MathTIMES Questions

 

Practice Problems: Measures of Central Tendency 

1.  A high school teacher at a small private school assigns trigonometry practice problems to be worked via the net. Students must use a password to access the problems and the time of log-in and log-off are automatically recorded for the teacher. At the end of the week, the teacher examines the amount of time each student spent working the assigned problems. The data is provided below in minutes.

Data	Array
15	49
28	48
25	43
48	39
22	34
43	33
49	28
34	27
22	25
33	25
27	22
25	22
22	22
20	20
39	15

2.  Bailey has been playing golf on the weekends for the past three years. Recently, she started keeping track of her recorded scores. Her scores for June and July at her favorite 9-hole (par 36) golf course are provided below.

Data	Array
45	56
49	49
42	45
56	42
41	42
36	41
34	41
38	41
41	41
40	40
42	40
41	39
39	39
38	38
40	38
39	36
36	36
41	34

 

 3. Midterm exam scores for a small advanced neuroanatomy class are provided below. Scores represent percent of items marked correct on the exam.

Data	Array
87	99
99	94
75	93
87	88
94	87
75	87
35	87
88	75
87	75
93	35

4. A group committed to quality television has been concerned about a new talk show. For two weeks, they decide to count the number of words that must be "bleeped" as too obscene for television and the number of physical altercations. They hope that after recording this data that they will be able to argue that the show is inappropriate for television particularly during the day. The data for number of words censored is provided below.

Data	Array
342	349
267	342
321	321
157	289
33	267
254	254
166	166
132	157
289	132
349	33

Solving linear equation in two variables

College Algebra:

Solving Linear Equation in two Variables

by Malen dO. Peňalosa BSED 2-A

Partido State University

College of Education

Goa, Camarines Sur
penalosamalen.gmail.com
penalosamalen.yahoomail.com

Learning Objectives

you should be able to: Know if an ordered pair is a solution to a system of linear equations in two variables or not. Solve a system of linear equations in two variables by graphing. Solve a system of linear equations in two variables by the substitution method. Solve a system of linear equations in two variables by the elimination by addition method.

System of Linear Equations

A system of linear equations is two or more linear equations that are being solved simultaneously.  

Solution of a System

In general, a solution of a system in two variables is an ordered pair that makes BOTH equations true.  In other words, it is where the two graphs intersect, what they have in common.  So if an ordered pair is a solution to one equation, but not the other, then it is NOT a solution to the system. A consistent system is a system that has at least one solution. An inconsistent system is a system that has no solution.   The equations of a system are dependent if ALL the solutions of one equation are also solutions of the other equation.  In other words, they end up being the same line. The equations of a system are independent if they do not share ALL solutions.  They can have one point in common, just not all of them.

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There are three possible outcomes that you may encounter when working with these systems: one solution no solution infinite solutions

One Solution
If the system in two variables has one solution, it is an ordered pair that is a solution to BOTH equations.  In other words, when you plug in the values of the ordered pair it makes BOTH equations TRUE. 
 

If you do get one solution for your final answer, is this system consistent or inconsistent?
If you said consistent, give yourself a pat on the back! 

If you do get one solution for your final answer, would the equations be dependent or independent?
If you said independent, you are correct!
 

The graph below illustrates a system of two equations and two unknowns that has one solution:

 

No Solution If the two lines are parallel to each other, they will never intersect.  This means they do not have any points in common.  In this situation, you would have no solution.   If you get no solution for your final answer, is this system consistent or inconsistent? If you said inconsistent, you are right!  If you get no solution for your final answer, would the equations be dependent or independent? If you said independent, you are correct!   The graph below illustrates a system of two equations and two unknowns that has no solution:   <!--[if !vml]--><!--[endif]-->

 

Infinite Solutions If the two lines end up lying on top of each other, then there is an infinite number of solutions.  In this situation, they would end up being the same line, so any solution that would work in one equation is going to work in the other.   If you get an infinite number of  solutions for your final answer, is this system consistent or inconsistent? If you said consistent, you are right!  If you get an infinite number of  solutions for your final answer, would the equations be dependent or independent? If you said dependent, you are correct!   The graph below illustrates a system of two equations and two unknowns that has an infinite number of solutions:

