Solving a system of non-linear equations means that at least one of the equations is not linear. For example, if one equation has an exponent it may be a parabola or a circle. With this no shape that is not linear it involves slightly different expectations.

Solving a system of non-linear equations is similar to solving a system with linear equations with one difference. The difference is that with nonlinear equations you can have more than one solution. What this means is that the lines that are the equations can intersect in more than one place. However, it is also possible they do not intersect. How many solutions depends on the lines involved.

For example, if one equation is a circle/parabola and the other is a line there can be 0-2 solutions. If one equation makes a circle and the other makes a parabola there can be up to 4 solutions. Two circles or two parabolas can make a multitude of solutions

The steps for solving a nonlinear system are the same. Therefore, in this post, we will demonstrate how to solve a system of non-linear equations using the substitution and elimination methods.

Substitution

The substitution method is when we plug one equation into the variable of the other equation. Below is our system of equations.

The first thing to notice is that the top equation would make a circle if you graphed it. That is why this is a non-linear system. To solve we take the second equation and substitute it for y. Below we find the values for x

Now we complete the system by finding the values for y.

THerefore are ordered pairs are (0, -3) and (1,0). These are the two points at which the equations intersect if you were to graph them.

Elimination

Elimination involves making the coefficients of one of the variables opposite so that when they are added together they cancel each other out. By removing one variable you can easily solve for the other. Below is the system of equation we want to solve.

The top equation makes a circle while the bottom one makes a parabola. This means that we can have as many as four solutions for this system. To solve this system we will multiply the bottom equation by -1. This will allow us to remove the x variable and then solve for y. Below are the steps.

Now we simply solve for y.

Now we can take these values for y to solve for x.

The order pairs are as follows

• (-2,0)
• (2,0)
•   (√3, -1)
• (-√3, -1)

Conclusion

From this, you can see that non-linear equations can be solved using the same approaches. Understanding this is key to many other fields of math such as data science and machine learning.

This post will look at how to solve radical equations. The concepts are mostly similar to solving any other equation in terms of isolating terms etc. However, for people who are new to this, it may still be confusing. Therefore, we will go through several examples.

Example 1

Our first example is a basic radical equation that includes a constant outside the radical. Below is the equation.

Solving this problem requires to main steps.

1. Isolate the radical
2. Remove the radical by squaring it

Doing these two steps will lead to our answer. We will have two answers but the reason for this will become clear as we solve the equation.

First, we will isolate the radical by subtracting 1 from both sides

Now, to remove the radical we will square both sides. This new equation will need to be simplified and will become a quadratic equation.

With our new quadratic equation we will factor this and as expected get two answers.

Index Other than 2

For a radical that has an index other than  2, the process involves raising the radical to whatever power will cancel out the radical. Below is an example that has an index of 3. We will first subtract the constant from both sides.

In order to remove the index of 3, we need to raise each side of the equation to the power of 3. After doing this, we solve a simple equation.

Radicals as Fractions

One a number is a raised to a power that is a fraction it is the same as a radical. Below is an example.

This means that the steps we took to solve equations with radicals can be mostly used to deal with equations with powers that are a fraction.

below is an equation. To solve this equation you must raise each side of the equation to the power of the denominator in the fraction.

As you can see both sides were raised to the 4th power because that is the number in the denominator of the fraction. On the left side of the equation, the 4th power cancels out the fraction. Now you can simply solve the equation like any other equation.

Hopefully, this is clear.

Conclusion

Solving radical equatons in not that diffcult. Usually, the ultimate goal is to remove the radical. The difference between this and solving for other equations is that with radical equations you want to first isolate the radical, remove the radical, and then solve for the unknown variable.

Roots, radicands, and radicals are yet another way to express numbers in algebra. In this post, we will go over some basic terms to know.

Roots

A square root is a number that is multiplied by itself to get a new number. Below is an example

In the example above 5 is the square root of 25. This means that if you multiply 5 by its self you would get 25.

Another term to know is the square. The square is the result of multiplying a number by its self. In the expression above 25 is the square of 5 because you get 25 by multiplying 5 by its self.

Square Roots

Square roots, in particular, have a lot of other ways to be expressed. To understand square roots you need to know what roots, radicands, and radical sign are. Below is a picture of these three parts.

The radical sign is simply a sign like multiplication and division are. The radicand is the number you want to simplify by finding a number that when multiplied by itself would equal the value in the radicand. We also call this new number the square root. For example,

What the example above means is that the number you can multiply by itself to get 100 is 10.

