Tag Archives: math

Solving a System of Equations with Three Variables

A system of equations can be solved involving three variables. There are several different ways to accomplish this when three variables are involved. In this post, we will focus on the use of the elimination method.

Our initial system of equations is below


The values eq1,eq2 and eq3 just mean equation 1, 2, 3

To solve this system we need to first solve two equations as a system and create a fourth equation we will call eq4. We then take eq1 and eq3 to create a new system of equations that creates eq5.

It is important to note that for the first two two-variable system of equations you create that you eliminate the same variable it both systems. So fare our example when we take equation 1 and 2 to create equation 4 and then take equation 1 and 3 create equation 5 we must solve for y in both situations or else we will have problems. In addition, you must make sure that all three equations appear at least once in the two two-variable systems of equations. For our purpose, we will use eq1 twice and eq2 and eq3 once.

Eq4 and eq5 are used to find the actual values we need for all three variables. This will make more sense as we go through the example. Therefore, we are going to solve first for y for eq1 and eq2.

To eliminate y we need to multiple eq2 by 2 and then combine the equations. Below is the process and the new eq 4


We will come back to eq4. For now, we will create eq 5 by eliminating y from eq1 and eq3.


We are essentially done using equations 1, 2, and 3. They will not reappear until the end. We will now use equations 4 and 5 to find our answers for two of the three variables.

We now will use eq 4 and 5 to eliminate the variable x. Eliminating x will allow us to solve for z. Doing means we will multiply eq4 by -1.


We know z = -3 we can plug this value into either eq4 or 5 to find the answer for x.


Now that we know x and z we can plug the two numbers into one of the three original equations to find the value for y. Notice how the first variable we eliminated becomes the last one we solve for.


We now know all three values which are

(4, -1, -3)

What this means is that if we were to graph this three equations they would intersect at (4, -1, -3). A solving a system of equations is simply telling us where the lines of the equations intersect.


Solving a system of equations involving three variable is an extension of the two variable system that has already been covered. It provides a mathematician with a tool for solving for more unknown variables. There are practical applications of this as we shall see in the future,


System of Equations and Uniform Motion

This post will provide examples of the use of a system of equations to solve uniform motion applications. A system of equations is used to solve for more than one variable. In the context of uniform motion, the basic equation is as follows

distance  = rate * time

We will look at the following examples

  • Two objectives moving in the same direction
  • Affect of a headwind/tailwind

Objects Moving in the Same Directions

Below is the problem followed by the solution.

Dan leaves home and travels to Springfield at 100 kph. About 30 minutes later Sue leaves the house and also travels the same way to Springfield driving 125 kph. How long will it take Sue to catch Dan?

The easiest way to solve this is to create a table with all of the information we have. The table is below.

Names Rate * Time = Distance
Dan 100 j 100j
Sue 125 k 125k

We first need to recognize that they will drive the same distance this leads to one of our equations


However, we are not done. We also need to realize that Sue leaves half an hour later, which leads to the second equation


We can now solve our system of equations


We know Dan travels for 2.5 hours before Sue catches him but we need to determine how long Sue drives before she catches Dan. We will take our answer J and plug it into the original equation for k.


It will take Sue 2 hours to catch up with Dan.

Affect of a Headwind/Tailwind

In transportation, it is common for a plan or ship to be able to travel faster with a tailwind or downstream than with a headwind or upstream The example below shows you how to determine the speed needed to travel a certain distance in the same amount of time as well as the speed of the wind/current.

A plane can travel 548 miles in 1.5 hours with a tailwind but only 494 hours when flying into a headwind. Find the speed of the plane and the wind.

We will have two variables because there are two things we want to know

  • p = the speed of the plane
  • w = the speed of the wind

The tailwind makes the plane go faster, therefore, the speed of the plane will be the plane speed + the wind speed

The tailwind slows the plane down. Therefore, the tailwind will be the speed of the plane minus the windspeed.

Below is a table with all of the available information

Rate * Time = Distance
Tailwind p + w 1.5 548
Headwind p – w 1.5 494

The initial system of equations is as follows


To solve this system of equations we will use the elimination method as shown below.


The plane travels 347.33 mph. We now take the value of p plug it into one of our equations to find the speed of the wind.


The speed of the wind is 18 mph. We know the plane travels 347 + 18 = 365 mph with a tailwind and 347-18 = 329mph with a headwind.


