Consider the following problem:

The instructions on a can of paint read that a half gallon of paint will cover approximately 175 square feet. How many half gallon cans of paint must be purchased to apply 2 coats of paint to cover 600 square feet per coat?

Solution:

We can think of this problems in terms of direct variation. The amount of paint needed is directly proportional to the area that needs to be painted.

Let y = amount of paint in gallons

x = area to be painted in square foot.

Therefore, we use the equation for direction variation, y = kx.

Substitute ½ for y and 175 for x and solve for k.

½ = k(175) (Divide both sides by 175)

1/350 = k

To apply 2 coats over 600 square feet means we need enough paint to cover 1,200 square feet, therefore

y = (1/350)(1200)

y = 3.42 gallons.

We need to purchase 7 cans of paint because 7(1/2) = 3.5.

Consider the following problem:

The flight of two planes is being tracked on a rectangular coordinate system. The flight of the first plane is defined by the equation 3x + 4y = 12 and the flight of the second plane is defined by the equation y = -(3/4)x + 10. If the planes continue along the same paths, is there any danger of them colliding? (Assume the planes are flying at the same altitude)

Solution:

To determine if the planes will collide, we need to find the point of intersection of the two lines.

Substitute -(3/4)x + 10 for y in the first equation and solve for y.

3x + 4[-(3/4)x + 10] = 12

3x - 3x + 40 = 12

40 = 12 is a false statement. Therefore, there is no solution. There is no chance of the two planes colliding if they keep flying along the same paths.

Another way to solve this is to get the slope of each line. If the slopes are the same without the same y – intercept, then the lines are parallel and therefore the planes cannot collide.

3x + 4y = 12

4y = 12 - 3x

y = 4 - (3/4)x. Notice the slope of both lines is -3/4. The y – intercepts are different. Therefore there is no chance of the planes colliding if they continue along their same path.

Here's the preface and about the author section of my second book. If anyone is interested in the book, let me know.

Preface

Algebra Simplified Intermediate & Advanced picks up where my first book, Algebra Simplified Basic & Intermediate left off. It is intended to assist students in intermediate and advanced topics studied in a 2^nd year high school algebra course or an intermediate college algebra course. The material is presented in textbook style format with each concept illustrated through numerous examples. The examples are solved methodically to explain each concept as simply as possible. Important notes and tips for easier learning are presented in bold throughout the book. The goal is provide readers sufficient detail in the examples so they can solve similar problems on their own, which are presented at the end of each section.

Specific topics covered in this book include division of polynomials, finding roots of polynomials, factoring techniques such as quadratic formula and completing the square, radicals, rational exponents, complex numbers, logarithms, conic sections, composition of functions, inverse functions, arithmetic and geometric sequences, matrices and basic probability and counting techniques At the end of each chapter there is a section titled Key Terms and Concepts to Review. Answers to problem sets along with key steps in the solution of each problem are also found at the end of each chapter. The book concludes with a glossary of terms used throughout the book.

About the Author

Kerry Kauffman has a Bachelor of Science Degree in Statistics from Lehigh University. He has tutored students in algebra, geometry, trigonometry, statistics, calculus, number theory and other courses at the high school and college level since 2000. He has worked on tutoring websites tutorsteach.com, tutornation.com and liveperson.com and has assisted students taking online college courses from Harvard University, Strayer University, Kaplan University and University of Phoenix among others. He is the author of Algebra Simplified Basic & Intermediate, which was published in 2011.

Suppose you have a right triangle ABC with m<A = 44, C is a right angle, length of side b = 10. What is the m<B, length of sides a and c?

For a problem such as this we will use trig functions.   We know that sine (represented by sin) is the opposite side divided by the hypotenuse.

So sin 44 = a/c. But notice this doesn't help at all since we have 2 unknowns. But we can use cosine (represented as cos) since cosine is adjacent side divided by the hypotenuse.

So cos 44 = 10/c.

Using a scientific calculator we get 0.7193 = 10/c.

Therefore c = 13.9 (rounded to 1 decimal place)

To get the length of side a we could use Pythagorean theorem. But it's best to not use this because if side c (which we just calculated) is incorrect, then the calculation for the length of side a will be incorrect as well.

sin 44 = a/13.9

Using a scientific calculator we get .6947 = a/13.9.

Therefore a = 9.7 (rounded to 1 decimal place)

This is good and a systems of equations are used. A little fun for Sunday!

When thinking of an asymptote, we are really thinking of a limit.  For example, suppose we want the asymptote of the function

f(x) = 1/x, a very basic function but used to illustrate this point.

First we know that x cannot equal 0 because a function is undefined with a 0 denominator.  So as x approaches 0, the function is undefined.

