Math Help with Mr. Pi the Math Guy

Classifying a Quadrilateral | Geometry How To Help

2012-01-21T09:47:00.000-08:00

If you are reading this lesson, then you are pretty far through your geometry course. The focus of that course has probably concentrated on proving lines parallel and triangles congruent. In a traditional textbook, this is usually the first lesson in a unit about quadrilaterals. The two examples of this lesson model how to classify quadrilaterals with two methods: by all names that apply or the most precise name that applies. To be able classify quadrilaterals properly; you must know the definitions of quadrilateral and the special quadrilaterals.

Definitions

Quadrilateral – four-sided figure

Parallelogram – a quadrilateral with both pairs of opposites sides parallel

Square – a parallelogram with four congruent sides and four right angles

Rhombus – a parallelogram with four congruent sides

Kite – a quadrilateral with two pairs of adjacent sides congruent and no opposite sides congruent

Rectangle – a parallelogram with four right angles

Trapezoid – a quadrilateral with exactly one pair of sides parallel

Isosceles Trapezoid – a trapezoid with the non-parallel sides congruent

Example 1 – Classifying a Quadrilateral

By appearance alone, classify the ABCD in as many ways as possible.

It is not usual for a person that is studying geometry to be asked to judge a shape by appearance alone, but in this case, that is what the directions ask you to do. If you are a verbal learner, then you are probably capable of just using the definitions to come up with your answer. A visual learner might find the above image with all the special quadrilaterals paired with their definitions. No matter how you learn best there are two names for ABCD:

1. Quadrilateral

2. Parallelogram

All too often, students tend to only name it a parallelogram because it has two pairs of opposite sides parallel. The directions to classify it in as many ways possible and by definition, it is also can be named a quadrilateral. It is very important to read and follow the directions to a problem carefully.

Example 2 – Connecting Algebra and Geometry

Determine the most precise name for the quadrilateral with vertices:

A(-2,1)

B(7,4)

C(4,-1)

D(-2,-3)

The connection between algebra and geometry in this example is the because the shape is given as vertices on the coordinate plane and to prove what this shape is relies on two skills: finding the slope of a line and finding the distance of a line segment. Finding the slope of a line is a skill taught in the typical Algebra 1 course and finding the length of a line segment is usually taught early on in a Geometry course.

To help the visual learner, I like to graph the points on graph paper and create the shape by connecting the vertices. I have done that in the above image. From the appearance of the shape it appears to be a trapezoid, because side AB appears to be parallel to side CD. Even though it appears to be a trapezoid, a proper coordinate proof finds the slope and distance of every side.

Slopes of Sides

After calculating the slopes, you can see that this is definitely a trapezoid because sides AB and CD have the same slope and the other sides do not, thus this quadrilateral has exactly one pair of parallel sides. Next, I have to prove the lengths of the sides. If sides BC and DA are the same length, then it would be an isosceles trapezoid, but based on the graph it does not appear those two sides are the same length.

Lengths of the Sides

Since no side is the same length, this is not an isosceles trapezoid and the most precise name for this quadrilateral is trapezoid.

Example 3 – Using Properties of Special Quadrilaterals

For the given kite, find the values of the variables and then find the lengths of the sides.

This example introduces how to use the geometry of special quadrilaterals to write an equation that needs to be solved. Since the shape is a kite, by definition, there are two pairs of adjacent sides congruent. From the diagram, it is apparent that the top two sides are congruent and the bottom two sides are congruent. Since these sides are congruent we can write equations setting the algebraic expressions equal to each other:

2x – 12 = 2y – 10

3x – 2 = 2x + 7

To solve this system if equations, I will solve the blue equation first, because it has only one variable.

(1) 3x – 2 = 2x + 7

(2) x – 2 = 7

(3) x = 9

I change equation (1) into equation (2) by subtracting 2x from both sides. I change equation (2) into equation (3) by adding 2 to both sides and the value of x is 9. Now that I know the value of x, I substitute 9 into x in the orange equation.

(1) 2x – 12 = 2y – 10

(2) 2(9) – 12 = 2y – 10

(3) 18 – 12 = 2y – 10

(4) 6 = 2y – 10

(5) 16 = 2y

(6) y = 8

To solve the orange equation, I first substitute 9 in for x, which gives equation (2). I simplify 2*9 to give equation (3), then simplify 18 – 12 to give equation (4). To get equation (5), I add 10 to both sides. Finally, I divide both sides by 2 and use the symmetric property of equality to turn the equation around to arrive at equation (6) and the second answer to this problem.

x = 9 and y = 8

To find the lengths of the sides, I substitute 9 for x and 8 for y into the algebraic expressions representing the lengths of the sides of the kite.

