Don't forget my book if you are having trouble in algebra.

http://www.lulu.com/spotlight/KKauffman1969

## Tuesday, February 28, 2012

In the system x = 2y + 5 it's easiest to subsitute 2y + 5 for x in the second equation.

2x + 3y = 10

But in the system x + 6y = 12 it's easiest to eliminate one of the two variables first.

5x + 6y = 20

A rule of thumb that I use to determine which method to use is if either equation in the system is already solved for one variable in terms of the other (see first system above), use the substitution method.

Solving the first sytem using substitution, we get

2(2y+ 5) + 3y = 10

4y + 10 + 3y = 10

7y + 10 = 10

7y = 0

y = 0

Substitute 0 for y in the first equation to get x = 2(0) + 5 = 5

Therefore, the solution is x = 5, y = 0.

Solving the second system, it's easiest to eliminate the y variable by subtracting the second equation from the first. This gives us

-4x = -8

x = 2

Substituting 2 for x in the first equation of the system gives us

2 + 6y = 12

6y = 10

y = 5/3

Therefore, the solution is x = 2, y = 5/3

## Monday, February 27, 2012

the first quartile, the median and the third quartile.

## Saturday, February 25, 2012

## Friday, February 24, 2012

For example |x + 5| = 10, the positive case is x + 5 = 10 and the negative case is x + 5 = -10. Therefore, the solutions are x = 5 or x = -15.

The graph of the function f(x) = |ax + b| where a and b are real numbers will always fall in the first and second quadrants since absolute value is always positive.

The graph of the function f(x) = c|ax + b| where a and b are real numbers and c is negative will be in the third and fourth quadrants.

## Thursday, February 23, 2012

For n = 1, 1/1 = 1

For n = .1, 1/.1 = 10

For n = .0001, 1/.0001 = 10,000

For n = .0000001, 1/.0000001 = 10,000,000

Notice as n approaches 0, 1/n approaches infinity.

Can we say then that 1/0 is infinity since we never actually use 0 in the denominator of the fraction 1/n?

## Tuesday, February 21, 2012

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## Sunday, February 19, 2012

Now supposed you have 5 books on the shelf and you want to randomly choose 2 of those books. How many ways can you choose 2 books from the 5 if order that the books are chosen does not matter? What I mean by order not mattering is choosing "Moby Dick" first and "A Tale of Two Cities" second would be the same as choosing "A Tale of Two Cities" first and "Moby Dick" second.

The formula for combinations is n!/r!(n-r)! , where n is the number objects to choose from and r is the number of objects chosen.

So in this example we'd have 5!/(2! * 3!) = 120/12 = 10

To simplify this, 5!/(2! * 3!) = (5 * 4 * 3 * 2 * 1)/(2 * 1 * 3 * 2 * 1)

Notice 3 * 2 * 1 in the denominator cancels with 3 * 2* 1 in the numerator which leaves us with

(5 * 4)/(2 * 1). The 2 in the denominator divides into the 4 in the numerator twice to leave us with

(5 * 2) = 10

## Friday, February 17, 2012

Any even number is divisible by 2.

Any number ending in 0 is divisible by 10 or 5.

Any number ending in 5 is divisible by 5.

If the sum of the digits of a number is divisible by 3, the number is divisible by 3.

If the last number digits of a number is divisible by 4, then the number is divisible by 4.

If the sum of the digits of a number is divisible by 9, the number is divisible by 9.

If a number is divisible by both 2 and 3, it is divisible by 6.

## Wednesday, February 15, 2012

For those having problems figuring out the graphical solution of a linear inequality, I have a tip that will help. Suppose you have the equation 3x + y < 6. First subtract 3x from both sides of the equation to get y < -3x + 6. From here you can graph the equation by plotting a point at (0, 6) and using the slope of -3 for another point. Draw a dotted line through the points as shown in the graph below. The solution set will be all points either above or below this dotted line. Pick a point that is NOT on the line and test in the equation to see if it is a true statement. The easiest point to pick is (0, 0). Substituting this into the equation we get 0 < -3(0) + 6, which is 0 < 6. This is a true statement. Therefore all the points on the same side of the line as (0, 0) is also in the solution set. Notice this region is shaded in yellow.

