# Linear Systems

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 Overview of Linear Systems Linear Systems (All) Translate 2 by 2 Linear Systems Translate 3 by 3 Linear Systems Graphical Solution of Linear Systems Solution of Linear Systems by Substitution Solution of Linear Systems by Elimination Solution of Linear Systems by Cramer's Rule Solution of Linear Systems by Gauss-Jordan Method Solution of Linear Systems by Matrix Inverse Method Word Problems on 2 by 2 Linear Systems Word Problems on 3 by 3 Linear Systems

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

## Overview of Linear Systems

#### Objectives

Students will:

(1.) Discuss linear systems.
(2.) Classify linear systems.
(3.) Translate linear systems from English to Math
(4.) Solve linear systems.
(5.) Check the solution of linear systems.
(6.) Discuss the application of linear systems.
(7.) Solve linear systems using technology including 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

Then, ask them to list more terms.

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.

##### 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.

The standard form of a linear system is: $ax + by = c$
where:
x and y are the variables
a is the coefficient of the x-variable
b is the coefficient of the y-variable
c is the constant
Expressing a linear system in standard form means that the linear system is lined up

##### Examples of Linear Systems
2 × 2 Linear System 3 × 3 Linear System Underdetermined Linear System Overdetermined Linear System
$2x - 5y = 1 \\[3ex] 4x + 3y = 15 \\[3ex]$ x and y are lined up
$3x = - y + 5 \\[3ex] -10 = - 6x - 2y \\[3ex]$ x and y are are not lined up
So, we need to line them up

$3x + y = 5 \\[3ex] 6x + 2y = 10$
$-\dfrac{1}{4} - \dfrac{2}{3} x = \dfrac{1}{4} y \\[5ex] \dfrac{2}{5} y - \dfrac{23}{60} = \dfrac{1}{2}x \\[5ex]$ 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

$12 * -\dfrac{1}{4} - 12 * \dfrac{2}{3} x = 12 * \dfrac{1}{4} y \\[5ex] -3 - 8x = 3y \\[3ex]$ Line them up

$-3 = 3y + 8x \\[3ex] 8x + 3y = -3 \\[3ex]$ 2nd Equation: LCD of 5, 60, 2 = 60
Multiply each term by 60

$60 * \dfrac{2}{5} y - 60 * \dfrac{23}{60} = 60 * \dfrac{1}{2}x \\[5ex] 24y - 23 = 30x \\[3ex]$ Line them up

$-23 = 30x - 24y \\[3ex] 30x - 24y = -23 \\[3ex]$ So, we have:

$8x + 3y = -3 \\[3ex] 30x - 24y = -23$
$y = 0.3x \\[3ex] 0.07x + 0.12y = 3.18 \\[3ex]$ 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

$10 * y = 10(0.3x) \\[3ex] 10y = 3x \\[3ex]$ Line them up

$3x = 10y \\[3ex] 3x - 10y = 0 \\[3ex]$ 2nd Equation: Multiple of 10 to multiply = 100
Multiply each term by 100

$100(0.07x) + 100(0.12y) = 100(3.18) \\[3ex] 7x + 12y = 318 \\[3ex]$ So, we have

$3x - 10y = 0 \\[3ex] 7x + 12y = 318$
$\dfrac{1}{x} + \dfrac{1}{y} = 7 \\[5ex] \dfrac{6}{x} - \dfrac{2}{y} = 10 \\[5ex]$ There are fractions.
However, the variables are written as denominators.
In this case, use appropriate substitutions.
Use the substitution
Let $c = \dfrac{1}{x}\: and\: d = \dfrac{1}{y}$

$\dfrac{1}{x} + \dfrac{1}{y} = 7 \\[5ex] \dfrac{6}{x} - \dfrac{2}{y} = 10 \\[5ex]$ becomes:

$c + d = 7 \\[3ex] 6c - 2d = 10$

Observations from the Examples

(1.) If the variables are not lined up, we need to line them up
In other words, if the linear system is not in standard form, express it in standard form.

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

(3.) 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? ☺)

(4.) 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? ☺)

(5.) 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 to '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.

$2x- 5y = 1 \\[3ex] 4x + 3y = 15 \\[3ex]$ Compare 2 and 4 (2 × 2 = 4)
Compare −5 and 3 (But −5 × 2 ≠ 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).
Therefore, there is no relationship on the LHS (Left Hand Side).

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.

