What are two symbolic techniques used to solve linear equations? Which do you feel is better? Explain why. • Post an example for your class to solve.

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There are more than two techniques used to solve linear equations, but two of the different methods that I chose to solve with are by using graphing, and by using elimination. Graphing can be used by solving for the intercepts of the equation, by solving with x equal to zero, and with y equal to zero. This is the quickest way to find the x and y intercepts, and draw a line in-between them. Another method is to make sure that the represented b in this equation is the y intercept. The slope is the m in the equation. The second method is to solve by elimination. This method makes it so that either the x or y variable in the two linear equations can be multiplied to equal the variable in the other equation. This makes one of the variables equal to the other side, and then can be combined to solve for the second variable. The value that you come up with is then plugged into the original equation and then solved for the other variable. A third method is to solve by using substitution. This method sets one equation equal to a single variable such as y, then plugging that value in for the y in the other equation. This method is the most difficult. I prefer solving using graphing. This is how I would most easily solve two linear equations. My Classmates can solve this problem: 3x+4=y, and 4x-6=y

Do the equations x = 4y + 1 and x = 4y – 1 have the same solution? How might you explain your answer to someone who has not learned algebra?

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They do not have the same solution simply because you can’t add and subtract a number from the same equation and get the same number. It just doesn’t work, especially in a line graph. If I was to explain this to someone who didn’t know what algebra was I would first take away the x value and just leave it blank. I would then assign a number to y, like 3. I would multiply it out, and show them that when 4y+1 = 13 it is not the same answer as when 4y-1 = 11. This would probably be the easiest way to explain that adding and subtracting that same number isn’t going to give you a similar value. If they still didn’t understand what I meant by the explanation I have used, I would then substitute 4 in for the 4y. I would ask them if 4+1 is the same as 4-1. I would illustrate perhaps by saying would you like 4 dollars plus 1 dollar or 4 dollars minus 1 dollar. Any logical person will choose to take the 4+1 dollars and more simply understand that the x value represented in the equation is just the answer after adding or subtracting the 1.

How many solution sets do systems of linear inequalities have? Do solutions to systems of linear inequalities need to satisfy both inequalities? In what case might they not?

There can be an infinite number of solutions to a linear inequality. The only real case where there is one solution is when the inequality has two cases and they are both and such as: x ≤ 3 AND x ≥ 3. In this inequality there is only one answer and that answer is 3. Solutions to systems of linear inequalities usually need to satisfy the condition of both inequalities if they are connected by an

AND. In cases where they only need to satisfy one of the inequalities is when they are connected by an OR. For instance in the case x < 3 OR x > 2, there is any number of solutions to this problem. In fact in that particular problem any real number would satisfy the conditions of the line graph. In the equation x < 2 AND x > 3, there is no real solutions because no number can satisfy the conditions of both inequalities.

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Provide one real-world example of when graphing could be useful. Do you think you would ever use graphing in your life to solve problems? Explain why or why not.

Graphing is useful in everyday life. I use graphing in my business to track the amounts of hits that I get on my website, as well as the amount of traffic that is generated from Google or Bing. I use graphing currently in my everyday life and I find it essential to my business. Graphing is something that is essential to many different forms of business and must be used on a daily basis. It makes everyday life simpler, and easier to understand predictions throughout time. Graphs are used for literally any type of data, and can help to track any historical progress throughout a varying time period. The easiest way to track I feel is to make your time variable the x variable, any other stats that you may be wanting to track your y variable.

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What concept learned in this course was the easiest for you to grasp? Why do you think it was easy for you? Which was the hardest? What would have made it easier to learn?

The concept in this class that was the easiest to grasp was definitely the standard form of the graphing line. I feel that the explanation of y=mx+b is the simplest way to describe a line. It is so simple in that b will always be the y intercept, and m will always be the slope which is the rise over the run. If the slope is negative, you can go down and over or up and back it is a constant piece of math which will never change and is one of the most useful tools for daily business endeavors. The hardest thing for me to graph is the inequalities of a line. It is very difficult to grasp the concept of flipping a sign when it needs to be flipped. I think through the DQ’s in this class that I have been able to learn a lot as time has gone on, and the concept of inequalities has been made easier to learn. I think that the Dq’s are the most helpful when it comes to learning new material.