A system of equations is proven to have a practical application. The assumption of a uniform speed is somewhat unrealistic in most instances. However, this assumption simplifies the calculation and prepares us for more complex models in the future.

Solving a System of Equations with Direct Translation

In this post, we will look at two simple problems that require us to solve for a system of equations. Recall that a system of equations involves two or more variables that must be solved. With each problem, we will use the direct translation to set up the problem so that it can be solved.

Direct Translation 

Direct translation involves reading a problem and translating it into a system of equations. In order to do this, you must consider the following steps

  1. Determine what you want to know
  2. Assigned variables to what you want to know
  3. Setup the system of equations
  4. Solve the system

Example `1

Below is an example  followed by a step-by-step breakdown

The sum of two numbers is zero. One number is 18 less than the other. Find the numbers.

Step 1: We want to know what the two numbers are

Step 2: n = first number & m =  second number

Step 3: Set up system


Solving this is simple we know n = m – 18 so we plug this into the first equation n + m = 0  and solve for m.


Now that we now m we can solve for n in the second equation


The answer is m = 9 and n = -9. If you add these together they would come to zero and meet the criteria for the problem.

Example 2

Below is a second example involving a decision for salary options.

Dan has been offered two options for his salary as a salesman. Option A would pay him $50,000 plus $30 for each sale he closes. Option B would pay him $35,000 plus $80 for each sale he closes. How many sales before the salaries are equal

Step 1: We want to know when the salaries are equal based on sales

Step 2: d =  Dan’s salary & s = number of sales

Step 3: Set up system


To solve this problem we can simply substitute d  for one of the salaries as shown below


You can check to see if this answer is correct yourself. In order for the two salaries to equal each other Dan would need to sale 300 units. After 300 units option B is more lucrative. Deciding which salary option to take would probably depend on how many sales Dan expects to make in a year.


Algebraic concepts can move beyond theoretical ideas and rearrange numbers to practical applications. This post showed how even something as obscure as a system of equations can actually be used to make financial decisions.

Solving a System of Equations by Substitution and Elimination

A system of equations involves trying to solve for more than one variable. What this means is that a system of equations helps you to see how to different equations relate or where they intersect if you were to graph them.

There are several different ways to solve a system of equations. In this post, we will solve y using the substitution and the elimination methods.


Substitution involves choosing one of the two equations and solving for one variable. Once this is done we substitute the expression into the equation for which we did not solve a variable for. When this is done the second equation only has one unknown variable and this is basic algebra to solve.

The explanation above is abstract so here is a mathematical example


We are not done. We now need to use are x value to find our y value. We will use the first equation and replace x to find y.


This means that our ordered pair is (4, -1) and this is the solution to the system. You can check this answer by plugging both numbers into the x and y variable in both equations.


Elimination begins with two equations and two variables but eliminates one variable to have one equation with one variable. This is done through the use of the addition property of equality which states when you add the same quantity to both sides of an equation you still have equality. For example 2+2 = 2 and if at 5 to both sides I get 7 + 7 = 7. The equality remains.

Therefore, we can change one equation using the addition property of equality until one of the variables has the same absolute value for both equations. Then we add across to eliminate one of the variables. If one variable is positive in one equation and negative in the other and has the same absolute value they will eliminate each other. Below is an example using the same system of equations as the previous example.


You can take the x value and plug it into y. We already know y =1 from the previous example so we will skip this.

There are also times when you need to multiply both equations by a constant so that you can eliminate one of the variables


We now replace x with 0 in the second equation


Our ordered pair is (0, -3) which also means this is where the two lines intersect if they were graphed.


Solving a system of equations allows you to handle two variables (or more) simultaneously. In terms of what method to use it really boils down to personal choice as all methods should work. Generally, the best method is the one with the least amount of calculation.

Relations and Functions

In mathematics, a relation is a connection between two distinct pieces of data or variables. For example, student name and ID number would be a relation commonly found at a school. What this means is that you can refer to a student by there name and get their ID number and vice versa.  These two pieces of information are connected and refer to each other. Another term for relation is ordered pair, however, this is more commonly use for coordinate graphing. Below is an example of several student names and ID numbers

Student Name (x values) ID Number (y values)
Jill Smith 12345
Eve Jackson 54321
John Doe 24681

Table 1

Two other pieces of information to know are domain and range. The domain represents all x values. In our table above the student names are the x values (Jill Smith, Eve Jackson, John Doe). The range is all of the y-values, THese are represented by ID number in the table above (12345, 54321, 24681).

