Tag Archives: algebra

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.

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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.

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.

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.