Linear Equations

Delving into the rich tapestry of the Islamic Golden Age, I'm struck by the timeless impact of algebra on our lives. The genius of scholars like Al-Khwarizmi has transcended centuries, shaping the very fabric of mathematics and its application in our daily lives.

This journey through history to the present showcases how the elegant simplicity of linear equations can decode complex modern challenges, be it in economics or technology. It's a vivid reminder that the language of numbers is universal, echoing through time from ancient scripts to today's digital algorithms.

In this article I will guide you through the essence of linear equations. We'll examine how these mathematical principles help us solve real-world problems and understand the intrinsic properties that make them so powerful!

Islamic Golden Age and Algebra

The formal development and understanding of linear equations, as we know them today, emerged during the Islamic Golden Age in the 9th to 13th centuries. Scholars such as Al-Khwarizmi, Al-Mahani, and Al-Khayyam made significant contributions to the field of algebra, including the study of linear equations.

Linear Equations in Real-world Problems

Let's delve into the world of linear equations through the eyes of Alex, who runs a vibrant car rental service in sunny California. Alex's business, "Sunset Cruisers" operates on a simple yet clever pricing model that beautifully illustrates the concept of linear equations in action.

As clients come to rent one of Alex's iconic convertibles, they encounter the company's straightforward pricing strategy: a base fee of $20 for the day, plus a variable charge of $0.3 per mile driven. This is where linear equations become the silent engine of the business.

To model this financially, Alex sets up an equation where C represents the total rental cost, and x symbolizes the miles driven. The resulting equation is C = 0.3x + 20.

This clever equation allows Alex to quickly calculate costs for any journey, whether it's a short scenic drive along the coast or a long adventure through the winding roads of the state parks.

In Sunset Cruisers' equation, the $20 represents the fixed cost, while the $0.3 per mile is the variable cost. This linear equation, empowers Alex to manage his business with precision and provide clear pricing to his adventure-seeking customers.

Note : When dealing with real-world applications, there are certain expressions that we can translate directly into math. Here are some examples

• One number exceeds another by “a”: x, x+a
• Twice a number: 2x
• One number is “a” less than twice another number: x, 2x?a
• The product of a number and “a”, decreased by “b”: ax?b
• The quotient of a number and the number plus “a” is five times the number: x/(x+a)= 5x
• The product of five times a number and the number decreased by “b” is “c”: 5x(x-b) = c

Let’s take another real case example

In the bustling cityscape, two renowned cell phone companies, Apex Mobile and Orange Telecom, vie for the patronage of tech-savvy consumers with their competitive monthly plans.

• Apex Mobile, renowned for its expansive network, offers a plan that includes a $34 monthly fee with an additional cost of $.05 per minute of talk time
• On the other side of the spectrum, Beacon Telecom, with its impeccable customer service, provides a slightly higher monthly fee of $40 but compensates with a lower talk-time rate of $.04 per minute.

1/ Write a linear equation that models the packages offered by both companies.

2/ If the average number of minutes used each month is 1160, which company offers the better plan?

3/ How many minutes of talk time would yield equal monthly statements from both companies?

Solution

1/ Write a linear equation that models the packages offered by both companies.

• The model for Apex Mobile can be written as A=0.05x+34
• The model for Orange Telecom can be written as B=0.04x+40

2/ If the average number of minutes used each month is 1160, which company offers the better plan?

• The model for Apex Mobile can be written as A=0.05(1160)+34 , A would be equal to $92
• The model for Orange Telecom can be written as B=0.04(1160)+40, B would be equal to $86.4

Orange Telecom offers a lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Apex Mobile when the average number of minutes used each month is 1160.

3/ How many minutes of talk time would yield equal monthly statements from both companies?

We can find this point by setting the equations equal to each other and solving for x.

• 0.05x+34=0.04x+40
• 0.01x = 6
• x = 600

Check the x-value in each equation.

• 0.05(600)+34 = 64
• 0.04(600)+40 = 64

Therefore, a monthly average of 600 talk time minutes renders the plans equal.