The index is trickier to understand. It tells you how many times to multiply the number by its self to get the radicand. If no number is there you assume the index is 2. Below is an expression with an index that is not 2.

What this expression is saying is that you can multiply 2 by its self 3 times to get eight as you can see below.

2 * 2  = 4 * 2 =  8

Additional Terms

There are some basic terms that are needed to understand using radicals. Generally, when every we are speaking of multiplying two times we call it square. Multiplying three times is referred to as cub or cubic. Anything beo=yond 3 is called to the nth powered. For example, multiplying a number by its self 4 times would be called to the 4th power, 5 times to the 5th power etc. However, some people referred to the square as the 2nd  power and the cube as the 3rd power if this is not already confusing. Below is a table that clarifies things

Conclusion

There are many more complex ideas and operations that can be performed with radicands and radicals. One of the primary benefits is that you can avoid dealing with decimals for many calculations when you understand how to manipulates these terms. As such, there actual are some benefits in understanding radicands and radicals use.

Polynomial is an expression that has more than one algebraic term. Below is an example,

Of course, there is much more to polynomials then this simple definition. This post will explain how to deal with polynomials in various situations.

Add & Subtract

To add and subtract polynomials you must combine like terms. Below is an example

All we did was combine the terms that had the y^2 in them. This, of course, applies to subtraction as well.

In the example above, the terms with “a” in them are combined and the terms with “n” in then are combined.

Exponents

When dealing with exponents when multiplication is involved you add the exponents together.

Notice how like terms were dealt with separately.

Since we add exponents during multiplication we subtract them during division.

The 2 in the numerator and denominator cancel out and  7 – 5 = 2.

When parentheses are involved it is a little more complicated. For example, when the exponent is negative you multiply the exponents below

For negative exponents, you fill the numerator and denominator around and make the negative exponent positive.

There are other concepts involving polynomials not covered here. Examples include long division with polynomials and synthetic division. These are fascinating concepts however in the books I have consulted, once these concepts are taught they are never used again in future chapters. Therefore, perhaps they are simply interesting but not commonly used in practice.

Conclusion

Understanding polynomials is critical to future success in algebra. As concepts become more advanced, it will seem as if you are always trying to simplify terms using concepts learned in relation to polynomials.

There are several different ways to solve a system of equations. Another common method is Cranmer’s Rule. However, we cannot learn about Cranmer’s rule until we understand determinants.

Determinants are calculated from a square matrix, such as 2×2 or 3×3. In a 2×2 matrix, the determinant is calculating by taking the product of the diagonal and finding the difference.

Here is how this look with real numbers

Determinant for 3×3 Matrix

To find the determinants of a 3×3 matrix it takes more work.  By address, it is meant the row number and column letter. To calculate the determinant you must remove the row and column of that contains the variable you want to know the determinant too. Doing this creates what is called the minor. Below is an example with variables.

As you can see, to find the determinant of a1 we remove the row and the column that contains a1. From there, you do the same math as in a 2×2 matrix. When using real numbers you may need to add the row letter and column number to figure out what you are solving for. Below is an example with real numbers.

Find the determinant of c2

The number at c2 is -3. Therefore, we remove the row and the column that contains -3 and we are left with the minor of c2 shown below.

IF we follow the steps for a 2×2 matrix we can calculate the determinant of c2 as follows.

4(4) – (-2)(-2) =
16 – 12 =
4

The answer is 4.

Expand by Minors

Knowing the minor is not useful alone, The minor of different columns can be added together to find the determinant for a 3×3 matrix. Below is the expression for finding the determinant of 3×3 matrix.

What is happening here is that you find the determinant of a1 and multiply it by the value in a1. You do this again for b1 and c1. Lastly, you find the sum of this process to evaluate the determinant of the 3×3 matrix. Below is another matrix this time with actual numbers. We are going to expand from the first row and first column

All we do not is obtain the determinant of each 2×2 matrix and multiply it by the outside value before adding it all together. Below is the math.

 2(-4 – 0) + 3(-6 – 0) – 1(-3 – (2-2)) = 
-18 – 18 + 1= 
-25

Conclusion

This information is not as useful on its own as it is as a precursor to something else. The knowledge acquired here for finding determinants provides us with another way to approach a system of equations using matrices.

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.

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

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.

Conclusion

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.

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

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.

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.

Conclusion

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.

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.

Conclusion

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.

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.

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

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

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

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.

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

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

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

Conclusion

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.