 


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<!--[if !vml]--><!--[endif]--> Example 1:  Determine whether each ordered pair is a solution of the system.            (3, 1) and (0, -1)

 

Let’s check the ordered pair (3, 1) in the first equation:

2x-3y=3            Plug in 3 for x and 1for y

2(3)-3(1)=3      True Statement

3=3

So far so good,  (3, 1) is a solution to the first equation 2x - 3y = 3. Now, let’s check (3, 1) in the second equation:

Hey, we ended up with another true statement, which means (3, 1) is also a solution to the second equation 4x - 2y = 10. Here is the big question, is (3, 1) a solution to the given system????? Since it was a solution to BOTH equations in the system, then it is a solution to the overall system.   Now let’s put (0, -1) into the first equation:

 

This is a true statement, so (0, -1) is a solution to the first equation 2x - 3y = 3. Finally, let’s put (0, -1) into the second equation:

 

This time we got a false statement, you know what that means.  (0, -1) is NOT a solution to the second equation 4x - 2y = 10. Here is the big question, is (0, -1) a solution to the given system????? Since it was not a solution to BOTH equations in the system, then it is not a solution to the overall system.

Three ways to Solve Linear Equations in two variables

 

three ways to solve systems of linear equations in two variables: graphing  substitution method elimination method

 Solve by Graphing

Step 1: Graph the first equation.

(3, 1) and (0, -1)

Step 2: Graph the second equation on the same coordinate system as the first.

 

You graph the second equation the same as any other equation.  Refer to the first step if you need to review how to graph a line. The difference here is you will put it on the same coordinate system as the first.  It is like having two graphing problems in one.

Step 3: Find the solution.

 

If the two lines intersect at one place, then the point of intersection is the solution to the system. If the two lines are parallel, then they never intersect, so there is no solution.   If the two lines lie on top of each other, then they are the same line and you have an infinite number of solutions.  In this case you can write down either equation as the solution to indicate they are the same line.

Step 4: Check the proposed ordered pair solution in BOTH equations.

You can plug in the proposed solution into BOTH equations.  If it makes BOTH equations true then you have your solution to the system.  If it makes at least one of them false, you need to go back and redo the problem.

   

<!--[if !vml]--> Example 2:  Solve the system of equation by graphing.

Step 1: Graph the first equation.

 

x-intercept

The x-intercept is (3, 0). y-intercept

The y-intercept is (0, 3). Find another solution by letting x = 1.

 

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Step 2: Graph the second equation on the same coordinate system as the first.

  
 

x-intercept

y-intercept

 

 

The y-intercept is (0, -1). Find another solution by letting x = 2.

 

 

Another solution is (2, 1). Solutions:

Step 3: Find the solution.

 

We need to ask ourselves, is there any place that the two lines intersect, and if so, where? The answer is yes, they intersect at (2, 1).

 

Step 4: Check the proposed ordered pair solution in BOTH equations.

 

You will find that if you plug the ordered pair (2, 1) into BOTH equations of the original system, that this is a solution to BOTH of them. The solution to this system is (2, 1).

  
  
Solve by Substitution method

Step 1: Simplify if needed.

 

This would involve things like removing ( ) and  removing fractions. To remove ( ): just use the distributive property.  To remove fractions: since fractions are another way to write division, and the inverse of divide is to multiply, you remove fractions by multiplying both sides by the LCD of all of  your  fractions.

 

Step 2:  Solve one equation for either variable.

 

It doesn't matter which equation you use or which variable you choose to solve for. You want to make it as simple as possible.  If one of the equations is already solved for one of the variables, that is a quick and easy way to go. If you need to solve for a variable, then try to pick one that has a 1 as a coefficient.  That way when you go to solve for it, you won't have to divide by a number and run the risk of having to work with a fraction (yuck!!).

 

Step 3: Substitute what you get for step 2 into the other equation.