As x gets larger approching infinity, notice 1/x gets smaller

x = 10, 1/x = .1
x = 100, 1/x = .01
x = 10000, 1/x = .0001

Therefore 1/x approaches 0 as x approaches infinity

When x is very small and negative, 1/x approaches infinity as it does when x is very small and positive.

x = .1, 1/x = 10
x = .01, 1/x = 100
x = .0001, 1/x = 10000
x = -.1, 1/x = -10
x = -.01, 1/x = -100
x = -.0001, 1/x = -10000

As x values get more ad more negative, 1/x approaches 0.

x= -10, 1/x = -.1
x= -100, 1/x = -.01
x= -10000, 1/x = -.0001

The asymptotes are the lines x = 0 and y = 0.

The idea of a limit is a topic seen in precalculus and in calculus.

Lim      1/x = infinity
x-> 0 (positive)

This means the limit as x approaches 0 from the positive side equals infinity

Lim     1/x = -infinity
x -> 0 (negative)

The limit as x approaches 0 from the negative side equals negative infinity.

Lim    1/x = 0    and     Lim   1/x = 0
x -> infinity                 x-> negative infinity

When given the lengths of two sides of a right triangle, we can obtain the third side by using the Pythagorean Theorem.

a^2 + b^2 = c^2, where c is the length of the hypotenuse.

A Pythagorean triple consists of 3 integers which satisfies the Pythagorean Theorem.

There are 12 such triples in which c < 100. For all practical purposes you need not memorize all of theses, but many times the tiples up to (11, 60, 61) are seen frequently. It helps to memorize those.

( 3 , 4 , 5 )	( 5, 12, 13)	( 7, 24, 25)	( 8, 15, 17)
( 9, 40, 41)	(11, 60, 61)	(12, 35, 37)	(13, 84, 85)
(16, 63, 65)	(20, 21, 29)	(28, 45, 53)	(33, 56, 65)
(36, 77, 85)	(39, 80, 89)	(48, 55, 73)	(65, 72, 97)

Suppose you are at a casino and want to know which game has the  highest probability of winning. It's actually the dice game Craps, which  is played as follows.   The player throws two dice. If the sum is 7 or 11, then he wins. If the sum is 2, 3 or 12, then he loses. If the sum is anything else, then he continues throwing until he either throws that number again (in which case he wins) or he throws a 7 (in which case he loses). To find the probability of winning, first finding the  probability of winning on the first roll by getting a 7 or 11.    The number of ways to get a sum of 7 on two dice is   (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) which is 6.   The number of possible outcomes when rolling two dice is 36. Therefore the probability of getting 7 is 1/6.   There are 2 ways in which get get a sum of 11, (6,5) and (5,6). Therefore the probability of getting 11 is 2/36 = 1/18.    The probability of winning on first roll = 1/6 + 1/18 = 2/9 Next we find the probability of losing on the first roll, which is getting the sum of 2,3 or 12 on two dice.   There is only 1 way to get a sum of 2 on a pair of dice, (1,1) so the probability of getting a 2 is 1/36.   There are 2 ways to get a sum of 3 on a pair of dice (1,2), (2,1), so the  probability of getting a 3 is 1/18.   There is only 1 way to get a sum of 12 on a pair of dice (6,6), so the probabiliy of getting a 12 is 1/36.    Therefore, the probability of losing on the first roll is 1/36 + 1/36 + 1/18 = 1/9.   The probability of rolling a sum of 4,5,6,8,9 or 10 which enables you to continue rolling until you either win or lose is 1 - 2/9 - 1/9 = 2/3.   You can also list the probability of obtaining each sum and adding, but there is no need to do this. We simply subtract the probabilities of obtaining all other possible sums (2,3,7,11,12) from 1.   Now focus on the individual probabilities depending on which sum has  been obtained which enables you to keep rolling until you win or lose. The probability of getting 4 as is = 1/12 since there are 3 ways (1,3), (3,1) (2,2) of getting a 4 and 36 possible outcomes.   The probability of getting 7 = 1/6 So the probability that the game continues = 1 - 1/12 - 1/6 = 3/4 So the probability of winning in this case is (1/12) + (3/4)(1/12) + (3/4)^2(1/12) + ..... = (1/12)[1 + (3/4) + (3/4)^2 + ..... to infinity] (factor out 1/12) = (1/12)[1/(1 - 3/4)] = (1/12)(4) = 1/3 Therefore the probability of getting a 4 and then eventually winning is = (1/12) (1/3) = 1/36.   Notice that this is the same as the probability of winning when rolling a 10 first because getting a 10 is equally likely as getting a 4 (3 possible  outcomes, (4,6), (6,4), (5,5).   Similarly the probability of getting a 5 or 9 is the same and is 1/9 since there are 4 possible ways to get a 5 and 4 possible ways to get a 9 ( (1,4), (4,1), (2,3), (3,2) and (4,5), (5,4), (3,6), (6,3), respectively))   The probability of continuing after is 1 - 1/9 - 1/6 = 13/18.      Using the same reasoning as above the probability of winning is (1/9)[1/(1 - 13/18)] = (1/9)(18/5) = 2/5.   The probability of getting a 5 or 9 and eventually winning is is (1/9)(2/5) = 2/45      The probability of getting a sum of 6 or 8 is 5/36 ( (1,5), (5,1), (2,4), (4,2), (3,3) and (2,6), (6,2), (3,5), (5,3), (4,4) respectively )    The probability of continuing after is 1 - 5/36 - 1/6 = 25/36 The probability of winning is  (5/36)[1/(1 - 25/36)] = (5/36)(36/11) = 5/11    The probability of getting a 6 or 8 and eventually winning is (5/36)(5/11)= 25/396    Since we went through all the possibilities, the total probability of winning = 2/9 + 2[1/36 + 2/45 + 25/396]    (probability of winning on 1st roll + probability of winning rolling 4 or 10 first + probability of winning rolling 5 or 9 first + probability of winning rolling 6 or 8 first)    = 0.493   So even the game with the best chance of winning, you will lose slightly more often than not.