The table above is nice and neat. However, sometimes the information is not organized into neat rows but is scrambled with the names and ID numbers not lining up. Below is the same information as the table 1 but the ID numbers are scrambled. The arrows tell who the ID number belongs to who.  This is known as mapping.

Student Name ID Number
Jill Smith ↘ 24681
Eve Jackson→ 54321
John Doe↗ 12345

If we find the ordered pair, domain and range it would be as follows.

  • Ordered pair = {(Jill Smith, 123450, (Eva Jackson, 54321), (John Doe,  24681)}
  • domain = {Jill Smith, Eva Jackson, John Doe}
  • Range = {24681, 54321, 12345}

Understanding Functions

A function is a specific type of relation. What a function does is assigns to each element in a domain. Below is an example of a function

f(x) = 2x + 7

Functions are frequently written to look the same as an equation  as shown below

y = 2x + 7

PLugging in different values of x in your function will provide you with a y as shown below


Here our x-value is 2 and the y-value is 11.

Of course, you can graph function as any other linear equation. Below is a visual.



This post explained the power of relations and functions. Relations are critical in computer science in particular relational databases. In addition., Functions are a bedrock in statistics and other forms of math. Therefore it is critical to understand these basic concepts of algebra.

Absolute Value Equations & Inequalities

The absolute value of a number is its distance from 0.  For example, 5 and -5 both have an absolute value of 5 because both are 5 units from 0. The symbols used for absolute value are |  | with a number or variable placed inside the vertical bars. With this knowledge lets look at an example of an absolute value.


The answer is +5 because both 5 and -5 are 5 units from 0.

In this post, we will look at equations and inequalities that use absolute values.

Solving one Absolute Value Equations

It is also possible to have inequalities with absolute values. To solve these you want to isolate the absolute value and solve the positive and also the negative version of the answer. Lastly, you never manipulate anything inside the absolute value brackets. you only manipulate and simplify values outside of the brackets. Below is an example.


As you can see absolute value inequalities involves solving two equations. Below is an example involving multiplication.


Notice again how the values inside the absolute value were never changed. This is important when solving absolute value inequalities.

Solving Two Absolute Values Equations

Solving two absolute values is not that difficult. You simply make one of the absolute values negative for one equation and positive for another. Below is an example.


Absolute Value Inequalities

Absolute value inequalities require a slightly different approach. You can rewrite the inequality in double inequality form and solve appropriately when the inequality is “less than.” Below is an example.


You can see that we put the absolute value in the middle and simply solved for x. you can even write this using interval notation as shown below.


“Greater than” inequalities are solved the same as inequalities with equal signs. You use the “or” concept to solve both inequalities.


The interval notation is as follows


We use the union sign in the middle is used in place of the word “or”.


This post provided a brief overview of how to deal with absolute values in both equations and inequalities.

Solving Compound Inequalities

Compound inequalities are two inequalities that are joined by the word “and” or the word “or”. Solving a compound inequality means finding all values that make the compound inequality true.

For compound inequalities join ed by the word “and” we look for solutions that are true for both inequalities. Fo compound inequalities joined by the word “or” we look for solutions that work for either inequality.

It is also possible to graph compound inequalities on a number line as well as indicate the final answer using interval notation. Below is a compound inequality with the line graph solution


Solving the answer is the same as a regular equation. Below is the number line for this answer.


The empty circle at -8 means that -8 is not part of the solution. This means all values less than -8 are acceptable answers. This is why the line moves from right to left. All values less than -8 until infinity are acceptable answers. Below is the interval notation.


The parentheses mean that the value next to it is not included as a solution. This corresponds to the empty circle over the -8 in the lin graph. If the value should be included such as with a less/greater than sign you would use a bracket.

Double Inequality

A double inequality is a more concise version of a compound inequality. The goal is to isolate the variable in the middle. Below is an example


This is not complex. We simply isolate x in the middle using appropriate steps. The number line and interval notation or as follows


[-4, 2/3)

This time there is a bracket next to -4 which means that -4 is also a potential solution. In addition, notice how the -4 has a filled circle on the number line. This is another indication that -4 is a solution.

Practical Application

You have signed up for internet access through your cell phone. Your bill is a flat $49.00 per month please $0.05 per minute for internet use. How many minutes can you use internet per month if you want to keep your bill somewhere between $54-$74 per month?