Steps to transform real-world problems into linear equations

• Identify known quantities and assign a variable to represent the unknown quantity
• If there is more than one unknown quantity, find a way to write the second unknown in terms of the first. (if we can find a relationship between x and y, we can express one variable in terms of the other and reduce the problem to a single unknown.)
• Write an equation interpreting the words as mathematical operations.
• Solve the equation. Be sure the solution can be explained in words, including the units of measure.

One and multiple variables linear equation

A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1.

Linear equations in one variable may take the form of ax+b=0, where “a” and “b” are constants, and “x” is the variable. The goal is to find the value of “x” that satisfies the equation. It can take also a different form as we will see in the article.

In Linear equations with multiple variables, you have an equation that involves more than one variable. Here’s an example of a linear equation with multiple variables: 3x + 5y + 7z = 10

Forms of Linear Equations

• Slope-Intercept form: It’s in the form of y = mx+b, where “m” is the slope, “b” is the “y” intercept, and “y” is …
• Standard form: It’s in the form of ax + by = c, where “a”, “b” (“a” and “b” cannot be fractions, and “a” cannot be negative) and “c” are constant, “x” and “y” are unknown variables.
• Point-slope form: It’s in the form of y1 — y2 = m(x1-x2), where “m” is the slope and (xi,yi) are points in the line.

Proprietes of linear equation

• The highest power of the variable is 1.
• The graph of a linear equation is a straight line with a constant slope (consistent change in the vertical (y-axis) coordinate for every unit change in the horizontal (x-axis) coordinate).
• Linear equations can be manipulated using addition, subtraction, multiplication, and division operations without altering the solutions.
• Linear equations exhibit the consistency property, which means that if two equations have a common solution, any linear combination (sum or difference) of these equations will also have the same solution.

Types of Linear Equations

Identity Equations: An identity equation is true for all values of the variable. When you simplify an identity equation, you end up with a statement that is always true, such as 3x = 2x+x. In this case, any value of “x” you choose will satisfy the equation.

Conditional Equations: A conditional equation is true for only some values of the variable but not all like an identity equation. When you solve a conditional equation, you obtain a specific solution or a set of solutions that make the equation true. For example, 5x+2=3x?6 is a conditional equation. Solving it gives x = ?4, which is the specific solution that satisfies the equation.

Inconsistent Equations: An inconsistent equation has no solution. When you attempt to solve an inconsistent equation, you will encounter a contradiction, and there will be no value of the variable that makes the equation true. An example of an inconsistent equation is 5x?15=5(x?4).

• 5x?15 = 5x?20
• We will subtract 5x from both sides, -15 ≠ -20

Techniques for solving linear equations:

Addition or Subtraction Method: This consists of isolating the variable term by adding or subtracting terms from both sides of the equation.

• Example: 6x + 6 = 12, we subtract 6 from both sides 6x = 6, then divide both sides by 6 and we have x=1

Multiplication or Division Method: The same principle for addition or subtraction, consists of isolating the variable term by multiplication or division.

• Example: (1/2)x = 6, we can multiply both sides by 2 , x = 12

Substitution: Consists of expressing one variable in terms of another and substituting it into the equation.

• Example: 2x + y = 4, x — 2y = 4, we can solve the second equation for x = 2y +4, then we substitute x in the first equation 2(2y +4) + y = 4, y would be equal to -4/5, then we can substitute y in one of the original equation to find x, if we choose the first one we would have 2x(-4/5) = 4, we can add 4/5 in both sides to isolate x and we would have 2x = 4 + 4/5 and it is equal to 2x = 24/5. Finally, we will divide both sides to isolate furthermore x and we would have x = 12/5.

Graphical Method: Let’s take an equation of x+2 = 0, first thing to do is to solve for the variable.