3x – 2 = 3(9) – 2 = 27 – 2 = 25

2x + 7 = 2(9) + 7 = 18 + 7 = 25

2x – 12 = 2(8) – 12 = 18 – 12 = 6

2y – 10 = 2(8) – 10 = 16 – 10 = 6

So, the lengths of the sides are 25, 25, 6, 6 and x = 9, while y = 8.

As you can see, solving geometry problems can rely heavily on Algebra skills.

Conclusion – Classifying Quadrilaterals

To be able to classify and solve problems with quadrilaterals, you need to the definitions of the special quadrilaterals. It is also important to keep your algebra skills sharp, because many times, solving problems with quadrilaterals requires solving equations or find the slope of lines.

Video Lesson on Classifying Quadrilaterals

Solving Two-Step Equations | Algebra How To Help

2011-12-29T10:26:00.001-08:00

Solving two-step equations at the algebra 1 level is usually pretty easy for most students. That is because it is generally a review of a pre-algebra lesson. Understanding this lesson requires you to understand solving one step equations. If you are looking for lesson on how to solve one-step equations check these two blog posts:

Solving One-Step Equations with Addition and Subtraction

Solving One-Step Equations with Multiplication and Division

Learning how to solve multi-step equations is a building process and solving two-step equations is really the third step. Once solving one-step equations is mastered, solving two-step equations can be learned and mastered. Finally, learning how to solve multi-step equations can be introduced.

Solving Two-Step Equations

When solving any equation, you must keep in mind the overall goal: isolate the variable on one side of the equation with a coefficient of 1. If that sounds too mathematical for you, how about this: the goal is to get the variable by itself. To achieve this goal, we use a combination of operations to “undo the order of operations”. I put, undo the order of operations, in quotes, because that is how to solve two-step equations:

1. Undo any Addition or Subtraction

2. Undo any Multiplication or Division

As you should know, the last two steps of the order of operations are:

1. Complete any multiplication or division from left to right

2. Complete any addition or subtraction from left to right

I like to think of it as put on a shirt and blazer. When I get dressed, I have to put on the shirt before I put on the blazer or I would look kind of weird. After a long day at school and I get home, I take the blazer off first and then my shirt before sitting down and doing homework with my children. Let me show you building an equation to solve.

Building a Two-Step Equation to Solve

Here I am going to build an equation to be solved. I will use the order of operations to build it. See image 1 for an illustration. I start with the equation

x = 6 (1).

Using the multiplication property of equality, I multiply both sides by 3 to give the equivalent equation

3x = 18 (2).

Next I use the subtraction property of equality to subtract 6 from both sides to give

3x – 6 = 12 (3).

Equation (3) is an example of a two-step equation I will be modeling how to solve in this blog post. In this example, I built an equation to be solved. To solve the equations in this post I will be using the opposite of the order of operations.

Example 1 - Solving Two-Step Equations with Integers

See image 2 for an illustration of this explanation. Starting with the equation

-5x + 42 = -8 (1),

the goal is to isolate the variable. The variable is on the left of the equation, therefore, using the reverse of the order of operations, I will have to undo the addition of 42 and the multiplication of -5. First, using the subtraction property of equality, I subtract 42 from both sides of equation (1) to give

- 5x = -50 (2).

Following the reverse of the order of operations I next undo the multiply by -5 by dividing both sides of equation (2) by -5 to give the solution

x = 10 (3).

Hopefully, you found this example easy to understand. Use the video below to practice these skills. In the video, I model how to solve several two-step equations. Then, you are asked to solve some problems on your own.

Example 2 – Solving Two-Step Equations with Decimals

See Image 3 for an illustration for this explanation.

Given the equation:

15.61 = -7.43 + 0.2x (1).

In equation (1), the variable is on the right hand side of the equation, whereas in Example 1 the variable is on the left hand side. This is a minor difference, but there is a more important difference: the order of the variable term and the constant term. Error Alert: because of this switch of the constant term and variable term, many students start this problem by subtracting 7.43 from both sides. THIS IS WRONG!!! The students that make this error get stuck on the sign that is between the constant and the variable term. It is the sign in front of the constant that determines if you add or subtract to both sides.

To solve this equation correctly, I must add 7.43 to both sides of the equation using the addition property of equality to get

23.04 = 0.2x (2).

Next I divide each side of the equation by 0.2 using the division property of equality to give

115.2 = x (3).

Finally, I use the symmetric property of equality to turn the equation around so the variable comes first

X = 115.2 (4).

As you can see from the first two examples, whether you are working with integers or rational numbers, the steps for solving two-step equations remains the same. As you will see, adding fractions to the mix does not change the steps.

Use the video below to practice these skills. In the video, I model how to solve several two-step equations. Then, you are asked to solve some problems on your own.

Example 3 – Solving Two-Step Equations with Fractions

(1).