Below is the solution using a double inequality


The answer indicates that you can spend anywhere from 100 to 500 minutes on the internet through your phone per month to stay within the budget. You can make the number line and develop the interval notation yourself.


Compound inequalities are useful for not only as an intellectual exercise. They can also be used to determine practical solutions that include more than one specific answer.

Solving Inequalities

Inequalities are equations that use symbols related to less than, greater than, etc. This allows for the solution to be a range of values rather than only one specific one as in many standard equations where you solve for x.

Unique Property of Inequalities

The rules for solving inequalities are mostly the same as for solving a regular equation with one exception. If you multiply or divide both sides of an inequality by a negative number you need to flip the inequality sign.  Below is an example of the sign flipping


If you look at the final answer you can see that the x must be greater than -2. This makes sense as -5 * -2 would come to 10 which is not less than 10. Naturally,  any number that is larger than -2 would only be worst. Below is a word problem that employs an inequality.

Single Inequality

You have $8,000 to buy math textbooks for your classroom. Each math book cost $127.06. What is the maximum number of math books you can buy?

In the problem above, the keyword is maximum. In other words, there is a range of potential answers from 1 book to whatever the max is. This indicates that this problem is an inequality. Therefore,

  • Let 127.06 be the price of a math book
  • x the number of math books we can buy
  • < use less than because we do not want to exceed our budget of 8000

Below is the solution to the problem.


You can buy up to 62 books and be less than or equal to 8000. We round down to 62 because we must stay under $8,000 in spending.

Below is another example but slightly more complex as it contains additional information.

Complex Single Inequality

You are planning a three-day camping trip for your students. Currently, there is $420 of money available. The students can earn $22.50 per hour through tutoring. The trip will cost $525 for transportation,  $390 for food, and $47.50 per night for the campground.  How many hours do the students need to tutor in order to have enough money for the trip?

This problem has three pieces of information on the left of the inequality

  • Transportation (525)
  • Food (390)
  • campground per night (47.5 * 3)

The information to the right is the following

  • The money available (420)
  • The earning rate per hour (22.50)
  • The variable for the hours to tutor (x)

We use the less than or equal to inequality <

Below is the solution


The students need to tutor for at least 28hours and 20 minutes in order to meet the expenses for the trip.


Inequalities are another useful tool taught in algebra. The applications are limitless. The key to appreciating inequalities is being able to determine when they can be used to solve real-world problems.

Uniform Motion Equations

A uniform motion equation involves trying to make calculations when an object(s) is moving at a constant rate. The formula for this type of equation is below.

rate * time = distance

Generally, you want to make a table that includes all of the known information. This allows you to determine what the unknown information is that needs to be solved. Below is a table that you can use.

Rate            * Time            = Distance

Let’s go through some examples

Example 1

Dan and William are riding bicycles. Dan’s speed is 4 kph faster than William’s speed. It takes William 1.5 hours to reach the beach while it takes Dan 1 hour. Find the speed of both bicyclists.

Here is what we know

  • Dan is 4 kph faster than William
  • It takes Dan 1 hour to get to the beach
  • William is 4 kph slower than Dan
  • It takes William 1.5 hours to get to the beach

We will now setup our table

Rate            * Time            = Distance
 Dan  r + 4  1  1(r + 4)
 William  r  1.5  1.5r

We will now solve this equation by placing Dan’s information on one side of the equation and William’s information on the other side of the equation. Below is the solution


We now know what r is so we need to plug this into the table to get the answers

Rate            * Time            = Distance
 Dan  8 + 4 = 12  1  1(8 + 4) = 12
 William  8  1.5  1.5(8) = 12

The speed of Dan was 12kph while the speed of William was 8kph. This first example was two people traveling the same distance. The next example will be two people travel a different distance.

Example 2

Jenny is traveling to meet her brother. She travels from Saraburi to Chang Mai while her brother travels from Chang Mai to Saraburi. They meet in Bangkok. The distance from Saraburi to Chang Mai is 620km. It takes Jenny 2 hours to get to Bangkok while it takes the brother 7.5 hours to get there. Jenny’s brother’s average speed is 30kph faster than hers. Find the average speed for both people.

The table below captures all of our information

Rate            * Time            = Distance
 Jenny r  2  2r
 Brother  r + 30  7.5  7.5(r + 30)

To solve this problem we combine the information about Jenny and her brother and set this information to equal 620 which is the total distance. Below is the solved equation.