Today I was working with a student on graphing inequalities. I noticed he was having a difficult time figuring out which way to shade the number line. A simple way to know is if the variable is written before the inequality sign, shade in the direction that the sign is pointing. For example, if the inequality x < 5, we have an open circle at 5 since 5 is not included in the solution and we shade to the left. If the inequality is x > 5, we would shade to the right since the point of the sign it to the right. Think of the sign as an arrow... ----------> or <-------------. In the example -2 < x < = 1 (< = means less than or euqal to) there it an open circle at -2, a closed circle at 1 and everything in between is shaded.

## Monday, February 13, 2012

## Sunday, February 12, 2012

I recently had a question from a question about squaring a binomial. The specific problem was

(2x + 7)^2. He said the answer was 4x^2 + 49. This is incorrect and a very common mistake that I see. What he did was simplify (2x)^2 + (7)^2, but (2x + 7)^2 is not the same as (2x)^2 + (7)^2.

If we take 4^2, for example, that's 4 times 4. So (2x + 7)^2 is (2x + 7)(2x + 7). From here we have to use the FOIL method to multiply, which is (2x)(2x) + (2x)(7) + (7)(2x) + (7)(7).

Multiply and simplify to get 4x^2 + 28x + 49.

(2x + 7)^2. He said the answer was 4x^2 + 49. This is incorrect and a very common mistake that I see. What he did was simplify (2x)^2 + (7)^2, but (2x + 7)^2 is not the same as (2x)^2 + (7)^2.

If we take 4^2, for example, that's 4 times 4. So (2x + 7)^2 is (2x + 7)(2x + 7). From here we have to use the FOIL method to multiply, which is (2x)(2x) + (2x)(7) + (7)(2x) + (7)(7).

Multiply and simplify to get 4x^2 + 28x + 49.

I will try to post interesting math problems and troublesome topics for students.

Quick mental math. 11 X 22 = 242, 11 X 27 = 297, 11 X 45 = 495..... These are examples of multiplication of a 2 digit number by 11. To get the answer in your head, the first digit in the answer is the first digit in the number being multiplied by 11, the middle digit is the sum of the 2 numbers and the last digit is the last digit in the number. Another example.. 11 X 67.. first digit is 6 the sum of the digits is 13, so carry the one to make the first digit 7 and the middle digit 3, then drop the 7 to get 737.

Feel free to ask me any math question that comes up.

Quick mental math. 11 X 22 = 242, 11 X 27 = 297, 11 X 45 = 495..... These are examples of multiplication of a 2 digit number by 11. To get the answer in your head, the first digit in the answer is the first digit in the number being multiplied by 11, the middle digit is the sum of the 2 numbers and the last digit is the last digit in the number. Another example.. 11 X 67.. first digit is 6 the sum of the digits is 13, so carry the one to make the first digit 7 and the middle digit 3, then drop the 7 to get 737.

Feel free to ask me any math question that comes up.

As many of you know, I have been a private math tutor since April 2000. Over the years I have worked with local students in their homes and libraries and have also worked with students from all over the country through online chat, webcam and phone. During this time, I have seen many textbooks which students simply cannot understand on their own. The instructors should help assist these students, but many times, particularly with online classes, the students are left to figure mostly everything out on their own. I often thought of writing self help books on mathematical topics. A few friends have mentioned similar things to me over the years. Last year I was working with a student to help him brush up on his algebra skills. He bought the "Algebra For Dummies" book. We worked through this book and to both of us it seemed to be lacking in organization. For instance, a very basic equation such as x + 2 = 4 was near the middle of the book, while more difficult topics on factoring, for instance, we closer to the beginning. This did not make sense. Furthermore, the answer key in the back of the book gave explanations on how to solve the easier problems, but many of the more difficult problems contained only the answer. This again seemed quite backward to us. That is when I decided to take a shot at writing a self help math book. Several months later, through the help of a friend who has some books published and a talented artist who created my cover design for me, I self published my first book, "Algebra Simplified Basic and Intermediate". I thought writing the book would be the hardest part, but as I am finding out, it's marketing the book that is the most difficult. I have posted the link to the book on Facebook, on math forums, contacted high school guidance counselors and a university professor that I know, as well as some friends to help promote the book. In the meantime, I am in the process of writing the next book, more intermediate and advanced topics seen in a high school algebra course. The front cover of the book is showed below. If anyone is interested in the book or know anyone that might be interested, you can find the book as an ebook as a pdf file and as a paperback at http://www.lulu.com/spotlight/KKauffman1969

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