In my 13 years teaching experience, these types of two-step equations cause even some of the best students to shut down. The steps are no different than I used in Examples 1 and 2, but because there are fractions involved, they shut down and do not even try. With today’s technology, everyone should be able to do calculations involving fractions. Back to the algebra…

The first step to solve this equation is to undo the addition by subtracting from each side to give

(2).

The next step can be done as I model in Example 5 of my post on solving one-step equations with multiplication and division. To get rid of the fractional coefficient, I can multiply both sides of the equation by the reciprocal or I can use two-steps using a combination of multiplication and division. To stay with the theme of using only two-steps total, I will solve this by multiplying both sides by the reciprocal of which is . This action isolates the variable giving a solution of

(3).

Use the video below to practice these skills. In the video, I model how to solve several two-step equations. Then, you are asked to solve some problems on your own.

Conclusion – Solving Two-Step Equations

You will know you are working with an equation that involves two-steps because there will be two operations to “undo”. It does not matter what kind of numbers are involved to solve two-step equations:

1. Undo any Addition or Subtraction

2. Undo any Multiplication or Division

Solving One-Step Equations with Multiplication and Division | Algebra How To Help

2011-12-18T17:48:00.000-08:00

Solving One-Step Equations
As I wrote in my first post on solving one-step equations with addition and subtraction, it does not matter which algebra textbook you are using, if you are studying algebra at any level, you need to know how to solve equations and this is the starting point, well it is the second step in the process of learning how to solve multi-step equations. The first step in learning how to solve equations was covered in my previous post on solving one-step equations with addition and subtraction.

Review of Basic Terms

When solving equation, you have to keep the overall goal in mind, to isolate the variable on one side of the equation. An equation has three parts: left, middle and right. The equal sign is 'the middle' of an equation. For example,

x = 8.

x is on the left of the equation and 8 is on the right of the equation. The equation x = 8 is an example of an equation where the variable has been isolated on the left of the equation.

If you are given an equation such as

2x = -26.

The variable is not isolated because it has a coefficient of 2. To solve this equation, divide 2 into both sides. When showing your work, you should write a fraction bar under each side of the equation and write a 2 in both denominators, thus showing each side being divided by 2. On the left, the 2's divide out and leave the variable x and on the right-26 divided by 2 is -13. Therefore, x = -13.

Solving One-Step Equations with Multiplication

Example 1

If an equation involves the division of the variable by a constant such as x/ 7 = -4, then the multiplication would be used to solve this equation because it is the opposite operation to division. To correctly show work, you use the multiplication property of equality and multiply both sides of equation 1 by 7. On the left side of the equation the 7's divide out to give x and on the right -4 times 7 gives -28.

Example 2

Similar to example 1, example 2 involves multiplication to solve the original equation. It differs from example because the side opposite the variable is a mixed number,

x/-3 = 4 1/3.

It is still solved with the multiplication property of equality, but the math on the right causes some students problems. The solution of this equation is found by multiplying both sides by -3. As you can see, in the image, the -3s divide out on the left of the equation and give the variable x. On the left hand side of the equation gives -3 * 4 1/3. The image gives a much clearer idea of calculations. The mixed number must be changed to an improper fraction before the multiplication can take place and changes to -3 * 13/3 which gives -13. Therefore, x = -13.

Solving One-Step Equations with Division

Example 3

Recognizing when to use and using division is easier for students. I am pretty sure it has to do with the fact the equations do not always start off looking like fractions. The equation in example 3 is

2x = -26.

If you read the left of the equation with the operation, 2 times x, instead of 2x, the opposite of times is divide. So, to solve this equation, you must divide by sides by 2. On the left of the equation, the 2s divide out leaving x and on the right -26 divided by 2 is -13. Thus, x = -13.

Example 4

When solving equations, it is good to compare new problems to to previously solved problems and to compare how they are similar and different.

-0.6x = 42

Looking back at example 3, it should be obvious the two problems are similar because the variable is being multiplied by a number. The difference is that the coefficient of the x is a rational number in example 4, whereas in example 3 it was a whole number. So, solving the equation in this example will involve the division property of equality to divide each side by -0.6. On the left, the -0.6s divide out leaving the variable x and the left hand gives -70 because 42 divided by -0.6 is -70, thus x = -70.

Should I Take 1 Step or 2 Steps?

Image 1

The equation in example 5 contains a fractional coefficient. In other words, a fraction in front of the variable. There are two ways that a fractional coefficient can be model, as can be seen in Image 1, because the fractional coefficient can be written as a number times the variable in the numerator then divided by another number, these types of equations can be solved with two steps that include one step of multiplication and one step of division.

I don't know how you are, but I do not like to show more than is necessary, so I want to let you know there is way to solve these equations with only one step! Yep! Less work to do! To use one step, all you have to do is multiply both sides of the equation by the fractional coefficient's inverse.