$4c + 3d + e = 16 \\[3ex] c - 3d + 2e = -29 \\[3ex] 11c - 2d + 3e = -19 \\[3ex]$ Let us compare the variable, c and the constant
Equations 1 and 2:
Compare 4 and 1 and 16 and −29

$4 * \dfrac{1}{4} = 1 \\[5ex] 16 * \dfrac{1}{4} \ne -29 \\[5ex]$ No similar relationships.
Equations 1 and 3:
Compare 4 and 11 and 16 and −19

$4 * \dfrac{11}{4} = 11 \\[5ex] 16 * \dfrac{11}{4} \ne -19 \\[5ex]$ No similar relationships.
Equations 2 and 3:
Compare 1 and 11 and 16 and −19

$1 * 11 = 11 \\[3ex] 16 * 11 \ne -19 \\[3ex]$ 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.

$2x - 5y = 1 \\[3ex] 4x - 10y = 2 \\[3ex] -6x + 15y = -3 \\[3ex] \dfrac{2}{7}x - \dfrac{5}{7}y = \dfrac{1}{7} \\[5ex]$ 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 one 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.

$c - 2d - e = -4 \\[3ex] -c + 6d - 3e = 12 \\[3ex] 2c - 11d + 5e = -22 \\[3ex]$ Let us compare the variable, d and the constant
Equations 1 and 2:
Compare −2 and 6 and −4 and 12

$-2 * -3 = 6 \\[3ex] -4 * -3 = 12 \\[3ex]$ 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!

$2x - 5y = 1 \\[3ex] 4x - 10y = 7 \\[3ex]$ 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.
Multiplying Equation (1.) by 2 to give Equation (2.) works for the LHS.
But, it does not work for the RHS.

$2 * 2 = 4 \\[3ex] -5 * 2 = -10 \\[3ex] But\;\; 1 * 2 \ne 7 \\[3ex]$ For 3 x 3 systems
There is at least one 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.

$2c - 5d - 5e = -7 \\[3ex] 6c - 15d - 15e = -5 \\[3ex] 2c - 2d - 5e = 4 \\[3ex]$ 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

$2 * 3 = 6 \\[3ex] -5 * 3 = -15 \\[3ex] -5 * 3 = -15 \\[3ex] -7 * 3 \ne -5 \\[3ex]$ 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.
(2.)
Nature of Solution
A unique solution (only one solution) Many solutions No solution
(3.)
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.
(4.)
Nature of Slope
Recall: $y = mx + b$
The slope-intercept equation of a straight line
m = slope
b = y-intercept
Different slopes

$2x - 5y = 1 \\[3ex] 4x + 3y = 15 \\[3ex]$ Let us re-write them in the slope-intercept form
Equation (1.)

$2x - 1 = 5y \\[3ex] 5y = 2x - 1 \\[3ex] y = \dfrac{2}{5}x - \dfrac{2}{5} \\[5ex]$ Slope of Equation (1.) = $\dfrac{2}{5}$

Equation (2.)

$3y = -4x + 15 \\[3ex] y = -\dfrac{4}{3}x + \dfrac{15}{3} \\[5ex] y = -\dfrac{4}{3}x + 5 \\[5ex]$ Slope of Equation (2.) = $-\dfrac{4}{3}$
Same slope

$2x - 5y = 1 \\[3ex] 4x - 10y = 2 \\[3ex]$ The equations can be reduced to just a single equation.
Because the equations are the same, the slopes are the same.
Same slope

$2x - 5y = 1 \\[3ex] 4x - 10y = 7 \\[3ex]$ Let us re-write them in the slope-intercept form
Equation (1.)

$2x - 1 = 5y \\[3ex] 5y = 2x - 1 \\[3ex] y = \dfrac{2}{5}x - \dfrac{1}{5} \\[5ex]$ Slope of Equation (1.) = $\dfrac{2}{5}$

Equation (2.)

$10y = 4x - 7 \\[3ex] y = \dfrac{4}{10}x - \dfrac{7}{10} \\[5ex] y = \dfrac{2}{5}x - \dfrac{7}{10} \\[5ex]$ Slope of Equation (2.) = $\dfrac{2}{5}$
(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 the 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.) = $-\dfrac{7}{10}$
(6.)
Determinants
(For Precalculus with Trigonometry Students)
Prerequisite: Cramer's Rule
The determinant of the denominator is non-zero. The determinant of the denominator is zero.
AND
The determinant of each numerator is zero.
The determinant of the denominator is zero.
AND
The determinant of any of the numerators is non-zero.
(7.)
This is okay. However, please note that it may not be applicable to all linear systems.
In other words, you may not be able to use it for some questions.

