Types of Solutions to Systems of Linear Equations

In this post, we will learn that any system of linear equations has one of three possible types of solution. Each type of solution has distinctive characteristics. These will be presented, so we can understand the type of system or situation we are dealing with when we observe a system of linear equations. The three types of solution are:

1. One unique solution
2. Infinitely many solutions
3. No solution

The type of solution can be determined by characteristics of lines used in analytic geometry. Those are the slope, the y-intercept, and the graph of the lines themselves. These are discussed in more detail below. However, it is important to understand that a system of equations has NO solution when the number of unknowns exceeds the number of equations. Therefore, a linear system that has 2 equations and 3 unknowns cannot be solved. Moreover, systems of 2 linear equations with 2 unknowns will be used in the examples below.

One Unique Solution

A system of linear equations has a unique solution when the graph of the lines intersect at a single point. This point of intersection is the solution of the system. Based on this, it can be inferred that the lines have distinct slope and y-intercept. A system of linear equations with an unique solution is shown below.

The graph generated in a TI-89 Titanium calculator shows the intersection of the lines at point (1, 3). This is the solution of the system of equations.

Infinitely Many Solutions

A system of linear equations has an infinite number of solutions when the graph of the lines are the same. In other words, the equations are the same. They are just written differently. Since the equations are the same, they obviously have the same slope and the same y-intercept. A system of linear equations with an infinite number of solutions is shown below.

The graph generated in a TI-89 Titanium calculator seems to show a single line. However, it shows the graph of both equations. This is because the equations are the same and their graphs touch each other at every point, hence infinitely many solutions. In fact, if the second equation is divided by -3 on both sides, the first equation is obtained.

No Solution

It was discussed previously that systems of linear equations with more unknowns than equations have no solution. Additionally, a linear system has no solution when the graph of the lines are parallel. In other words, they have no intersection point. These lines have the same slope, but different y-intercept. A system of linear equations with no solution is shown below.

The graph generated in a TI-89 Titanium calculator shows two parallel lines. This means the lines will never intersect and the system has no solution.

Conclusion

When dealing with system of linear equations, no matter the size, they will have one of these three types of solution. 

Solving Simple Systems of Linear Equations (Part 3)

Previously, the methods of elimination of variables and substitution of variables were introduced (Part 1 - click here). Subsequently, these methods were used to solve a couple of linear systems of 2 equations with 2 unknowns (Part 2 - click here). Now in this post, we take it a little further. We will find the solution to a system of 3 linear equations with 3 unknowns. 

Example:

The methods of elimination and substitution of variables can be applied to all kinds of systems of linear equations, provided they have a solution. However, as these systems become larger in number of equations and unknowns, these methods become cumbersome. Instead one can use other methods to handle large systems of linear equations such as the application of matrices or even numerical methods. These will be discussed in future posts.

Finding Composite Functions

A composite function is one that obtained by plugging in an inner function into an outer function. In other words, a function takes the place of the independent variable in an outer function. To understand this concept, consider two functions f and g. Then, the composite function is denoted by f ◦ g which is defined by:

In other words, the function g is plugged into the function f.

To understand how to find composite functions, we will look at a few examples. 

Example 1:

It is also possible to have a composite function of three or more functions. The same rules of the previous examples apply. An example is shown next.

Example 2:

Finding the Inverse of a Function

First of all, it is important to know that not all functions have an inverse. Only one-to-one functions have an inverse function. 

But what is a one-to-one function?

A one-to-one function is one that never takes the same value twice. For example, the function f(x) = x is one-to-one, because f(1) = 1, and no other value of x can have a value of f(x) that is equal to 1. 

On the other had, the function f(x) = x² is not one-to-one, because f(-2) = 4 and f(2) = 4. As noted, there are two values of x that take the value of 4.

Now back to the inverse of a function:

The inverse of a function f(x) is denoted by f^-1(x). It is important to understand that -1 is not an exponent, but just notation. It denotes that it is the inverse function of f(x). Now, we will look at a couple of examples to learn how to calculate the inverse of a function.