 

This is why it is called the substitution method.  Make sure that you substitute the expression into the OTHER equation, the one you didn't use in step 2.

Step 4:  Solve for the remaining variable .

 

Solve the equation set up in step 3 for the variable that is left. If your variable drops out and you have a FALSE statement, that means your answer is no solution.  If your variable drops out and you have a TRUE statement, that means your answer is infinite solutions, which would be the equation of the line.

Step 5:  Solve for second variable.

 

If you come up with a value for the variable in step 4, that means the two equations have one solution.  Plug the value found in step 4 into any of the equations in the problem and solve for the other variable.

 

Step 6: Check the proposed ordered pair solution in BOTH original equations.

 

You can plug in the proposed solution into BOTH equations.  If it makes BOTH equations true, then you have your solution to the system.  If it makes at least one of them false, you need to go back and redo the problem. You will find that if you plug the ordered pair (4, 3) into BOTH equations of the original system, that this is a solution to BOTH of them. (4, 3) is a solution to our system.    <!--[if !vml]--><!--[endif]--> Example 1: Solve the system of equations by the substitution method: <!--[if !vml]--><!--[endif]-->   Step 1: Simplify if needed.   Both of these equations are already simplified.  No work needs to be done here.   Step 2:  Solve one equation for either variable.   It does not matter which equation or which variable you choose to solve for.   Just keep it simple. Since the x in the second equation has a coefficient of 1, that would mean we would not have to divide by a number to solve for it and run the risk of having to work with fractions (YUCK).  The easiest route here is to solve the second equation for x, and we definitely want to take the easy route.  You would not be wrong to either choose the other equation and/or solve for y, again you want to keep it as simple as possible. Solving the second equation for x we get:     Step 3: Substitute what you get for step 2 into the other equation AND Step 4:  Solve for the remaining variable .   Substitute the expression y + 1 for x into the first equation and solve for y: (when you plug in an expression like this, it is just like you plug in a number for your variable)      Step 5:  Solve for second variable.   Plug in 3 for y into the equation in step 2 to find x’s value.     Step 6: Check the proposed ordered pair solution in BOTH original equations.   You will find that if you plug the ordered pair (4, 3) into BOTH equations of the original system, that this is a solution to BOTH of them. (4, 3) is a solution to our system.    Example 5:  Solve the system of equations by the substitution method: <!--[if !vml]--><!--[endif]-->   Step 1: Simplify if needed.   This equation is full of those nasty fractions.  We can simplify both equations by multiplying each separate one by it’s LCD, just like you can do when you are working with one equation.  As long as you do the same thing to both sides of an equation, you keep the two sides equal to each other. Multiplying each equation by it's respective LCD we get:   Step 2:  Solve one equation for either variable.   Note how the second equation is already solved for y.  We can use that one for this step. It does not matter which equation or which variable you choose to solve for.  But it is to your advantage to keep it as simple as possible.  Second equation solved for y:   <!--[if !vml]--><!--[endif]--> Y=-3x+4      *Solved for y   Step 3: Substitute what you get for step 2 into the other equation  AND Step 4:  Solve for the remaining variable.   Substitute the expression -3x + 4 for y into the first equation and solve for x: (when you plug in an expression like this, it is just like you plug in a number for your variable)      Wait a minute, where did our variable go???? As mentioned above if your variable drops out and you have a FALSE statement, then there is no solution.  If we were to graph these two, they would be parallel to each other.   Step 5:  Solve for second variable.   Since we did not get a value for x, there is nothing to plug in here.   Step 6: Check the proposed ordered pair solution in BOTH original equations.   There are no ordered pairs to check.   The answer is no solution.

Solve by the Elimination by Addition Method

Step 1: Simplify and put both equations in the form Ax + By = C if needed.

This would involve things like removing ( ) and  removing fractions. To remove ( ): just use the distributive property.  To remove fractions: since fractions are another way to write division, and the inverse of divide is to multiply, you remove fractions by multiplying both sides by the LCD of all of  your  fractions.

