Tag Archives: math

The Quadratic Formula

The quadratic formula is used for solving quadratic equations. The actual creation of this formula is somewhat complex. Creating it requires the use of completing the square as well as square root property. Below is what the equation looks like.

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For our purposes, we will go through an example that solves a quadratic equation using the quadratic formula. In addition, we will also explore the idea of the discriminant as it relates to quadratic formulas.

Example

The mechanics of solving a quadratic formula using this approach is similar to most other methods. You simply plug in the substitutes in the equation to get your actual answer. Below is an example,

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We will now plug in the values and determine x.

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Discriminant 

The discriminant of a quadratic equation is used to determine the type if the answer you would get if you solve the equation. THere are three types of answers that you can get when solving a quadratic equation.

  • Two real solutions-This happens when the discriminant results are positive.
  • One real solution-Happens when the discriminant results are zero
  • Two complex-Happens when the discriminant is negative

A complex solution involves the use of an imaginary number. This happens when the square root number is negative, which is technically impossible. To deal with this in math the letter i is used instead of the negative sign below is an example.

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The actual formula for calculating the discriminant is already in the quadratic formula. You simply calculate only the information under the square root. This is shown below.

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IN our first example, we got two real solutions. We will now confirm this by calculating the discriminant.

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Are answer is positive, which means that we can expect to calculate to real solutions for this particular problem.

Conclusion

The quadratic formula provides another way to solve a quadratic equation. This is probably the easiest method to learn as it is simply a matter of plugging numbers into the formula. This may explain why the quadratic formula is frequently the first method algebra students learn for solving quadratic equations.

The discriminant is a shortcut calculation that allows you to determine the quality of the solutions you would get if you solve the equation.

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Quadratic Equations and the Square Root Property

A quadratic equation is an equation that includes a variable raised to the second power. Below is a common format for a quadratic equation.

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This characteristic makes it difficult to rely on linear equation tricks of addition, subtraction, and multiplying to isolate the variable. One trick that we often have to use now is factoring as shown below.

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An alternative way to solve quadratic functions is through having knowledge of the square root property which is shown below.

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Below is the same example as our first example but this time we use the square root property.

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This trick works for numbers that cannot be factored.

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This leads us to the point that the square root property is used for speed or when factoring is not an option.

With this knowledge, all the other possible ways to solve a linear equation can be used to solve a quadratic equation

Division

In the example below is a quadratic formula in which you have to divide to isolate the variable. From there you solve like always1.pngFraction

To remove a fraction you must multiply both sides by the reciprocal as shown below.

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When we got the square root of 18 we had to further simplify the radical by finding the factors of 18. In the second to last line if you multiply these numbers together you will get 18 because  9 * 2 = 18. Furthermore, if you square root 9 you get 3 but you cannot square root 2 and get a whole number. This is why the final answer is 3 * the square root of 2.

Conclusion

 

Quadratic formulas are common in algebra and as such there are many different ways to solve them. In this post, we looked at an alternative to factoring called the square root property. Understanding this approach is valuable as you can often solve quadratic equations faster and or they can be used when factoring is not possible.

Solving Single Radical Equations

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.

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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

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

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

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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.

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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.

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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.

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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.

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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.

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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

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

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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.

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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,

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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.

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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

Number Power Example
2 square n2
3 cube n3
4 4th power n4
5 5th power n5

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.

Polynomials

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

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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 example1

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

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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.

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Notice how like terms were dealt with separately.

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

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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

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

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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.

Cramer’s Rule

Cramer’s rule is a method for solving a system of equations using the determinants. In order to do this, you must be familiar with matrices and row operations. Generally, it is really difficult to explain that is a simple matter but there are two main parts to completing this

Part 1:

  • Evaluate the determinants using the coefficients aka D
  • Evaluate the determinants using the constants in place of x aka Dx
  • Evaluate the determinant using the constants in place of y aka Dy

Part 2:

  • Find x by  Dx / D
  • Find y by Dy / D

This is modified if the system is 3 variables. Below we will go through an example with 2 variables.

Example

Here is our problem

2x + y = -4
3x – 2y = -6

Below is the matrix of the system of equations

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We will first evaluate the Determinant D using the coefficients. In other words, we are going to calculate the determinant for the first two columns of the matrix. Below is the answer.