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Week 2 Concept Check Post your 50-word response to the following: • How do you know when an equation has infinitely many solutions? • How do you know when an equation has no solution?

An equation has an infinite number of solutions when the variable equals itself. For example, x=x, both sides are equal. No matter what values are given for x the terms of the equation will always be met. If an equation has no solution you will get an untrue answer such as 6=2 when you see something like 6=2, it has no solution because 6 cannot equal 2. Only 6=6 and 2=2.

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Exercise: Week 4 Concept Check Post your 50-word response to the following: Explain in your own words why the line x = 4 is a vertical line.

If the value of x=4 then no matter what the value of y is the value of x will always be 4. When you draw an x and y axis graph, each point on the line represents the place where an x and y value meet. x=4 means that, no matter what the value of y, the x value will always be 4. Every y value on the line meets the matching x value at 4, resulting in a vertical line.

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Exercise: Week Six Concept Check Post your 50-word response to the following: How can you determine if two lines are perpendicular? Post in your Individual Forum

Non vertical lines are perpendicular if the product of the slopes equal -1. So essentially, (M1) (M2) will equal -1. Also if you can make a little box (or 90 degree angle) then lines will are perpendicular. If one line is horizontal and another is vertical they will also be perpendicular.

How do you think you will use the information you learned in this course in the future? Which concepts will be most important to you? Which will be least important? Explain your answers.

Math is a very useful tool. It has many practical applications from basic graphing all the way to complex equations involving physics and many other things. You can use it to determine the family income needs as well as professional or job related tasks. Math can be used to track statistics for literally anything.

            In my job I am able to track through the internet demographically who buys what products. This is something that I can track with a simple scatter graph. In my personal life I can use some of the geometric formulas I learned to determine what the layout and required materials I need for my hobbies such as sewing.

            I probably will not use some of the more intricate formulas such as solving inequalities. These types of formulas simply do not serve a purpose in my job or personal life. But I believe that it is good to know these formulas and problem solving skills. 

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Using the readings discussed in this course, provide one real-world application of the information learned that has been the most valuable to you. Why has it been valuable?

Learning about the differences between scatter plots and line graphs has been some of the most practical information that I have learned during this class. Scatter plots express data in such a way that any amount of random data can be shown and relationships can be drawn in one graph. Scatter plots can be used practically to show whether or not a group of people meet certain criteria.  Line graphs express data so that a difference, or slope, can be shown from period to period. Line graphs can be used to express information from year to year, or week to week to show improvements so that values can be tracked. Different information can be classified through different types of graphs, this is why line and scatter graphs are useful in real-world situations.

What concept learned in this course was the easiest for you to grasp? Why do you think it was easy for you? Which was the hardest? What would have made it easier to learn?

The easiest concept to grasp in this math class was the explanation used to determine when to flip the sign in an inequality. Whenever you have to divide or multiply by a negative number the sign in an inequality changes or flips. This is the easiest explanation of when to flip the sign of inequalities; I was never sure when I studied the topic in high school. One of the hardest concepts to grasp was the use of PEMDAS. I never understood why there must be order to solving equations throughout the equation. It was difficult for me to grasp the order, but once I read up on the topic I understood that there must be an order so that regardless of the writing of the equation it will always be solved the same way. It would have been easier to understand if the concept that it is just a certain order to solve math all of the same way was explained. 

Using the readings discussed in this course, provide one real-world application of the information learned that has been the most valuable or meaningful to you. Why has it been valuable?

Learning about the differences between scatter plots and line graphs has been some of the most practical information that I have learned during this class. Scatter plots express data in such a way that any amount of random data can be shown and relationships can be drawn in one graph. Scatter plots can be used practically to show whether or not a group of people meet certain criteria.  Line graphs express data so that a difference, or slope, can be shown from period to period. Line graphs can be used to express information from year to year, or week to week to show improvements so that values can be tracked. Different information can be classified through different types of graphs, this is why line and scatter graphs are useful in real-world situations.

How many solution sets do systems of linear inequalities have? Do Solutions to systems of linear inequalities need to satisfy both inequalities? In What case might they not?

A solution set may have any number of solutions, the most common answers for this are; 0,1 or infinite. An example where only one is had is when x≤3 and x≥3. An example of a solution where none are found is when given x>3 and x<2. Both of these cannot be satisfied at the same time, so there is no solution set for this number. If a system does not fit into both of the inequalities and the word “and” is used, there will be no solutions to the given problem. The last option is when you are given something like x>3 and x>4. This is a solution where any number greater than 4 will satisfy the conditions, and will be considered infinite. 

