Quotient Power Property

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Solve a system of equation with elimination

Mixture problem and system of equations

Solving a system of equations with substitution

Using the concept of system of equations in the context of uniform motion problems

Application of solving a system of equations

Solving linear inequalities

Solving double inequalities

Uniform motion equations

Calculating Confidence Intervals for Proportions

Solving mixture problems

Solving linear inequalities in word problems.

Calculating simple interest

Solving Linear equations involving word problems

Calculating standard deviation

Solving a linear equation

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.

A logarithm is the inverse of exponentiation. Depending on the situation one form is better than the other. This post will explore logarithms in greater detail.

**Converting Between Exponential and Logarithmic Form**

There are times when it is necessary to convert an expression from exponential to logarithmic and vice versa. Below is an example of who the expression is rearranged form logarithm to exponential.

The simplest way to explain I think is as follows

- for the logarithm, the exponent (y) and the base (a) are on opposite sides of the equal sign
- For the exponent form, the exponent (y) and base (a) are on the same side of the equal sign.

Here is an example using actual numbers

As you can see the exponent 3 and the base 2 are on opposite sides of the equal sign for the logarithmic form but er together for the exponential form.

When the base is e (Euler’s Number) it is known as a natural logarithmic function. e is the base rate growth of a continual process. The application of this is limitless. When the base is ten it is called a common logarithmic function.

**Logarithmic Model Example**

Below is an example of the application of logarithmic models

*Exposure to noise above 120 dB can cause immediate pain and damage long-term exposure can lead to hearing loss. What aris the decimal level of a tv with an intensity of 10^1 watts per square inch. *

First, we need the equation for calculating the decibel level.

Now we plug in the information into the word problem for I and solve

Our tv is dangerously loud and should include a warning message. We dropped the negative sign because you cannot have negative decibel level.

**Conclusion**

Logarithms are another way to express exponential information and vice versa. It is the situation that determines which to use and the process of concert an expression from one to another is rather simple. In terms of solving actual problems, it is a matter of plugging numbers into an equation and allowing the calculator to work that allows you to find the answer.

There are times when we want to understand growth that is not constant. An example of this would be the growth of a virus. As time goes by the virus growth rate increases more and more. Another example would be in the world of finance when we are dealing with interest.

In situations like the ones mentioned above, it is critical to understand the use and application of exponential models. This post will go through examples of the use of exponential models.

**Finance Example**

One common exponential model in finance is for compounded interest. The equation is as follows…

Below is a simple word problem that calls for this equation

*You invest $10,000 in a mutual fund to prepare for retirement. The interest rate is 5% compounded monthly, how much will be in the account when you plan to retire in 25 years. *

Below is what we now

- balance = ?
- principal = $10,000
- rate = 0.05
- years= 25
- times in year = 12 * 25 = 300

Now, we simply plug this information into the equatiom to get the answer.

The answer is shown above. The initial investment would grow to almost $35,000 dollars over 25 years.

**Continuous Growth**

In some fields, such as the life sciences, you want to now the growth of a virus or bacteria. Unlike in finance where the balance grows several times a year, a bacteria is growing continuously. This leads to a slightly different exponential model as shown below.

e is an irrational number that serves as the base. With this information, we can address the problem below

*A student starts their experiment with 10 bacteria. He knows the bacteria grow 100% every hour. He will come back and check in 12 hours. How many bacteria will he find?*

Here is what we know

- final size =?
- initial size = 10
- rate = 1/hour
- time = 12

We plug this into the equation to get the answer

As you can see, the growth of the bacteria is almost incomprehensible in such a short time. This is the power of exponential growth.

**Conclusion**

Exponential models provide another way to find answers to questions people have. Whether the growth is over a certain number of times or continuously the model can be adjusted to deal with either of this situations.

One method for solving quadratic equations is called completing the square. This approach is a little confusing but we will try to work through it together in this post.

**What is Completing a Square**

Completing the square is used when your quadratic equation is not a perfect square. Below is an example of a perfect square quartic formula =. The first is in the standard quadratic form and the second is after it has been simplified.

However, not all equations are this easy, consider the example below.

There is no quick way to factor this as there is no perfect square. We have to use something called the binomial square pattern.

This is where it gets confusing but essential what the binomial square pattern is saying is that if you want to find the third term (b squared) you must take the second term and multiple it by1/2. We multiplied by 1/2 because this is the reciprocal of multiplying by 2 as shown in the equation. Lastly, we square this value. Below is the application of what we just learned from our problem equation.

By taking the second term, multiplying by 1/2 and squaring it we were able to create the trinomial we needed to create the perfect square. By doing this we also solved for x if this was a full equation.

**Examples**

When using the completing the square approach with a quadratic formula there are some additional steps. We will work through an example below

We are missing the third term and we need to find this first. Our second term is 8 so we will plug this in to find the third term

We take this number 16 and add it to both sides which is a rule whenever manipulating an equation. Therefore, we get the following.

We can now factor the left side as shown below.

To remove the square we need to square root both sides. In other words, we are employing the use of the square root property.

This leads to our two answers as shown below

There are variations of this but they all involve just moving some numbers around before the steps shown here. As such, there is not much need to discuss them.

**Conclusion**

Completing the square provides a strategy for dealing with quadratic formulas that do not have a perfect square. Success with this technique requires identifying the terms you know and do not know and taking the appropriate steps to calculate the third term for the trinomial.

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.

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,

We will now plug in the values and determine x.

**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.

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.

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

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.

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

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.

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

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

This trick works for numbers that cannot be factored.

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 always**Fraction**

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

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.

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.

- Isolate the radical
- 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.

**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 | n^{2} |

3 | cube | n^{3} |

4 | 4^{th} power |
n^{4} |

5 | 5^{th} power |
n^{5} |

**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.

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

3x – 2y = -6

Below is the matrix of the system of equations

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.

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.

Lastly, here is the determinant of y

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.

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.

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

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.

-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 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

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

x + 2y = 1

Here is our system

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.

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.

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.

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

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

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.

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

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.

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.

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.

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

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.

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.

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.

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

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.

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

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

We now substitute and solve

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

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.

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