I was thinking about another book and a few friends came up with some ideas.  The consensus is a book featuring the top mistakes seen in algebra. It would be written like a conversation between me and the student. The student would tell me how he or she is solving the problem and I will reply with their errors and the correct way to solve it. That project will not be now, want to keep promoting the other books by contacting colleges and homeschool organizations across the country.

Here's an small portion of the chapter on logarithms from my second book.

Properties of Logarithms and Logarithmic Equations

Recall that a logarithm is an exponent. Since there are properties for exponents, we would expect there to be properties associated with logarithms. The seven properties of logarithms are as follows:

1. log _b 1 = 0 because b⁰ = 1

2. log _b b = 1 because b¹ = b

3. log _b b^x = x because b^x = b^x

4. b ^log _b^x = x because log _b x is the exponent that b is raised to get x.

5. log _b xy = log _b x + log _b y

6. log _b (x/y) = log _b x – log _b y
7. log _b x ^y = y log _b x

• Note that an easy way to remember rules 5 and 6 is to think of the rules for exponents. If you multiply like bases, add the exponents. In rule 5 we have the log of a product, which is the same as the sum of the logs. If you divide like bases, subtract the exponents. In rule 6 we have the log of a quotient, which is the same as the difference of the logs.

Below are some examples using the properties of logarithms.

Examples: Write each as the sum or difference of logarithms.

1. log ₃ (3 ∙7)

log ₃ (3 ∙7) = log ₃ 3 + log ₃ 7 (the log of a product is the sum of the logs, property 5)
log ₃ (3 ∙7) = 1 + log ₃ 7 (property 2)

2.. log (1000xyz)

log (1000xyz) = log 1000 + log x + log y + log z (property 5)
log (1000xyz) = 3 + log x + log y + log z (property 3, log 1000 = log 10 ³ = 3)

 

3. log ₃ (x/81)

log ₃ (x/81) = log ₃ x – log ₃ 81 (the log of a quotient is the difference of the logs, property 6)

log ₃ (x/81) = log ₃ x – 4 (log ₃ 81 = log ₃ 3 ⁴ = 4)

4. log (2xy/7z)

log (2xy/7z) = log(2xy) – log(7z) (the log of a quotient is the difference of the logs, property 6)

log (2xy/7z) = log 2 + log x + log y – log(7z) (apply property 5 to log(2xy))

log (2xy/7z) = log 2 + log x + log y – log 7 – log z (apply property 6 to log(7z))

Examples: Write each of the following without a radical or an exponent.

1. log (1/3)⁴

log (1/3)⁴ = 4log(1/3) (property 7, log _b x ^y = y log _b x)

2. log √17

log √17 = log(17) ^½ (recall that √x = x ^½)

log √17 = (1/2)log(17)

Examples: Write each as the sum and/or difference of logarithms.

1. log x³y⁴z⁵

log x³y⁴z⁵ = log x³ + log y⁴ + log z⁵ (property 5)

log x³y⁴z⁵ = 3log x + 4log y + 5log z (the log of a power is the power times the log, property 7)

2. ln (4x²y/z)

ln (4x²y/z) = ln (4x²) + ln y – ln z (properties 5 and 6)

ln (4x²y/z) = ln 4 + ln x² + ln y – ln z (property 5 to ln (4x²))

ln (4x²y/z) = ln 4 + 2 ln x + ln y – ln z (property 7 to ln x²)

3. log ⁵√(x³y/z²)

log ⁵√(x³y/z²) = log (x^3/5y^1/5/ z^2/5)  

log ⁵√(x³y/z²) = log x^3/5 + log y^1/5 – log z^2/5 (properties 5 and 6)

log ⁵√(x³y/z²) = (3/5)log x + (1/5)log y – (2/5)log z (property 7)