What are two symbolic techniques used to solve linear equations? Which do you feel is better? Explain why. Please be sure that you pay attention to the bolded word. Be specific in your explanation and use examples to support your response:

Two symbolic techniques used to solve linear equations are; graphically, and algebraically. Personally I prefer to use graphically the most because it is the fastest method to solve the equation, also it is easy to solve with a calculator.

To solve an equation graphically you must first set it into slope intercept form. This is achieved by setting the equation equal to Y. In this given problem, x + y = 10, and 3x+2y=20 we will solve for y. You get -x+10 = y and -3/2x+10=y. When graphing these two numbers check to see where they intersect, which is at (0,10). Where they intercept is the answer to the problem.

The second method to solve by is algebraically. This method uses substitution and can be a bit confusing if a clear paper trail is not kept when doing your homework. Given the same problem, x+y=10, and 3x+2y=20 we will take the first half of the problem and set it all equal to y. We get –x+10=y. Given this information, take the y value and substitute it into the other problem (3x+2y=20). Substituted we get, (3x-2x+20=20) Simplified the equation comes out to (x+20=20). Therefore the value of x is 0. We then take this value of x and substitute it into the first half of the 


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linear equation. Where (x+y=10) when substituted this turns into (0+y=10) Therefore the value of Y is 10. Given this information we can conclude that the answer to this linear equation is (0,10) (x,y).

A third valuable method is to solve by elimination. Given the same problem, x+y=10, and 3x+2y=20, we will solve by eliminating the variable x. The goal here is to make it so that the x values will cancel out when the problem is combined together. The first step is to multiply the first linear equation by -3. This is done here -3(x+y=10) we get, -3x-3y=-30. This new linear equation is combined with the second one, (3x+2y=20). Together when combined This new equation comes out to (3x-3x+2y-3y=-10) When canceled out it becomes, -y=-10. y therefore equals 10. We plug this value back into either one of the original linear equations, and can conclude that x is again 0.

In conclusion I feel that the first method, solving graphically, is a far faster method to solve the problem. Not only that, but you can see just by the explanation of the two types that substituting and solving algebraically takes longer to explain, and longer to follow through with. 

MAT 116 Week 8 Exercise Concept Check

A square bracket means that the interval includes that endpoint. Since in this problem both brackets are square, both endpoints are included. This interval goes from -4 to 10 and includes the endpoints. On a number line, one can fill in those endpoints. Alternatively, instead of filling in the endpoints, one can draw a bracket at -4 and another bracket at 10. If the interval had not included parentheses, one would not have filled in the circles (or if one used brackets/parenthesis, he or she would have put a parenthesis on the number line). A straight line and mark a line from -4 to 10 with [at -4 and] at 10

Give an example of a division of a polynomial long division by a binomial and show all the steps when performing the division. The example must be your own and not from the text book. How is this similar to numerical division of real numbers? Give another

Polynomial division is very similar to long division that we learned in elementary school. Instead of simply using numbers, there are variables as well. The process, though, is effectively the same. We go through the problem term by term, just like in standard numerical long division. If we understand how to do one type of long division, it is quite easy to extend the technique to the other type of division. As long as we know how to multiply monomial terms with variables, the actual process is the same. The polynomial goes to the inside of the division symbol, and the binomial goes outside. We try to get the first term of polynomial from the binomial first term, and repeat this till we are able to get a polynomial of lesser degree than binomial (remainder) or till we get zero remainder. Example (x^2 + 5x + 6) divided by x + 5 We get X+5 /x^2 + 5x + 6 First we multiply by x to get x^2 + 5x We get X + 5/ x^2 + 5x + 6 -x^2 – 5x 6 We get the remainder as 6 Here quotient is x and the remainder is 6 If we divide same polynomial by x + 2 We get X+2 /x^2 + 5x + 6 First we multiply by x to get x^2 +2x, and then by 3 to get x + 6 We get X+5/ x^2 + 5x + 6 -x^2 – 2x X + 6 -x – 6 Zero remainder The quotient is x + 3 and remainder 0