Step 2: Multiply one or both equations by a number that will create opposite coefficients for either x ory if needed.

Looking ahead, we will be adding these two equations together.  In that process, we need to make sure that one of the variables drops out, leaving us with one equation and one unknown.  The only way we can guarantee that is if we are adding opposites.  The sum of opposites is 0.  If neither variable drops out, then we are stuck with an equation with two unknowns which is unsolvable. It doesn't matter which variable you choose to drop out.  You want to keep it as simple as possible.  If a variable already has opposite coefficients than go right to adding the two equations together.  If they don't, you need to multiply one or both equations by a number that will create opposite coefficients in one of your variables.  You can think of it like a LCD.  Think about what number the original coefficients both go into and multiply each separate equation accordingly.  Make sure that one variable is positive and the other is negative before you add.  For example, if you had a 2x in one equation and a 3x in another equation, we could multiply the first equation by 3 and get 6x and the second equation by -2 to get a -6x.  So when you go to add these two together they will drop out.

Step 3: Add equations.

Add the two equations together. The variable that has the opposite coefficients will drop out in this step and you will be left with one equation with one unknown.

Step 4: Solve for remaining variable.

Solve the equation found in step 3 for the variable that is left. If you need a review on solving linear equations, feel free to go to Tutorial 14: Linear Equations in On Variable. If both variables drop out and you have a FALSE statement, that means your answer is no solution.  If both variables drop out and you have a TRUE statement, that means your answer is infinite solutions, which would be the equation of the line.

Step 5: Solve for second variable.

If you come up with a value for the variable in step 4, that means the two equations have one solution.  Plug the value found in step 4 into any of the equations in the problem and solve for the other variable.

Step 6: Check the proposed ordered pair solution in BOTH original equations.

You can plug the proposed solution into BOTH equations.  If it makes BOTH equations true, then you have your solution to the system.  If it makes at least one of them false, you need to go back and redo the problem.

Example 7:  Solve the system of equation by the elimination method: <!--[if !vml]--><!--[endif]-->

Step 1: Simplify and put both equations in the form Ax + By = C if needed.

This equation is full of those nasty fractions.  We can simplify both equations by multiplying each separate one by it’s LCD, just like you can do when you are working with one equation.  As long as you do the same thing to both sides of an equation, you keep the two sides equal to each other. Multiplying each equation by it's respective LCD we get:

Step 2: Multiply one or both equations by a number that will create opposite coefficients for either x ory if needed.

Again, you want to make this as simple as possible.  Note how the coefficient on y in the first equation is 2 and in the second equation it is 5.  We need to have opposites, so if one of them is 10 and the other is -10, they would cancel each other out when we go to add them.  If we added them together the way they are now, we would end up with one equation and two variables, nothing would drop out.  And we would not be able to solve it. So I proposed that we multiply the first equation by 5 and the second equation by -2, this would create a 10 and a -10 in front of the y’s and we will have our opposites. Multiplying the first equation by 5 and the second equation by -2 we get:

Step 3: Add equations.

Step 4: Solve for remaining variable.

-x=-10

x=10

Step 5: Solve for second variable.

You can choose any equation used in this problem to plug in the found x value. I choose to plug in 10 for x into the first simplified equation (found in step 1) to find y’s value.

Step 6: Check the proposed ordered pair solution in BOTH original equations.

You will find that if you plug the ordered pair (10, 24) into BOTH equations of the original system, that this is a solution to BOTH of them. (10, 24) is a solution to our system.

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The following are webpages that can assist you in the topics that were covered on this page.

http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut19_systwo.htm This webpage will help you with solving linear equations in two variables. http://www.purplemath.com/modules/systlin1.htm This webpage goes over the basic definition of systems. http://www.purplemath.com/modules/systlin2.htm This webpage will help you with solving linear equations in two variables by graphing. http://www.purplemath.com/modules/systlin4.htm This webpage will help you with solving linear equations in two variables by substitution. http://www.purplemath.com/modules/systlin5.htm This webpage will help you with solving linear equations in two variables by addition/elimination.

  
  

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