Solve the Linear System and note the LHS = RHS
There is a unique solution for the variables. 0 = 0 0 = something else besides 0

##### Check for Understanding

Classify these linear systems.
indicate the number of solution(s).

No guess work.
Reasons may vary.
They could explain using any or all of the properties discussed earlier.
Guide them to learn and understand this concept.

 $2x - 5y = 1 \\[3ex] 4x + 3y = 15 \\[3ex]$ Answer: Consistent-Independent One solution $15x - 20y = 21 \\[3ex] 3x - 4y = 7 \\[3ex]$ Answer: Inconsistent No solution $2x - 5y = 10 \\[3ex] 6x - 15y = 30 \\[3ex]$ Answer: Consistent-Dependent Many solutions $y = 0.3x \\[3ex] 0.07x + 0.12y = 3.18 \\[3ex]$ Answer: Consistent-Independent One solution $3x = -y + 5 \\[3ex] -10 = -6x - 2y \\[3ex]$ Answer: Consistent-Dependent Many solutions $y = -\dfrac{3}{2}x + 3 \\[5ex] 3x - 4y = 24 \\[3ex]$ Answer: Consistent-Independent One solution $15y = -7 + 10x \\[3ex] 4x - 6y = -7 \\[3ex]$ Answer: Inconsistent No solution $\dfrac{2}{3}x + \dfrac{1}{4}y = -\dfrac{1}{4} \\[5ex] \dfrac{1}{2}x - \dfrac{2}{5}y = -\dfrac{23}{60} \\[5ex]$ Answer: Consistent-Independent One solution $y = 4x - 3 \\[3ex] y = 7 - x \\[3ex]$ Answer: Consistent-Independent One solution $y = \dfrac{1}{4}x + 3 \\[5ex] y = 5 - \dfrac{1}{4}x \\[5ex]$ Answer: Consistent-Independent One solution $5x + y = -4 \\[3ex] 2x - y = -3 \\[3ex]$ Answer: Consistent-Independent One solution $x = 4 - y \\[3ex] 5 = -y + 5x \\[3ex]$ Answer: Consistent-Independent One solution $-2y = 4 - x \\[3ex] 3x = -4 - 2y \\[3ex]$ Answer: Consistent-Independent One solution $-2y = -x + 8 \\[3ex] -3x = -5 - 6y \\[3ex]$ Answer: Inconsistent No solution $97 - 0.2n = 0.3m \\[3ex] 0.1m + 47 = 0.5n \\[3ex]$ Answer: Consistent-Independent One solution $\dfrac{1}{3}c + \dfrac{1}{4}d - \dfrac{4}{15} = 0 \\[5ex] -d + \dfrac{4}{5} + \dfrac{1}{6}c = 0 \\[5ex]$ Answer: Consistent-Independent One solution $4e - 16 = -4f \\[3ex] 4 = e + f \\[3ex]$ Answer: Consistent-Dependent Many solutions $2v = w \\[3ex] w - 6 = -4v \\[3ex]$ Answer: Consistent-Independent One solution $k - 2 - 5j = 0 \\[3ex] 10j - 6 -2k = 0 \\[3ex]$ Answer: Inconsistent No solution $-6q + 3 = -9p \\[3ex] -3p = 1 - 2q \\[3ex]$ Answer: Consistent-Dependent Many solutions $-\dfrac{1}{2}y = -4 - x \\[5ex] -1 = -4y + 8x \\[3ex]$ Answer: Inconsistent No solution $-\dfrac{3}{2}x + 3y = -1 \\[5ex] 2 + 6y = 3x \\[3ex]$ Answer: Consistent-Dependent Many solutions $\dfrac{1}{2}x + \dfrac{1}{7}y = 3 \\[5ex] \dfrac{1}{4}x - \dfrac{2}{7}y = -1 \\[5ex]$ Answer: Consistent-Independent One solution $\dfrac{1}{x} + \dfrac{1}{y} = 7 \\[5ex] \dfrac{6}{x} - \dfrac{2}{y} = 10 \\[5ex]$ Answer: Consistent-Independent One solution $c - 2d - e = -4 \\[3ex] -c + 6d - 3e = 12 \\[3ex] 2c - 11d + 5e = -22 \\[3ex]$ Answer: Consistent-Dependent Many solutions $2c - 18d = 4 \\[3ex] 4c + 45e = 23 \\[3ex] 36d + 35e = 13 \\[3ex]$ Answer: Consistent-Independent One solution $4c + 7d = -1 \\[3ex] 8c + 3e = -16 \\[3ex] 6d + 3e = 6 \\[3ex]$ Answer: Consistent-Independent One solution $2c - 5d - 5e = -7 \\[3ex] 6c - 15d - 15e = -5 \\[3ex] 2c - 2d - 5e = 4 \\[3ex]$ Answer: Inconsistent No solution $c + d + e = -10 \\[3ex] 2c + 5d + 2e = -35 \\[3ex] -c + 9d - 3e = -42 \\[3ex]$ Answer: Consistent-Independent One solution $c - 3d - e = -6 \\[3ex] -c + 8d -4e = 16 \\[3ex] 2c - 15d + 7e = -30 \\[3ex]$ Answer: Consistent-Dependent Many solutions $\dfrac{2}{c} + \dfrac{3}{d} - \dfrac{2}{e} = -1 \\[5ex] \dfrac{6}{c} - \dfrac{18}{d} + \dfrac{5}{e} = 5 \\[5ex] \dfrac{5}{c} + \dfrac{3}{d} - \dfrac{1}{e} = 1 \\[5ex]$ Answer: Consistent-Independent One solution $c + d + e - f = 5 \\[3ex] 4c + d - e + f = 12 \\[3ex] c - 4d + 3e + f = 6 \\[3ex] -c - d + e + 4f = 2 \\[3ex]$ Answer: Consistent-Independent One solution