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What we have just done we will do two more times. Once two find the determinant of x and once to find the determinant of y. When we say determinant x or y we are excluding that column from the 2×2 matrix. In other words, if I want to find the determinant of x I would exclude the x values from the 2×2 matrix when calculating. Below is the determinant of x.

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Lastly, here is the determinant of y

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We no have all the information we need to solve for x and y. To find the answers we do the following

  • Dx / D = x
  • Dy / D = y

We know these value already so we plug them in as shown below.

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You can plug in these values into the original equation for verification.

The steps we took here can also be applied to a 3 variable system of equation. In such a situation you would solve one additional determinant for z.

Conclusion

Solving a system of equations using Cramer’s rule is much faster and efficient than other methods. It also requires some additional knowledge of rows and matrices but the benefits far outweigh the challenge of learning some basic rules of row operations.

Evaluating the Determinant of a Matrix

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.

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Here is how this look with real numbers

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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.

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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

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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.

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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.

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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

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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.

Solving a System of Equations with Matrices: 2 Variables

This post will provide examples of solving a system of equations with 2 variables. The primary objective of using a matrix is to perform enough row operations until you achieve what is called row-echelon form. Row-echelon form is simply having ones all across the diagonal from the top left to the bottom right with zeros underneath the dia. Below is a picture of what this looks like

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It is not necessary to have ones in the diagonal it simply preferred when possible. However, you must have the zeros underneath the diagonal in order to solve the system. Every zero represents a variable that was eliminated which helps in solving for the other variables.

Two-Variable System of Equations

Our first system is as follows

3x + 4y = 5
x + 2y = 1

Here is our system

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Generally, for a 2X3 matrix, you start in the top left corner with the goal of converting this number into a 1.Then move to the second row of the first column and try to make this number a 0. Next, you move to the second column second row and try to make this a 1.

With this knowledge, the first-row operation we will do is flip the 2nd and 1st row. Doing this will give us a 1 in the upper left spot.

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Now we want in the bottom left column where the 3 is currently at. To do this we need to multiply row 1 by -3 and then add row 1 to row 2. This will give us a 0.

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We now need to deal with the middle row, bottom number, which is -2. To change this into a 1 we need to multiple rows to by the reciprocal of this which is -1/2.

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If you look closely you will see that we have achieved row-echelon form. We have all 1s in the diagonal and only 0s under the diagonal.

Our new system of equations looks like the following

1x + 2y = 1
0x +1y = -1 or y = -1

If we substitute -1 for y in our top equation we can solove for x.

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We now know that x = 3 and y = -1. This indicates that we have solved our system of equations using matrices and row operations.

Conclusion

Using matrices to solve a system of equations can be cumbersome. However, once this is mastered it can often be faster than other means. In addition, understanding matrices is critical to being able to appreciate complex machine learning algorithms that almost exclusively use matrices.

Augmented Matrix for a System of Equations

Matrices are a common tool used in algebra. They provide a way to deal with equations that have commonly held variables. In this post, we learn some of the basics of developing matrices.

From Equation to Matrix

Using a matrix involves making sure that the same variables and constants are all in the same column in the matrix. This will allow you to do any elimination or substitution you may want to do in the future. Below is an example

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Above we have a system of equations to the left and an augmented matrix to the right. If you look at the first column in the matrix it has the same values as the x variables in the system of equations (2 & 3). This is repeated for the y variable (-1 & 3) and the constant (-3 & 6).

The number of variables that can be included in a matrix is unlimited. Generally,  when learning algebra, you will commonly see 2 & 3 variable matrices. The example above is a 2 variable matrix below is a three-variable matrix.

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If you look closely you can see there is nothing here new except the z variable with its own column in the matrix.

Row Operations 

When a system of equations is in an augmented matrix we can perform calculations on the rows to achieve an answer. You can switch the order of rows as in the following.

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You can multiply a row by a constant of your choice. Below we multiple all values in row 2 by 2. Notice the notation in the middle as it indicates the action performed.

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You can also add rows together. In the example below row 1 and row 2, are summed to create a new row 1.