We can now place this information in our table.

Rate            * Time            = Distance
 Jenny 41.57  2  2(41.57) = 83.14
 Brother  41.57 + 30 = 71.57  7.5  7.5(41.57 + 30) = 536.78

Jenny average speed was 41.57kph while her brother’s speed was 71.57kph. If you add up the distance traveled it will sum to 620.

Our final example will look at determining the time travel when we know the rate of the two objects.

Example 3

A husband and wife both leave their home. The wife travels east and the husband travels west. Wife travels 80kph while the husband travels 100kph. How long will they travel before they are 360km apart?

Below is what we know

Rate            * Time            = Distance
Husband 100  t 100t
Wife 80 t  80t

To solve this we combine the wife and husband information on one side of the equation and put the total distance traveled on the other side. The solution is below.


We place our answer inside our table

Rate            * Time            = Distance
Husband 100  2 100(2) = 200
Wife 80 2  80(2) = 160

It takes two hours for the wife and husband to be 360km apart.


Understanding uniform equations involve determining first what you know and then determining what the problem wants you to figure out. If you follow this simple process and are able to identify when an equation involves a uniform application it should not be difficult to find the solution.

Algebraic Mixture Problems

There are many examples in the world in which you want to know the quantity of several different items that make up a whole. When such a situation arise it is an example of mixture problem.

In this post, we will look at several examples of mixture problems. First, we need to look at the general equation for a mixture problem.

number * value = total value

The problems we will tackle will all involve some variation of the equation above. Below is our first example

Example 1

There are times when you want to figure out how many coins are needed to equal a certain dollar amount such as in the problem below

Tom has $6.04 of pennies and nickels. The number of nickels is 4 more and 6 times the number of pennies. How many nickels and pennies does Tom have?

To have success with this problem we need to convert the information into a table to see what we know. The table is below.

Type Number * Value = Total Value
Pennies x .01 .01x
Nickels 6x+4 .05 .05(6x+4)

total 6.04

We can now solve our equation.


We know that there are 18.83 pennies. To determine the number of nickels we put 18.83 into x and get the following.


Almost 117 nickels

You can check if this works for yourself.

Example 2 

For those of us who love to cook, mixture equations can be used for this as well below is an example.

Tom is mixing nuts and cranberries to make 20 pounds of trail mix. Nuts cost $8.00 per pound and cranberries cost $3.00 per pound. If Tom wants to his trail mix to cost $5.50 per pound how many pounds of raisins and cranberries should he use? 

Our information is in the table below. What is new is subtracting the number of pounds from x. Doing so will help us to determine the number of pounds of cranberries.

Type Number of Pounds* Price Per Pound = Total Value
nuts x 8 8x
Cranberries 20-x 3 3(20-x)
Trail Mix 20 5.5 20(5.50)

We can now solve our equation with the information in the table above.


Once you solve for x you simply place this value into the equation. When you do this you see that we need ten pounds of nuts and berries to reach our target cost.


This post provided to practical examples of using algebra realistically. It is important to realize that understanding these basics concepts can be useful beyond the classroom.

Solving Equations with Fractions or Decimals

This post will provide an explanation of how to solve equations that include fractions or decimals. The processes are similar in that both involve determining the least common denominator.

Solving Equations with Fractions

The key step to solving equations with fractions is to make sure that the denominators of all the fractions are the same. This can be done by finding the least common denominator. The least common denominator (LCD) is the smallest multiple of the denominators. For example, if we look at the multiples of 4 and 6 we see the following.


You can see clearly that the number 12 is the first multiple that 4 and 6 have in common. You can find the LCD by making factor trees but that is beyond the scope of this post. The primary reason we would need the LCD is when we are adding fractions in an equation. If we are multiplying we could simply multiply straight across.

Below is an equation that has fractions. We will find the LCD


Here is an explanation of each step

  • A. This is the original problem. We first need to find the LCD
  • B. We then multiply each fraction by the LCD
  • C. This is the equation we solve for
  • D. We get the variable alone by subtracting 20 from each side
  • E. We have our new simplified equation
  • F. We further isolate the variable by dividing by 3 on both sides.
  • G. This is our answer

Solving Equations with Decimals

The process for solving equations with decimals is almost the same as for fractions. The LCD of all decimals is 100. Therefore, one common way to deal with decimals is to multiply all decimals by 100 and the continue to solve the equation.