Example 1:

Example 2:

The steps explained in the examples can be used as an algorithm to find the inverse of a function (when possible, of course).

Stoichiometry in Combustion (Part 2 - Example with Excess Air)

Determine the air-fuel ratio on a mass basis for the complete combustion of octane, C8H18, with 150% theoretical air (50% excess air).

Solution:

The first part of the solution is to obtain the equation for the complete combustion of octane with the theoretical amount of air. Find the coefficients of the equation.

The second part of the solution is using the left hand side of the chemical equation to obtain the new coefficients of the products when 150% theoretical air is used. Now, the air is multiplied by 1.5, and there is excess air that appears as uncombined oxygen in the products.

The third part of the solution is actually calculating the air-fuel ratio which is what the problem asked for.

The ratio of the amount of air in this reaction to the amount of fuel is about 22.7.

Bibliography:

Michael J. Moran, Howard N. Shapiro. Fundamentals of Engineering Thermodynamics. John Wiley & Sons, Inc., 2008.

Stoichiometry in Combustion

Stoichiometry is the study of the relationship between quantities of reactants and products in chemical equations. It is based on the principle of conservation of mass. Therefore, it involves balancing both sides of the equation by assuring there are the same number of atoms for all elements involved. 

In the field of mechanical engineering, stoichiometry is mostly used in the field of thermodynamics when studying combustion reactions. In this type of reactions, a fuel reacts chemically with oxygen in order to release energy and combustion products. When a combustion is complete, all the carbon present in the fuel is burned to carbon dioxide, all the hydrogen is burned to water, all the sulfur is burned to sulfur dioxide, and all other combustible elements are fully oxidize. If these conditions are not met, the combustion is said to be incomplete. Now, we will observe a couple of examples.

Example 1:

Balance the equation below for the complete combustion of methane.

Solution:

By applying the principle of conservation of mass, we can create equations to balance the number of atoms of the same elements on both sides in terms of the coefficients of each compound in the reaction.

By solving the system of four equations and four unknowns, the balanced chemical equation is:

Example 2:

For the balanced equation in the previous example, obtain the air-fuel ratio.

Solution:

The ratio of the amount of air in this reaction to the amount of fuel is about 17.24.

Bibliography:

Michael J. Moran, Howard N. Shapiro. Fundamentals of Engineering Thermodynamics. John Wiley & Sons, Inc., 2008.

The Relationship among Position, Velocity and Acceleration

Position, velocity and acceleration are concepts all of us deal with every day. We give directions to a friend about our location in town in order to meet up at a restaurant. We drive to work as we push the acceleration and brake pedals in order to change the velocity of the car. These are just every day examples of how these concepts affect our life. We do not give much thought to these actions, but at the mathematical level there is more than just doing our daily activities.

The relationship among position, velocity and acceleration is one of the fundamental applications of calculus. From the point of view of calculus, these quantities are related by differentiation and integration. This tells us that if we know at least one of them, we can calculate the other two. The figure below shows the relationship among these quantities.

The arrows in the figure above show the directions in which we have to either differentiate or integrate in order to obtain the other quantities. A simple example is shown next.

Solution:

Differentiate the position equation to obtain the velocity. Then evaluate at t = 10.

Differentiate the velocity equation to obtain the acceleration. Then evaluate at t = 10.

At t = 10, the velocity and acceleration are 97 and 10, respectively, in their corresponding units.

Derivation of Free Fall Equations

Rectilinear motion with constant acceleration is introduced early in physics courses. While there are many cases of this type of motion, we can consider free fall one of the best examples of the relationship among position, velocity and acceleration (click here). It is even more important, since it shows how physics and calculus go hand in hand. Therefore, we will observe how to derive three of the equations for this special case.

Free fall is simply the motion of a falling object under the influence of acceleration due to gravity, g, which is considered a constant. Therefore, we will start with the acceleration function and then apply integration to get the velocity and position as functions of time. The procedure is shown below:

Finally, we rewrite the equations in the most common form that appears in most textbooks for consistency.