Steps for Solving Equations | How to Solve Equations – Algebra

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Inverse Variation | Modeling and Identifying Inverse Variation

Inverse Variation can be modeled after any equation in the form xy = k where k \neq 0. The constant of variation is k.

Inverse Variation - Time to Roller Blade 36 Miles

Figure 1

The table to the left models the time it would take to roller blade 36 miles at various speeds. As the rate increases the amount of time to travel 36 miles decreases. If the rate data is examined more closely and focus is placed on the times of 3, 6 and 12 hours, it should be come clear as the rate is doubled, the amount of time is cut in half.

Since the product is 36 for each pair of numbers, this is an inverse variation problem and the product is the constant of variation for this data. Letting x be the rate and y be the time, one equation to model this data is xy = 36. A more useful equation is y = \frac{36}{x}. Figure 2 is the graph of the inverse variation modeled in this problem.

Inverse Variation y = 36/x

Figure 2

Critical Thinking – Inverse Variation

Why is y = \frac{36}{x} a more useful equation than xy = 36?

Write your responses in the comments section.

Modeling Inverse Variation

This problem involves finding the equation or function that models a given inverse variation. This is a two step process:

  1. Multiply the given x- and y-values
  2. Substitute the product into y = \frac{k}{x}

Before just memorizing this two step process, it would be better to understand what these two steps are ‘doing’ mathematically. In step 1, multiplying the given x- and y-values calculates the constant of variation. The second step is to substitute the constant of variation into one of the acceptable forms of the inverse variation equation.

Example 1 – Modeling Inverse Variation

Suppose that x and y vary inversely and x = 7 and y = 5. Write an equation that models the inverse variation.

xy = k   Definition of Inverse Variation

(7)(5) = k Substitution Property of Equality – Step 1 from above

35 = k Simplify

k = 35   Symmetric Property of Equality

y = \frac{k}{x}   Step 2 from above

y =  \frac{35}{x}   Final Answer

Identifying Inverse Variation

When given a table of values, it is often important to be able to identify the equation modeling the data. To determine if a table of values is indeed an inverse variation, all of the data pairs must have the same product. This is because of the definition of an inverse variation and how the constant of variation is calculated. So, to identify a table of values as an inverse variation, one must test the products of all the available data. If the products are all equal, then the data models an inverse variation and by find the products, the constant of variation is already known.

Example 2 – Identifying Inverse Variation

Is the relationship between the variables in figure an inverse variation? If so, then write the function that models the inverse variation.

Inverse Variation Data

Figure 3

As state above, first find the product of all pairs of numbers:

2(0.7) = 1.4

(0.35) = 1.4

14(0.1) = 1.4.

Since all pairs of corresponding numbers form the same product, this table of values models an inverse variation with the constant of variation being 1.4. Therefore the function to models this data:

y = \frac{1.4}{x}.

If I don’t write this at the end of every blog article, I should start. Use the comments section for any questions you may have.

How to Solve Logarithmic Equations

In this video there are two problems modeled. The first example works with common logs and the second example models an equation with a natural log. The keys to solving any type of logarithmic equation: being able to write the given equation in exponential form or being able to take the log of each side.

To be able to write any logarithmic function in exponential form knowing the definition of logarithm and natural logarithm is key. So here they are…

Definition: Logarithm – The log to the base b of a positive number y i s defined:

If y = b^x, then log_b \hspace{0.1 cm}y = x.

Definition: Natural Logarithm – If y = e^x , then log_e \hspace{0.1 cm}y = x which is also written as ln \hspace{0.1 cm}y = x. The natural logarithmic function is the inverse, written as y = ln \hspace{0.1 cm}x.

In other words, if y = e^x, then y = ln \hspace{0.1 cm}x.

Now that you have reviewed the definition of log and natural log, read through these two examples, then watch the embedded algebra 2 video.

Example 1 Solving a Logarithmic Equation

This a problem that I assigned to my Algebra 2 class for a review before a quiz on solving logarithmic equations:

7^{x-3} = 25 Given

log 7^{x-3} = log 25 Take the log of each side

(x-3)log 7 = log 25 Product Property

x-3 = \frac{log25}{log7} Division Property

x = \frac{log25}{log7}+3 Addition Property

x \approx 4.6542 Use a calculator

As you can see in this example, you can easily solve an equation with multiple log in it. All you need to do is use the properties of logarithms and that taking the log of each side is a legal mathematical move. The properties for common logs are used with natural logs.

Example 2 Solving a Natural Logarithmic Equation

Solve the following equation.

2 \cdot ln \hspace{0.1 cm}x + 3 \cdot ln \hspace{0.1 cm} 2 = 5 Given

ln \hspace{0.1 cm}x^2 + ln \hspace{0.1 cm}2^3 = 5 Power Property

ln \hspace{0.1 cm}x^2 + ln \hspace{0.1 cm}8 = 5 Simplify

ln \hspace{0.1 cm}8x^2 = 5 Product Property

8x^2 = e^5 Write in Exponential Form

x^2 = \frac{e^5}{8} Division Property

x = \sqrt{ \frac{e^5}{8}} Square Root Property

x \approx 4.3072 Use a calculator

Solving logarithmic equations with natural log is easy if you can use the properties of natural logs and you know how to write a natural logarithmic equation in exponential form.