The primary benefit of multiply by 100 is to remove the decimals because sometimes we make mistakes with where to place decimals. Below is an example of an equation with decimals.

1.pngHere is what we did

  • A. Initial equation
  • B. we distribute the 0.10
  • C. Revised equation
  • D. We multiply everything by 100.
  • E. Revised equation
  • F. Subtract 20 from both sides to isolate the variable
  • G. Revised equation
  • H. Divide both sides by 30 to isolate the variable
  • I. Final answer


Understand the process of solving equations with fractions or decimals is not to complicated. However, this information is much more valuable when dealing with more complex mathematical ideas.

Linear Equations

In this post, we will look at several types of equations that you would encounter when learning algebra. Algebra is a foundational subject to know when conducting most quantitative research.


An equation is a statement that balances two expressions. Often equations include a variable or an unknown value. By solving for the unknown value you are able to balance the equation.

There are many different types of equations such as

(1) Linear equation

(2) Quadratic equation

(3) Polynomial equation

See number 1 or 2. The rule for polynomial equation is that the exponent must be positive
(4) Trigonometric equation

(5) Radical equation

(6) Exponential equation

This post will focus on linear equations.

Linear Equation

A linear equation is an equation that if it is graph will render a straight line. It is common to have to solve for the variable in a linear equation by isolating as in the example below.


There are also several terms related to equations and the include the following

  • Conditional equation: An equation that is true for only one value of the variable. The example above is a conditional equation.
  • Identity: An equation that is true for any value of a variable. Below is an example

Any value of x will work with an identity equation.

  • A contradiction is an equation that is false for all values. Below is an example


No value of x will work with the equation above.


This post provided an overview of the types of equations commonly encountered in algebra.

Teaching Math

Probably one of the most dreaded subjects in school is math. Many students fear this subject and perhaps rightfully so. This post will provide some basic tips on how to help students to understand what is happening in math class.

Chunk the Material

Many math textbooks, especially at the college level, are huge. By huge we are talking over 1000 pages. That is a tremendous amount of content to cover in a single semester even if the majority of the pages are practice problems.

To overcome this, many have chapters that are broken down into 5 sub-sections such as 1.1, 1.2, etc. This means that in a given class period, students should be exposed to 2 or 3 new concepts. Depending on their background this might be too many for a student, especially if they are not a math major.

Therefore, a math teacher must provide new concepts only after previous concepts are mastered. This means that the syllabus needs to flexible and the focus is on the growth of students rather than covering all of the material.

Verbal Walk Through

When teaching math to a class, normally a teacher will provide an example of how to do a problem. The verbal walkthrough is when the teacher completes another example of the problem and the students tell the teacher what to do verbally. This helps to solidify the problem-solving process in the students’ minds.

A useful technique in relation to the verbal walkthrough is to intentional make mistakes when the students are coaching you. This requires the students to think about what is corrected and to be able to explain what was wrong with what the teacher did. The wisest approach is to make mistakes that have been experienced in the past as these are the ones that are likely to be repeated.

The verbal walkthrough works with all students of all ages. It can be more chaotic with younger children but this is a classic approach to teaching the step-by-step process of learning math calculations.

Practice Practice Practice

Daily practice is needed when learning mathematical concepts. Students should be learning new material while reviewing old material. The old material is reviewed until it becomes automatic.

This requires the teacher to determine the most appropriate mix of new and old. Normally, math has a cumulative effect in that new material builds on old. This means that students are usually required to use old skills to achieve new skills. The challenge is in making sure the old skills are at a certain minimum level that they can be used to acquire new skills.


Math is tough but if a student can learn it math can become a highly practical tool in everyday life. The job of the teacher is to develop a context in which math goes from mysterious to useful.

Algebraic Expressions and Equations

This post will focus mainly on expressions and their role in algebra. Expressions play a critical role in mathematics and we all have had to try and understand what they are as well as what they mean.

Expression Defined

To understand what an expression is you first need to know what operation symbols are. Operation symbols tell you to do something to numbers or variables. Examples include the plus, minus, multiply, divide, etc.

An expression is a number, variable, or a combination of numbers[s] and variable[s] that use operation symbols. Below is an example


Expressions consist of two terms and these are variables and constants. A variable is a letter that represents a number that can change. In our example above, the letter a is a variable.

A constant is a number whose value remains the same. In our example above, the numbers 2 and 4 are constants.