• Let’s take -2 for both sides x+2–2 = 0–2. So x would be equal to -2.
• So x is always -2 no matter how much y is, you can see the representation below

• Let’s try it for two variables ; suppose there are 2 equations : x+ y = 6 and x — y = 4
• We need to prepare a table containing the x and y coordinates.
• Let’s have the values 0,2, and 4 for x and find y for both equations

The intersection of the lines is (5,1) and that’s the answer to the equation

Elimination Method: Consist of solving a system of linear equation by eliminating one variable, let’s take an example of those 2 equations: 3x + 2y = 7 and 5x-3y = 37

• We will multiply the equation one by 2 and the equation 2 by 3 to eliminate the “y” and we would have 9x + 6y = 21 and 10x -6y =74
• Now we add the equation 1 to the equation 2 and we will get 19x = 95
• Finally, find x which is equal to 5
• Now that we know that x is equal to 5, we will add it to Equation 1 or Equation 2 to find y, let’s put it in Equation 1
• 3(5) + 2y = 7, So 2y = -8, y is equal to -4

Matrix Method: Consist of representing the system of equations in matrix operations to solve for the variables, we can use a method of reduction or a method of inversion, we will start by exploring the method of reduction than the method of inversion.

a/ Reduction

• Let’s take an example with those equations : x+y+z =2 , x+2y+z = 0, 3x+4y+4z = 7
• The first thing to do is to arrange these equations in the form of matrices, we should place the coefficients of x, y, and z in a matrix, the second matrix will contain the unknowns that is x, y, and z and the third matrix will contain the constant coefficients

• Now we will reduce the matrix of the coefficient “in the upper triangular matrix” that’s mean the 3 numbers in the bottom left representing a triangle should be 0

• To reduce the matrix we will be using row transformation, so the first operation for R2 would be Row 2 — Row 1

• The next transformation for Row 3 would be Row3–(3*Row1) to eliminate the 3

• The last change is for R3 and it would be Row 3 — Row 2 to eliminate the 1

• Now that we have our upper triangular matrix, we will start with the last row which has 2 zeros, we can say that 1z = 3 so z = 3
• For the second row we have 1y + 0z =-2, so y = -2
• For the first row we have 1x + 1y + 1z = 2, with substitution we have 1x -2 + 3 = 2, So x = 1
• Now we have the value of x, y, and z

b/ Inversion

• In the same first step as we did in the reduction method, we start by representing the equations in a form of a matrix

• We will name each matrix, the first one would be A, see second representing the unknown variable would be X and the last one would be B, so we would have A * X = B
• If we multiply both the sides by a inverse A^-1 A X = A^-1 B, we will get X = A^-1 B
• We need to find the inverse of matrix A with the adjoint method and multiply it by matrix B, let’s start by finding the determinant of matrix A

• |A| = (1*2*4)+(1*1*3)+(1*1*4) — [(3*2*1)+(4*1*1)+(4*1*1)] and it will be equal to |A| = (8 + 3 + 4) – (6 + 4 + 4) = -1, If the determinant is not equal to 0 (-1≠0) it means that the inverse exists.
• Now we need to find the minors and the co-factors of each element from the matrix
• Here is how to find the minors, for the co-factors we just multiply the result by (-1)^the position of the element in the case of “a” in the image below it would be (-1)^(1+1) (a12 a33) — (a23*a32)
• Note: a1,1 or a11 means that we need to block the first row and the first column, for a12 it would be the first row and the second column.

• The minors and co-factors of our matrix would be :

• AdjA would be equal to

• Find the inverse of the matrix with the adjoint method

• We will get

• Finally ?? we need to multiply the result by the matrix B to find X

• Like the last one we will find that x =1, y = -2 and z = 3

It feels like an article by itself… ??.

The article is coming to an end, and we've traversed through a panorama of algebra's legacy, from its profound roots in the Islamic Golden Age to its practicality in modern quandaries like rental services and telecom packages. We've unpacked the essence of linear equations, translating life's nuances into the language of algebra and using it as a beacon to navigate the seas of data-driven decisions. How often do you find yourself applying these mathematical principles in your personal or professional life?

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