 3 = -9p$ $$-3p = 1 - 2q$$ Answer: Consistent-Dependent $-\dfrac{1}{2}y = -4 - x$ $$-1 = -4y + 8x$$ Answer: Inconsistent $-\dfrac{3}{2}x + 3y = -1$ $$2 + 6y = 3x$$ Answer: Consistent-Dependent $\dfrac{1}{2}x + \dfrac{1}{7}y = 3$ $$\dfrac{1}{4}x - \dfrac{2}{7}y = -1$$ Answer: Consistent-Independent $\dfrac{1}{x} + \dfrac{1}{y} = 7$ $$\dfrac{6}{x} - \dfrac{2}{y} = 10$$ Answer: Consistent-Independent $c - 2d - e = -4$ $$-c + 6d - 3e = 12$$ $2c - 11d + 5e = -22$ Answer: Consistent-Dependent $2c - 18d = 4$ $$4c + 45e = 23$$ $36d + 35e = 13$ Answer: Consistent-Independent $4c + 7d = -1$ $$8c + 3e = -16$$ $6d + 3e = 6$ Answer: Consistent-Independent $2c - 5d - 5e = -7$ $$6c - 15d - 15e = -5$$ $2c - 2d - 5e = 4$ Answer: Inconsistent $c + d + e = -10$ $$2c + 5d + 2e = -35$$ $-c + 9d - 3e = -42$ Answer: Consistent-Independent $c - 3d - e = -6$ $$-c + 8d -4e = 16$$ $2c - 15d + 7e = -30$ Answer: Consistent-Dependent $\dfrac{2}{c} + \dfrac{3}{d} - \dfrac{2}{e} = -1$ $$\dfrac{6}{c} - \dfrac{18}{d} + \dfrac{5}{e} = 5$$ $\dfrac{5}{c} + \dfrac{3}{d} - \dfrac{1}{e} = 1$ Answer: Consistent-Independent $c + d + e - f = 5$ $$4c + d - e + f = 12$$ $c - 4d + 3e + f = 6$ $$-c - d + e + 4f = 2$$ Answer: Consistent-Independent

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## Linear Systems using the TI-84/84 Plus

This calculator will:
(1.) Solve several problems involving linear systems.

• Using the TI-84/84 Plus Graphing Calculator
• Click the "Matrices" folder
• Click the "Solving a linear system with a matrix inverse" link

## "Up to $3 * 3$ Linear Systems" Calculator

This calculator will solve these linear systems if there is a solution.
(1.) Solve $1 * 2$ linear systems.
(2.) Solve $2 * 1$ linear systems.
(3.) Solve $1 * 3$ linear systems.
(4.) Solve $3 * 1$ linear systems.
(5.) Solve $2 * 2$ linear systems.
(6.) Solve $2 * 3$ linear systems.
(7.) Solve $3 * 2$ linear systems.
(8.) Solve $3 * 3$ linear systems.
(1.) Type your equations in the textbox (the bigger textbox).
(2.) Type it according to the examples I listed.
(3.) Delete the "default" equations in the textboxes of the calculator.
(4.) Copy and paste the equations you typed, into the small textbox of the calculator.
(5.) Click the "Submit" button.
(6.) Check to make sure they are the correct equations you typed.
(7.) Review the answers. At least one of the answers is probably what you need.

• Using the "Up to $3 * 3$ Linear Systems" Calculator
• Type: $-14 = -20y - 7x$ as -14 = -20 * y - 7 * x
• Type: $x - 2y = 5p + 4$ as x - 2 * y = 5 * p + 4
• Type: $\dfrac{5}{c} + \dfrac{3}{d} - \dfrac{1}{e} = 1$ as 5/c + 3/d - 1/e = 1
• Type: $\dfrac{1}{4}x = -1 + \dfrac{2}{7}y$ as (1/4) * x = -1 + (2/7) * y

Solve

## "Up to $4 * 4$ Linear Systems" Calculator

This calculator will solve these linear systems if there is a solution.
(1.) Solve $1 * 2$ linear systems.
(2.) Solve $2 * 1$ linear systems.
(3.) Solve $1 * 3$ linear systems.
(4.) Solve $3 * 1$ linear systems.
(5.) Solve $2 * 2$ linear systems.
(6.) Solve $2 * 3$ linear systems.
(7.) Solve $3 * 2$ linear systems.
(8.) Solve $3 * 3$ linear systems.
(1.) Type your equations in the textbox (the bigger textbox).
(2.) Type it according to the examples I listed.
(3.) Delete the "default" equations in the textboxes of the calculator.
(4.) Copy and paste the equations you typed, into the small textbox of the calculator.
(5.) Click the "Submit" button.
(6.) Check to make sure they are the correct equations you typed.
(7.) Review the answers. At least one of the answers is probably what you need.

• Using the "Up to $4 * 4$ Linear Systems" Calculator
• Type: $-14 = -20y - 7x$ as -14 = -20 * y - 7 * x
• Type: $x - 2y = 5p + 4$ as x - 2 * y = 5 * p + 4
• Type: $\dfrac{5}{c} + \dfrac{3}{d} - \dfrac{1}{e} = 1$ as 5/c + 3/d - 1/e = 1
• Type: $\dfrac{1}{4}x = -1 + \dfrac{2}{7}y$ as (1/4) * x = -1 + (2/7) * y

Solve

### References

Chukwuemeka, S.D (2016, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.samuelchukwuemeka.com

Kaufmann, J., & Schwitters, K. (2011). Algebra for College Students (Revised/Expanded ed.). Belmont, CA: Brooks/Cole, Cengage Learning.

Lial, M., & Hornsby, J. (2012). Beginning and Intermediate Algebra (Revised/Expanded ed.). Boston: Pearson Addison-Wesley.

Sullivan, M., & Sullivan, M. (2017). Algebra & Trigonometry (7th ed.). Boston: Pearson.

Tan, S. (2015). Finite Mathematics for the Managerial, Life, and Social Sciences (Revised/Expanded ed.). Boston: Cengage Learning.