Expression vs Equation

An equation is when two expressions connected by an equal sign as shown below.


In the example above we have to expressions. To the left of the equal sign is 2a * 4 and to the right of the equal sign is 16. Remember that an expression can be numbers and or variables so 16 is an expression because it is a number.

Simplify an Expression

Simplifying an expression involves completing as much math as possible to reduce the complexity of an expression. Below is an example.


In order to complete this expression  above you need to know the order of operations which is explained below

Multiplication Division
Adition Subtraction

In the example above, we begin with multiplication of 8 and 4 before we do the addition of adding 2. It’s important to remember that for multiplication/division or addition/subtraction that you move from left to right when dealing with these operation symbols in an expression. It is also important to know that subtraction and division are not associative (or commutative) that is: (1 – 2) – 3 != 1 – (2 – 3).

Evaluating an Expression

Evaluating an expression is finding the value of an expression when the variable is replaced with a specific number.


Combining Like Terms

A common skill in algebra is the ability to combine like terms. A term is a constant or constant with one or more variables. Terms can include a constant such as 7 or a number and variable product such as 7a. The constant that multiplies the variable is called a coefficient. For example, 7a, 7 is the constant and the coefficient while a is the variable.

Combining like terms involves combining constant are variables that have the same characteristics for example

  • 3 and 2 are like terms because they are both constants
  • 2x and 3x are like terms because they are both constants with the same variable

Below is an example of combining like terms


In this example, we first placed like terms next to each other. This makes it easier to add them together. The rest is basic math.


Hopefully, the concept of expressions makes more sense. This is a foundational concept in mathematics that if you do not understand. It is difficult to go forward in the study of math.

Basic Algebraic Concepts

This post will provide insights into some basic algebraic concepts. Such information is actually useful for people who are doing research but may not have the foundational mathematical experience.


A multiple is a product of  and a counting number of n. In the preceding sentence, we actually have two unknown values which are.

  • n
  • Counting number

The can be any value, while the counting number usually starts at 1 and continues by increasing by 1 each time until you want it to stop. This is how this would look if we used the term n,  counting number, and multiple of n. 

n * counting number = multiple of n

For example, if we say that = 2 and the counting numbers are 1,2,3,4,5. We get the following multiples of 2.


You can see that the never changes and remains constant as the value 2. The counting number starts at 1 and increases each time. Lastly, the multiple is the product of n and the counting number.

Let’s take one example from above

2 * 3 = 6

Here are some conclusions we can make from this simple equation

  • 6 is a multiple of 2. In other words, if I multiply 2 by a certain counting number I can get the whole number of 6.
  • 6 is divisible by 2. This means that if I divide 2 into six I will get a whole number counting number which in this case is 3.

Divisibility Rules

There are also several divisibility rules in math. They can be used as shortcuts to determine if a number is divisible by another without having to do any calculation.

A number is divisible by

  • 2 when the last digit of the number 0, 2, 4, 6, 8
    • Example 14, 20, 26,
  • 3 when the sum of the digits is divisible by 3
    • Example 27 is divisible by 3 because 2 + 7 = 9 and 9 is divisible by 3
  • 5 when the number’s last digit is 0 or 5
    • Example 10, 20, 25
  • 6 when the number is divisible by 2 and 3
    • Example 24 is divisible by 6 because it is divisible by 2 because the last digit is for and it is divisible by 3 because 2 + 4 = 6 and six is divisible by 3
  • 10 when the number ends with 0
    • Example 20, 30 , 40, 100


Factors are two or more numbers that when multiplied produce a number. For example


The numbers 7 and 6 are factors of 42. In other words, 7 and 6 are divisible by 42. A number that has only itself and one as factors is known as a prime number. Examples include 2, 3, 5, 7, 11, 13. A number that has many factors is called a composite number and includes such examples as 4, 8, 10, 12, 14.

An important concept in basic algebra is understanding how to find the prime numbers of a composite number. This is known as prime factorization and is done through the development of a factor tree. A factor tree breaks down a composite number into the various factors of it. These factors are further broken down into their factors until you reach the bottom of a tree that only contains prime numbers. Below is an example



You can see in the tree above that the prime factors of 12 are 2 and 3. If we take all of the prime factors and multiply them together we will get the answer 12.



Understanding these basic terms can only help someone who maybe jumped straight into statistics in grad school without have the prior thorough experience in basic algebra.