Example 5

In the image, I solve the equation in two different ways. On the left, I solved it with one step and on the right I solved it with two steps.

On the Left - Notice how I multiplied each side of the equation by the fractions inverse. On the left, all of the numbers divide out to 1 and leave the variable isolated. Finding the solution is a matter of basic math.

On the Right - I multiplied both sides by 3 first to give 2x = 36, then divided both sides by 2 to get x = 18. I could have just as easily divided by 2 first.

Summing it all up, solving equations requires you to be flexible in how you deal with a given equation. Though not every equation is solved exactly the same, it does require the use of the opposite operation and the properties of equality to maintain the balance of the equation.

Solving One-Step Equations with Addition and Subtraction | Algebra How To Help

2011-12-17T14:30:00.000-08:00

Solving One-Step Equations

It does not matter which textbook your school has chosen to use, nor does it really matter the age of the text. Solving multi-step equations will be apart of what you have learned. I work with Prentice Hall's Algebra 1 (2009) and in the years before that I worked with Glencoe's Algebra 1 (1999). I tell you what books I have used because that should give you a clear picture that solving equations is an important part algebra and learning how to solve one-step equations is the building block to solving two-step and multi-step equations. Remember, the goal of solving multi-step equations is to isolate the variable correctly to get the answer.

Isolating the Variable

Example 1

To solve equations means to "isolate the variable", which means to get the 'letter on one side of the equal sign'. As in example 1 to the right, equation (1) is x + 4 = -6. On the left of the equal sign is x + 4 and on the right is -6. To 'get the letter by itself', undo the add 4 by subtracting 4 from both sides as shown in red. Equation (2) gives the variable now isolated with a value equal to -6. From this example, I hope you understand how I will tie the examples into what I am blogging about. Please notice how I use numbers next to the equations.

Solving One Step Equations with Addition and Subtraction

Example 2

Example 1 is a perfect example of solving a equation with subtraction. The reason subtraction was used is because the original equation involves addition. The focus is usually what operation, addition or subtraction, is between the variable and the number. This strategy will work most of the time, but not always.

Example 2 is another equation in which the focus can be on the operation between the variable and the number. The focus can be on the the sign between the number and variable when the number is positive. As in example 2, since 6 is added to the variable, to get rid of the 6, it is subtracted from the 6 on the left and it is subtracted from the right to maintain the balance of the equation. Because the 6 and -6 on the left are opposites, those numbers combine to give 0 and the variable is isolated. Another way teachers will describe what happens on the left is to say "the 6's cancel".

Example 3

Example 3's original equation start with subtracting a positive 6 from the variable. As in example 2, in equation 1, the number on the same side as the variable is positive, so the focus can be on the operation separating the variable and the number, in this case it is subtraction. In the pink, adding 6 to both sides is shown. The -6 and plus 6 cancel on the left and on the right, -11 + 6 is equal to -5 and I conclude equation 2 as x = -5.Just as I did in both examples 1 and 2, I performed the operation to both sides of the equation. In this example, I added 6 both sides by using the addition property of equality. As you continue with this lesson, the original equation will need to be modified before the focus of the opposite operation that is between the variable and the number can be addressed.

Addition is Subtraction and Subtraction is Addition

What? This topic can be be very confusing if you are unaware of two facts:

Adding a negative number is the same as subtracting a positive number
Subtracting a negative number is the same as adding positive number

Example 4 Incorrect and Correct Way To Solve an Equation

In the above image, I model how to solve the same equation two different ways. I solve equation 1by subtracting 4 from both sides. This is in correct because of the double sign between the variable and the number. The double sign means I should employ one of the rules listed above. In this case a negative 4 is being added, by rule 1, I should think of that as subtracting a positive 4. You can see how I changed my work with equation 1 as I changed it to equation 2. Once the equation is written with only one sign the focus can be on using the opposite operation, the opposite of subtracting 4 is adding 4 and that work is done in red.

Example 5

In example 5 the original equation is x + (-2) = 8. From equation 1 to equation 2 I apply rule #1 from above. Now that equation is written with only one sign between the variable and the number, using the opposite operation now applies, since 2 is being subtracted in the equation, I must add 2 to both sides of the equation. On the left, the -2 and +2 cancel because they are opposites. On the right, 8 + 2 is 10, therefore x = 10.

If it is obvious to you that you need to add to solve this equation, then rewriting the equation as equation 2 is not necessary to find the correct solution.

Example 6

The final example, rule #2 is illustrated. Equation 1 is x - (-5) = 7. Notice the "double negative" between the variable and the number. Some teachers, including will say, "a double negative makes a positive." We say that because it does and it offers an easy way to remember rule number 2. After using rule #2, the equation can be solve by subtracting 5 from both sides. This isolates the variable and gives x = 2.

Using the below video will give you some interactive practice with solving one-step equations with addition and subtraction.