Wednesday, May 13, 2009

Solving Quadratic Equations

Here are a few sites that may help you when solving quadratic equations:

Tuesday, May 5, 2009

Solving Systems of Equations Using Elimination

I have found several websites that might help you with this concept.

Purple Math has a great site that shows examples here.

Check out the Unusual Outcomes in this PDF file. The other information there is also very good. The explanations are detailed.

Use this PDF file to help you determine when to add and when to multiply and then add.

Here are some problems that you can try to determine if you are getting it. Their answers are provided as well.

Thursday, April 23, 2009

Polynomial Operations and Factoring

Here are some websites that might help clarify polynomial operations and factoring.

Polynomial addition and subtraction -

Adding polynomials -

Subtracting polynomials -

Multiplying polynomials - and

Factoring polynomials - and

In order for my students to earn extra credit, they must visit at least 2 of the sites and comment on both.

Wednesday, March 4, 2009

Pythagorean Theorem

The Pythagorean Theorem is used to find a missing side length of a right triangle. It only works for right triangles (one with a right angle).

The formula to use is a2 + b2 = c2. A really good explanation of the formula and its derivation can be found here.

In order to use the theorem, you must know the parts of the triangle. Parts a and b of the theorem are the legs of the triangle. These are the side lengths that make up the right angle. The third side (longest side) is noted as c in the theorem.

Now substitute the values given into the theorem (formula) and solve the equation.

Example: A triangle has a side length of 6 cm and a side length of 8 cm. What is the length of the hypotenuse?

62 + 82 = c2
36 + 64 = c2
100 = c2
√100 = √c2
10 = c

Example: A triangle has a side length of 6 cm and a hypotenuse length of 12 cm. What is the length of the missing side?

a2 + 62 = 122
a2 + 36 = 144
a2 + 36 – 36 = 144 – 36
a2 = 108
√a2 = √108
a = √108
a = 6√3

Picture of triangle above from

Wednesday, February 18, 2009

Rate-Time-Distance Word Problems Part 2

Now for the second part of the rate-time-distance problems. This example refers to opposite direction travel.

Let’s look at the following example:

Train 1 leaves the train station travelling south at a rate of speed that is 10 mph more than train 2 which is travelling north. After 3 hours, they are 450 miles apart. Find the rate of train 2.

We are going to use another table to organize the rate, time and distance information.






r +10


3 (r + 10)





Since the trains are moving in opposite directions, we are going to add their distances together to get 450 miles.

3 (r + 10) + 3r = 450
3r + 30 + 3r = 450
6r + 30 = 450
6r = 420
r = 70

Therefore the rate of train 2 is 70 mph.

Tuesday, February 17, 2009

Rate-Time-Distance Word Problems

Or otherwise known as those dreaded train problems. These are always tricky for algebra students. Ask someone you know which problems they remember from Algebra I and almost all of them will have something to say about the train problems.

Train problems can be separated into two categories: same direction travel and opposite direction travel. Each are handled differently, but using a chart makes it easier to set up the equations. You also have to remember that distance equals rate times time or d=rt.

Here is an example. A train leaves the train station at 2:00 p.m. Its average rate of speed is 90 mph. Another train leaves the same station a half hour later. Its average rate of speed is 120 mph. If the second train follows the same route on a parallel track to the first, how many hours will it take the second train to catch the first?













Since the trains are travelling in the same direction, their distances are equal when the 2nd one catches the 1st. Therefore, to solve this equation, we set the distance of Train 1 equal to Train 2.

90t = 120(t-0.5)
90t = 120t - 60
-30t = -60
t = 2

The first train travelled for 2 hours before the 2nd train caught up to it. The problem asks how long it takes the 2nd train to catch the first. So it takes the 2nd train a half hour less than it did the first train which is 1.5 hours.

Check back tomorrow for information on opposite direction travel.

Thursday, February 5, 2009

Slope Y-Intercept Form to Standard Form

My students are having some trouble with changing slope y-intercept form of a line to standard form and vice versa. You must first understand solving equations before this will make sense. Having a good grasp of variables on both sides of the equation will help since it involves moving a variable term from one side of the equation to the other.

Here is an example:

Rewrite y = 2x - 5 (slope y-intercept form) as standard form.

Move the 2x to the left side of the equation with the y. -2x + y = -5

That is all that is required for this one. It is now in standard form (Ax + By = C).

Another part of the assignment that we worked on required that the standard form be written with integers (no fractions).

Rewrite 3/5y = 2/3x + 2 in standard form using integers.

Move the x term to the left side of the equation. -2/3x + 3/5y = 2
Now to get rid of the fractions, you will multiply both sides of the equation by 15 (the least common multiple of 3 and 5). You could multiply by any multiple of 3 and 5 but generally we use the least common multiple. As long as this multiplication is done to BOTH sides, the equation will stay balanced. Hint: Remember that you must use the distributive property on the left side of the equation.

15[-2/3x + 3/5y] = 2(15)

-30/3x + 45/5y = 30

-10x + 9y = 30 (standard form using integers)

Wednesday, January 28, 2009

Problems with Starting Homework?

Here's a tip that I use for any job that I continue to put off (procrastinate about). Set a timer for 5, 10 or 15 minutes. You can do anything (even Algebra work) for those periods of time. Sit down and get started. When the timer goes off, stop all work and rest your brain for 10-15 minutes. If there's still more to do, set the timer again and go again. Continue this until you have completed that homework.

Don't forget that there are answers in the back of most books. Checking those answers to see if you are on the right track should not be considered cheating. I encourage my students to check those answers. This will help boost your confidence if you are getting the problems right. It will also encourage you to look back over notes, examples in the textbook, etc. if you are getting the answers incorrect.

Sunday, January 25, 2009

Algebra Games On-line

Hey guys! I'm trying to find some good online games for my Algebra classes. I found the following site and it looked like it might be interesting. Let me know what you think.

Ever Wonder About Dividing by Zero?

If you've ever wondered why you can't divide by zero, here is an explanation that uses what you learned in elementary school to help you understand this concept.

In elementary school, you learned that division is just multiplication fact families in an alternate order. In other words, 6/3 = 2 because 2 x 3 = 6.

We can use fact families to find out about dividing by zero.

0/7 = ?? The answer is 0, because 0 x 7 = 0.

What about 7/0? If you put this in your calculator, you'll get an error. Why?

Well, if we use fact families, 7/0 = ?? What number can you multiply by 0 to get 7? There isn't any number. So the answer is undefined.

Now you may be wondering about 0/0? The calculator still gives you an error. Shouldn't it have given you a 1 or maybe a 0? Isn't 7/7 =1? And 2/2=1?

Let's try fact families again to see what the answer should be. 0/0 = ??

What number multiplied by 0 = 0? Well, every number would work wouldn't it, so the answer is indeterminate. You can't determine what the answer should be.

Hopefully, this explanation has helped to clear up division by zero and why is doesn't work.

Thursday, January 22, 2009

New Blog Created

I've created this new blog to help answer some of those tricky algebra questions that high school students have. I will be including articles that seem to give lots of students trouble. I currently teach Algebra I and have been doing so for several years. I'd like to provide assistance to those who need it so here goes! Welcome and please visit often.