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You can even multiply a row by a constant and then sum it with another row to make a new row. Below we multiply row 2 by 2 and then sum it with row 1 to make a new row 1.

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The purpose of row operations is to provide a way to solve a system of equations in a matrix. In addition, writing out the matrices provides a way to track the work that was done. It is easy to get confused even the actual math is simple

Conclusion

System of equations can be difficult to solve. However, the use of matrices can reduce the computational load needed to solve them. You do need to be careful with how you modify the rows and columns and this is where the use of row operations can be beneficial.

System of Equations and Mixture Application

Solving a system of equations with a mixture application involves combining two or more quantities. The general setup for the equations is as follows

Quantity * value = total

This equation is used for both equations. You simply read the problem and plug in the information. The examples in this post are primarily related to business as this is one of the more practical applications of solving a system of equations for the average person. However, a system of equations for mixtures can also be used for determining solutions but this is more common in chemistry.

Example 1: Making Food 

John wants to make 20 lbs of granola using nuts and raisins. His budget requires that the granola cost $3.80 per pound. Nuts are $4.50 per pound and raisins are $1.00 per pound. How many pounds of nuts and raisins can he use?

The first thing we need to determine what we know

  • cost of the raisins
  • cost of the nuts
  • total cost of the granola
  • number of pounds of granola to make

Below is all of our information in a table

Pounds * Price Total
Nuts n 4.50 4.5n
Raisins r 1 r
Granola 20 3.80 3.8(20) = 76

What we need to know is how many pounds of nuts and raisins can we use to have the total price per pound be $3.80.

With this information, we can set up our system of equations. We take the pounds column and create the first equation and the total column to create the second equation.

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We will use elimination to solve this system. We will multiply the first equation by -1 and combine them. Then we solve for n as in the steps below

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We know n = 16 or that we can have 16 pounds of nuts. To determine the amount of raisins we use our first equation in the system.

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You can check this yourself if you desire.

Example 2: Interests

Below is an example that involves two loans with different interest rates. Our job will be to determine the principal amount of the loan.

Tom owes $43,080 on two student loans. The bank’s interest rate is 5.25% and the federal loan rate is 2.95%. The total amount of interest he paid last two years was 6678.72. What was the principal for each loan

The first thing we need to determine what we know

  • bank interest rate
  • Federal interest rate
  • time of repayment
  • Amount of loan
  • Interest paid so far

Below is all of our information in a table

Principal * Rate Time Total
Bank b 0.0525 1 0.0525b
Federal f 0.0295 1 0.0295f
Total 43080 1752.45

Below is our system of equation

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To solve the system of equations we will use substitution. First, we need to solve for b as shown below

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We now substitute  and solve

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We know the federal loan is $22,141.30 we can use this information to find the bank loan amount using the first equation.

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The bank loan was $20,938.70

Conclusion

Hopefully, it is clear by now that solving a system of equations can have real-world significance. Applications of this concept can be useful in the context of business as shown here.

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

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

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

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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.

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

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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.

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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.

Conclusion

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

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

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We can now solve our system of equations

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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.

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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

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

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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.

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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.

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

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

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

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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

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

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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.

Conclusion

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

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

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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.

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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

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.

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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

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We now replace x with 0 in the second equation

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

Conclusion

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

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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.

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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.

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.

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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.

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

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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.

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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.

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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.

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

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The interval notation is as follows

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

Conclusion 

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

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

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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.

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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

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

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[-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

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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.

Conclusion

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

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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.

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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

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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.

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

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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)
 620

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.

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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
 620

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
 360

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.

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We place our answer inside our table

Rate            * Time            = Distance
Husband 100  2 100(2) = 200
Wife 80 2  80(2) = 160
 360

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.

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.

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

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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.

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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.

Conclusion

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.

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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

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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

Conclusion

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.

Equations

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

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(2) Quadratic equation

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(3) Polynomial equation

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

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(5) Radical equation

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(6) Exponential equation
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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.

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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
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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

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No value of x will work with the equation above.

Conclusion

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.

Conclusion

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

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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.

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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.

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

Parentheses
Exponents
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.

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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

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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.

Conclusion

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.

Multiple

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.

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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

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

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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

 

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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.

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Conclusion

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.