## Applications of Linear Systems

#### Why Study Linear Systems?

Linear Systems are used:
(1.) Mixology: Linear Systems is used to calculate the exact volumes of liquids needed for a mixture of a specific volume and concentration.
(3.) 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?

Let us discuss each application.

##### Mixology

Mixology is the art of mixing drinks.
There are different kinds of drinks including alcoholic beverages (vodka and wine), water, fruit juices, and soda among others.
For every drink, two factors are important: Volume and Concentration.
The volume is indicated in every drink container.
The concentration (alcohol by volume or ABV) is indicated as a percentage in every alcoholic beverage container.
Let us focus on alcoholic beverages and other beverages.

We do know that:
(1.) Pure water is 100% water.
Pure water is 0% alcohol.
Water is 0% alcohol.
(2.) Juice is 0% alcohol.
(3.) Soda is 0% alcohol.

But, why do we mix drinks?

Let us analyze this scenario between Paul, Timothy, Rebecca, Lot, Nehemiah, and Noah.
Paul is the mentor to Timothy
Timothy is the mentee of Paul.
Rebecca is a friend of Timothy.
Lot is a friend of Timothy.
Nehemiah is a friend of Timothy.
Noah is a mixologist.

Sometimes, Timothy experiences stomach problems.

Paul: Son Timothy, do not drink only water.
Use a little wine because of your stomach and frequent ailments. (1 Timothy 5:23)

Timothy was celebrating his birthday.
Rebecca brought several bottles of water for him.
The water is 0% alcohol by volume.
Lot brought some bottles of alcoholic beverages (bottles of vodka and kegs of wine)
The vodka is 50% alcohol by volume.
The wine is 10% alcohol by volume.
Nehemiah brought some sweet drinks (bottles of fruit juices and cans of soda)
Fruit juice is 0% alcohol by volume.
Soda is 0% alcohol by volume.
Noah was also in attendance at Timothy's birthday celebration.

These are some responses from my students regarding this scenario.
Teacher: Vodka is a strong drink.
Assume Timothy wants something moderate.
What should the mixologist, Noah do?
Student: Timothy can take the wine.
It is not strong.
Teacher: What if he says that the wine is too light?
Assume he wants something moderate.
What options are available?
Student: He can mix the vodka with water.
Teacher: That is correct. First option.
Student: He can also mix the vodka with fruit juice and/or soda.
Teacher: Second option...correct.
Any other option?
Student: He can also mix it with the wine.
Teacher: That is correct.
Because the vodka is 50% alcohol and water is 0% alcohol, we expect the concentration of the mixture to be between 0% and 50% (both ends not included)
Similarly, because the vodka is 50% alcohol and fruit juice/water is 0% alcohol, we expect the concentration of the mixture to be between 0% and 50% (both ends not included)
Further, because the vodka is 50% alcohol and the wine is 10% alcohol, we expect the concentration of the mixture to be between 10% and 50% (both ends not included)
Assume Timothy wants a 300 milliliters of 20% alcohol
How much of each drink should be mixed?