Check Out the Video for Two More Examples of Solving Logarithmic Equations

Find the Domain of a Function and Its Inverse

In this post, I have embedded an Algebra 2 Video Math Lesson about the function f(x) = \sqrt {2x+2}. Before considering the function, the video defines f^{-1} as the inverse of f or as f inverse.

The video models how to:

  1. Find the domain and range of f(x) = \sqrt {2x+2}
  2. Find f^{-1}
  3. Find the domain and range of f^{-1}
  4. Determine if f^{-1} is a function

If you have any questions regarding the video lesson or other questions regarding finding the inverse of a function, use the comments section.

How To Find and Graph the Inverse of a Function

In this video Algebra 2 Math Lesson, I model how to graph a quadratic function and its inverse. The process starts with graphing a parabola in the form y=a^2+c. The vertex is given by (0,c). After graphing the parabola with four additional points, I create the inverse’s graph by moving the points about y = x line. Finally, I find the inverse of the original function.

If this video math lesson on finding the inverse of a function and then graphing it helped you, leave a comment. Heck, if it didn’t help leave a comment and let me know how to make it better.

How to Find the Inverse of a Relation and an Equation

This video lesson covers the definition of the inverse of a relation and reviews the concept of how to find the inverse of a relation.

Example 1 covers how to find the inverse of a relation from a table of values and provides a good visual of a relation that is a function and its inverse is not a function.

Before example 2, a discussion about how switching the x and the y in the equation is the best method for finding the inverse of a function. Example 2 models how to use the x and y switcheroo to find the inverse of a given function.

Using Properties of Parabolas to Graph a Parabola Problem 2

As I discussed in an earlier post on graphing a parabola, the following information allows to easily graph a parabola in standard form f(x)=ax^2+bx+c.

Properties of a Parabola or a Quadratic Function in Standard Form

Given that f(x)=ax^2+bx+c is a parabola when a\neq0.

  1. If a > 0, then the graph opens upward. If a < 0, then the graph opens downward.
  2. The axis of symmetry is given by the line -\frac{b}{2a}.
  3. The vertex is given by the ordered pair (-\frac{b}{2a},f(-\frac{b}{2a}))
  4. Finding 2 other points is easy. Choose any x-value and find the matching y-value. The second point is found by using the axis of symmetry and the location of the 1st points corresponding point.

In my first post the equation did not have a linear term and was in the form f(x)=ax^2+ c. The vertex could be found using (0,c) and the axis of symmetry (AOS) of x = 0. In terms of the coordinate plane, the y-axis is the AOS.

In this post, we change gears and will be graphing a parabola that is given in standard form f(x)=ax^2+bx+c. The point of this work is to sketch a graph using the core concepts behind the parabolic relation. Remember, because a parabola is symmetric every point except the vertex has a mirror image. Once you find the AOS and vertex, you will have to use the same process to find another point and then its mirror image. Let’s take this to the example.

Example Graphing a Quadratic Equation in Standard Form

Graph f(x)=2x^2+4x-3

Looking at the properties above in order, first we know this graph will open upwards, this is due to the fact the coefficient on the squared term is positive or a>0. The AOS is found by using the ratio -\frac{b}{2a}=-\frac{4}{2(2)}=-1 and can be written as x = -1. If you are trying to create the graph at the bottom of this post, you should draw a vertical dashed line at x = -1, because this represent the AOS.

Next we find the vertex. Since the equation or function is given as f(x)=2x^2+4x-3 and x = -1, it is natural to substitute -1 to get f(-1)=2(-1)x^2+4(-1)-3. Using the order of operations to simplify the right side, we arrive at f(-1)=-5. If you have no idea what is going on, here is an explanation. The AOS serves as the x-coordinate of the vertex. Since we know the x-value, we can find the y-value (***f(x) is the same as using the variable y because of ‘function notation’***). By plugging in the known x-value, we can find the related y-value and we now have an ordered pair. For this example the vertex is the ordered pair is (-1,-5). If you are trying to graph this problem, you should plot the point (-1,-5)

Lastly, we need to find two more points. To do this, pick any x-value to the left or right of the AOS. For this example I will use x = -3. Using the given function, f(x)=2x^2+4x-3, substituting -3 in for x gives f(-3)=2(-3)^2+4(-3)-3 applying the order of operations f(x)=3 and the ordered pair (-3,3). Again, because parabolas are symmetric, (-3,3) has a mirror image at (1,3). If you are graphing along with the post, plot the points and draw a curve to fit those points. Even though this is a sketch, you should try to make it neat.

Graphing a Parabola in Standard Form

Graphing a Parabola in Standard Form

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