Thread for Week 3 - Discussion Question #1 Take any number (except for 1). Square that number and then subtract one. Divide by one less than your original number. Now subtract your original number. Did you reached 1 for an answer? You should have. How do

Suppose we take the number 5 Squaring 5^2 = 5*5 = 25 Subtracting 1 25 -1 = 24 Dividing by 1 less means 5-1 = 4 24/4 = 6 Subtracting original number 6 – 5 = 1 Yes answer is 1. Taking variable y Squaring y^2 Subtracting 1 y^2 -1 Dividing by 1 less means y^2-1 / (y-1) = (y+1)(y-1)/(y-1) = y+1 Subtracting original number y+1 - y = 1 Yes answer is 1.   Here is a number game that uses the skills of simplifying rational expressions. Take any number (except for -4) and add 2. Next, multiply by 2 less than the number. Add 3 times the original number. Divide by 4 more than the original number. Finally, add 1. You should be back where you started! Here it is with symbols: Take any number (except for -4) and add 2. x x+2 Next, multiply by 2 less than the number. (x+2)(x-2) = x^2 - 4 Add 3 times the original number. x^2 + 3x - 4 = (x+4)(x-1) Divide by 4 more than the original number. (x+4)(x-1)/(x+4) = x-1 Finally, add 1. You should be back where you started!

Give an example of a sum of two rational expressions with different denominators, then perform the operation by showing all the steps, including how you found the common denominator. These rational expressions must have a variable in the denominator, such

Rational expression with different denominators. x/(x + 2) + x-2 /( x+3) we find the LCD which is (x+2)(x+3) here making these like fractions.. so that we can add x(x+3)/(x+2)(x+3) + (x-2)(x+2) / (x+3)(x+2) x^2 + 3x / (x+2)(x+3) + x^2 – 4 / (x+2)(x+3) x^2 + 3x + x^2 – 4 / (x+2)(x+3) 2x^2 + 3x – 4 / (x+2)(x+3) LCD is found by finding the factors of the denominators just as we do in the case of fractions. Example for classmates… Add x – 2 / x^2 – 4 + (x + 3) / (x – 2) Rational Expression can be used in real life where we need to solve for questions which involve fractions. Example: Speed, Distance Cases, or Time and work Cases. Here is a number game that uses the skills of simplifying rational expressions. Take any number (except for -4) and add 2. Next, multiply by 2 less than the number. Add 3 times the original number. Divide by 4 more than the original number. Finally, add 1. You should be back where you started! Here it is with symbols: Take any number (except for -4) and add 2. x x+2 Next, multiply by 2 less than the number. (x+2)(x-2) = x^2 - 4 Add 3 times the original number. x^2 + 3x - 4 = (x+4)(x-1) Divide by 4 more than the original number. (x+4)(x-1)/(x+4) = x-1 Finally, add 1. You should be back where you started! x

Why is it important to simplify radical expressions before adding or subtracting? How is adding radical expressions similar to adding polynomial expressions? How is it different? Provide a radical expression which consists of a sum or difference of two ra

Why is it important to simplify radical expressions before adding or subtracting? It is important to get radical expressions into their simplest form before you add or subtract them. You can only add and subtract like radicals (for example ). You cannot add them if they are different (for example ). How is adding radical expressions similar to adding polynomial expressions? How is it different? It is similar because in polynomial expressions, you can only add like terms. In radical expressions, you can only add like radicals. They are different, though, because instead of powers, you are dealing with roots. Provide a radical expression for your classmates to simplify. Here is one to simplify: The Answer is

Review section 10.2 (p. 692) of your text. Describe two laws of exponents and provide an example illustrating each law. Explain how to simplify your expression. How do the laws work with rational exponents? Provide the class with a third expression to sim

Describe two laws of exponents and provide an example illustrating each law. -In multiplying, we can add the exponents as long as the base is the same ex. 3^2 x 3^4= 3^2+4 -To raise a power to a power, multiply the exponents ex. (3^2)^4= 3^2*4 Explain how to simplify your expression. -Convert to an exponential expression, use math to simplify the exponent, and convert back to radical expression when appropriate How do the laws work with rational exponents? -When using radical exponents, simplify the exponent after coverting to exponential exression, rather than simplifying the expression. Provide the class with a third expression to simplify that includes rational (fractional) exponents. 4^2/3 *4^1/3