Let us continue with the scenario.

Timothy thanked all in attendance at the celebration.
He opened a bottle of vodka and took a shot of it.

Timothy: It's too strong, he exclaimed!
My brain cannot handle it.
I need something moderate.

Lot: Well, take the wine.
I brought wine too.

Timothy took a shot of the wine.

Timothy: Oh, this is too light. 😊
I prefer something a little strong.

Rebecca: Well, mix the vodka with water.
I brought some water.

Nehemiah: Or mix it with some fruit juices
Or soda: Sprite, Coca Cola, Pepsi, etc.
I brought fruit juices and soda cans

Timothy: How much should I mix?
I do not want to mess it up.

Noah Mixologist here
I can help.
What do you have?
What do you want?

Timothy: First Option:
Please mix the vodka with the water.
The vodka is 50% alcohol solution.
The water does not contain alcohol.
I want 400 milliliters of 15% alcohol solution.

Second Option:
Please mix the vodka with the fruit juice.
The vodka is 50% alcohol solution.
The fruit juice does not contain alcohol.
I want 300 milliliters of 18% alcohol solution.

Third Option:
Please mix the vodka with the soda.
The vodka is 50% alcohol solution.
The fruit juice does not contain alcohol.
I want 200 milliliters of 35% alcohol solution.

Fourth Option:
Please mix the vodka with the wine.
The vodka is 50% alcohol solution.
The wine is 10% alcohol solution.
I want 500 milliliters of 20% alcohol solution.

Noah: Okay, let me get to work...one at a time.

$50\% = \dfrac{50}{100} = 0.5 \\[5ex] 0\% = \dfrac{0}{100} = 0 \\[5ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] 15\% = \dfrac{15}{100} = 0.15 \\[5ex] 20\% = \dfrac{20}{100} = 0.2 \\[5ex] 18\% = \dfrac{18}{100} = 0.18 \\[5ex] 35\% = \dfrac{35}{100} = 0.35 \\[7ex] \underline{First\;\;Option:\;\;Vodka + Water} \\[3ex] Vodka:\;\; 50\%\;ABV \\[3ex] Water:\;\; 0\%\;ABV \\[3ex] Mixture:\;\; 400\;mL \;\;and\;\; 15\%\;ABV \\[3ex] Let: \\[3ex] volume\;\;of\;\;vodka = x \\[3ex] volume\;\;of\;\;water = y \\[3ex] \underline{Volume:} \\[3ex] x + y = 400 ...eqn.(1) \\[3ex] \underline{Concentration:} \\[3ex] 50\%\;\;of\;\;x + 0\%\;\;of\;\;y = 15\%\;\;of\;\;(x + y) \\[3ex] 0.5x + 0y = 0.15(400) \\[3ex] 0.5x = 60 ...eqn.(2) \\[3ex] From\;\;eqn.(2) \\[3ex] x = \dfrac{60}{0.5} \\[5ex] x = 120 \\[3ex] Substitute\;\;120\;\;for\;\;x\;\;in\;\;eqn.(1) \\[3ex] 120 + y = 400 \\[3ex] y = 400 - 120 \\[3ex] y = 280 \\[3ex]$ Noah: Timothy, I need to mix:
120 milliliters of 50% alcohol vodka with 280 milliliters of water to obtain 400 milliliters of 15% alcohol solution.
Next one:
Ask students to do the second option.
Then, check the solution.

Second Option: Mix the vodka with the fruit juice.

$\underline{Second\;\;Option:\;\;Vodka + Fruit\;\;Juice} \\[3ex] Vodka:\;\; 50\%\;ABV \\[3ex] Fruit\;\;Juice:\;\; 0\%\;ABV \\[3ex] Mixture:\;\; 300\;mL \;\;and\;\; 18\%\;ABV \\[3ex] Let: \\[3ex] volume\;\;of\;\;vodka = x \\[3ex] volume\;\;of\;\;juice = y \\[3ex] \underline{Volume:} \\[3ex] x + y = 300 ...eqn.(1) \\[3ex] \underline{Concentration:} \\[3ex] 50\%\;\;of\;\;x + 0\%\;\;of\;\;y = 18\%\;\;of\;\;(x + y) \\[3ex] 0.5x + 0y = 0.18(300) \\[3ex] 0.5x = 54 ...eqn.(2) \\[3ex] From\;\;eqn.(2) \\[3ex] x = \dfrac{54}{0.5} \\[5ex] x = 108 \\[3ex] Substitute\;\;108\;\;for\;\;x\;\;in\;\;eqn.(1) \\[3ex] 108 + y = 300 \\[3ex] y = 300 - 108 \\[3ex] y = 192 \\[3ex]$ Noah: Timothy, I need to mix:
108 milliliters of 50% alcohol vodka with 192 milliliters of fruit juice to obtain 300 milliliters of 18% alcohol solution.

Ask students to do the third option.
Then, check the solution.

Third Option: Mix the vodka with the sprite soda.

$\underline{Third\;\;Option:\;\;Vodka + Sprite\;\;Soda} \\[3ex] Vodka:\;\; 50\%\;ABV \\[3ex] Sprite\;\;Soda:\;\; 0\%\;ABV \\[3ex] Mixture:\;\; 200\;mL \;\;and\;\; 35\%\;ABV \\[3ex] Let: \\[3ex] volume\;\;of\;\;vodka = x \\[3ex] volume\;\;of\;\;soda = y \\[3ex] \underline{Volume:} \\[3ex] x + y = 200 ...eqn.(1) \\[3ex] \underline{Concentration:} \\[3ex] 50\%\;\;of\;\;x + 0\%\;\;of\;\;y = 35\%\;\;of\;\;(x + y) \\[3ex] 0.5x + 0y = 0.35(200) \\[3ex] 0.5x = 70 ...eqn.(2) \\[3ex] From\;\;eqn.(2) \\[3ex] x = \dfrac{70}{0.5} \\[5ex] x = 140 \\[3ex] Substitute\;\;140\;\;for\;\;x\;\;in\;\;eqn.(1) \\[3ex] 140 + y = 200 \\[3ex] y = 200 - 140 \\[3ex] y = 60 \\[3ex]$ Noah: Timothy, I need to mix:
140 milliliters of 50% alcohol vodka with 60 milliliters of sprite soda to obtain 200 milliliters of 35% alcohol solution.

Ask students to do the fourth option.
Then, check the solution.

Fourth Option: Mix the vodka with the wine.

$\underline{Third\;\;Option:\;\;Vodka + Wine} \\[3ex] Vodka:\;\; 50\%\;ABV \\[3ex] Wine:\;\; 10\%\;ABV \\[3ex] Mixture:\;\; 500\;mL \;\;and\;\; 20\%\;ABV \\[3ex] Let: \\[3ex] volume\;\;of\;\;vodka = x \\[3ex] volume\;\;of\;\;wine = y \\[3ex] \underline{Volume:} \\[3ex] x + y = 500 ...eqn.(1) \\[3ex] \underline{Concentration:} \\[3ex] 50\%\;\;of\;\;x + 10\%\;\;of\;\;y = 20\%\;\;of\;\;(x + y) \\[3ex] 0.5x + 0.1y = 0.2(500) \\[3ex] 0.5x + 0.1y = 100 \\[3ex] Multiply\;\;each\;\;term\;\;by\;\;10\;\;to\;\;clear\;\;the\;\;decimals \\[3ex] 5x + y = 1000 ...eqn.(2) \\[3ex] \underline{Elimination-by-Subtraction\;\;Method} \\[3ex] eqn.(2) - eqn.(1) \implies \\[3ex] (5x + y) - (x + y) = 1000 - 500 \\[3ex] 5x + y - x - y = 500 \\[3ex] 4x = 500 \\[3ex] x = \dfrac{500}{4} \\[5ex] x = 125 \\[3ex] \underline{Substitution\;\;Method} \\[3ex] Substitute\;\;125\;\;for\;\;x\;\;in\;\;eqn.(1) \\[3ex] 125 + y = 500 \\[3ex] y = 500 - 125 \\[3ex] y = 375 \\[3ex]$ Noah: Timothy, I need to mix:
125 milliliters of 50% alcohol vodka with 375 milliliters of 10% alcohol wine to obtain 500 milliliters of 20% alcohol solution.