Strangely enough, even as a math professor, I had never heard the term “two-step equation” until I was working on a side project for high school math! I had always tended to think of all equations like this as linear equations and that was that.

The phrase “two step equation” simply refers to linear equations that take two steps to solve. This comes from the idea of a one-step equation which, you guessed it – takes ONE step to solve! See, you got it already! Just to make it perfectly clear, let’s take a look at an example.

The equation 3x+5=14 is a two step equation. Why? In general, to solve any linear equation you want to isolate the variable. In other words, you want to get the variable by itself.

In order to do this, the first step you would take here is to subtract the 5 from both sides. Doing this will give you the equation 3x=9. Now, the only thing “happening” to the x is that it is being multiplied by 3. In order to undo this, you must divide by 3. Ahhh – the second step. Doing this gives you the wonderful equation x=3. Why is this wonderful? ‘cuz its the answer! Not only that, but you got it in TWO STEPS (did I drive that point home enough yet?).

Another example (without all the commentary – I promise) is 2x+6=24:

2x + 6 = 24
2x = 18 (subtract 6 from both sides)
x = 9 (divide by 2 on both sides).

In general, the steps to solve any “two-step” equation will be :

  1. Move the constant that is added or subtracted from the variable to the other side of the equation – If the number is added to the variable (like x), subtract it from both sides. If it is subtracted from the variable, add it to both sides.
  2. Divide both sides by the coefficient – The coefficient is the number that is multiplying the variable (often x). So, if you have a 5x, you will divide both sides by 5 at this step.

Once your equation is in the form “your variable = a constant (a number)”, you know you are done! Well, not completely… see, you aren’t really done until you have followed me on twitter. Doing that will make sure you don’t miss any new math tips that I post ;) .

, , ,

While there are plenty of methods to get a good hand drawn graph of a linear equation, many plot more points than necessary and (if you aren’t careful) misrepresent the location of two important points: the x and y intercepts.

Using the intercepts to plot your line takes care of both of these problems and more importantly: its fast! In this video, I will use two examples to show you exactly how it’s done.

http://www.mathbootcamps.com/media/fastgraphsoflinearequations.mp4
,

At this point, I haven’t even written out any equation or expression and you already know what I’m talking about. There are problems that come up in algebra where it seems like you should be able to cancel a variable out and move on. Yet, in these very situations your professor keeps telling you it can’t happen!

It may not always seem like it, but there really is a pattern and a simple rule of when you can and can’t cancel a term or a variable. Let’s focus on rational expressions specifically. Those are expressions (like the one below) where both the numerator and the denominator are polynomials.


\dfrac{x+5}{x^2+1}

When can you cancel terms in a rational expression?

You can cancel any matching factors that occur in both the numerator and the denominator. Let’s use numbers to understand this a little better:


\dfrac{4}{14}


In the fraction above, 2 is a factor of 4 since 2 x 2 is 4. Similarly, 2 is a factor of 14 since 2 x 7 = 14. Based on my rule, I should be able to cancel the 2′s and get a fraction that is still the same thing! Let’s see:



\dfrac{4}{14}=\dfrac{2(2)}{2(7)}=\dfrac{2}{7}

If you check on a calculator you will find that 4 divided by 14 is the same as 2 divided by 7 – so this rule seems to be working here. What is really going on?

Why does this work?

Every number other than zero, when divided by itself is 1. Since a fraction simply represents division (in one sense), \dfrac{2}{2} = 1 and we really just have 1\times\dfrac{2}{7}.

Trying it with Variables

Now that everything makes some sense with numbers, let’s try variables. For us, the variables represent numbers so they will behave in the same way. Take for instance the expression


  \dfrac{x^2-x}{x^3}, x\neq 0

The numerator has a couple of factors which we can find by factoring out an x (doesn’t that phrase make a lot more sense now?!): x^2-x = x(x-1). This shows us that x and x-1 are factors of the numerator. Looking at the denominator, we see x is also a factor since x^3=x(x^2). Remember, when both the numerator and the denominator share a factor, you can cancel that factor since it is really just 1.

  \dfrac{x^2-x}{x^3} = \dfrac{x(x^2-1)}{x(x^2)}= \dfrac{x^2-1}{x^2}

The only time you can’t cancel terms in the numerator and denominator is when they are both NOT factors.. That’s why you can’t cancel x in \dfrac{x-5}{x}. The x is not a factor of the numerator; its just a term being added. Cancelling the x here would be like cancelling the 5 in \dfrac{5+1}{5} and saying that is 1. If you check it in a calculator it certainly isn’t!

, ,

There are many many ways to solve a quadratic equation (any equation that can be written as ax^2+bx+c=0), but today I’m going to focus on using the quadratic formula. Surely by now, you have seen a nice song to help you memorize it right? And you saw this nice tool to check your answer with right?

First the formula itself! When ax^2+bx+c=0 (quick notes: these are all real numbers and a is not zero) then

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
.

Now, a, b, and c are all numbers from your equation. That means, as long as you can figure out what a,b,c are then you can just plug them into this equation and get the solution or solutions.

For example, take a look at this quadratic equation:

x^2-x-6=0.

In this example, a=1, b= -1, and c=-6. Why are b and c negative? The formula is based off the form ax^2+bx+c=0, so if you come across anything that is negative, keep the sign when you use the formula!

x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)}

=\dfrac{1\pm\sqrt{1+24}}{2}

=\dfrac{1\pm\sqrt{25}}{2}.

At this stage, the plus or minus symbol (\pm) tells you that there are actually two different solutions:

\dfrac{1+\sqrt{25}}{2}=\dfrac{1+5}{2}=\dfrac{6}{2}=3 and

\dfrac{1- \sqrt{25}}{2}=\dfrac{1-5}{2}=\dfrac{-4}{2}=-2.

Therefore the final answer would be x=3,-2.

That example worked out very nicely but sometimes there will be a bit more work to do. Take, for instance,

2x^2+2x-7=0

In this case, a=2,b=2, and c=-7. Therefore

x=\dfrac{-2\pm\sqrt{(-2)^2-4(2)(-7)}}{2(2)}

=\dfrac{-2\pm\sqrt{4+56}}{4}

=\dfrac{-2\pm\sqrt{60}}{4}

=\dfrac{-2\pm 2\sqrt{15}}{4}

Notice that 2 is a FACTOR of both the numerator and denominator, so it can be cancelled.

=\dfrac{-1\pm\sqrt{15}}{2}.

This answer can not be simplifed anymore without using decimals. Therefore the final answer is: x=\dfrac{-1+\sqrt{15}}{2},\dfrac{-1-\sqrt{15}}{2}.

It is also possible to discover that you have no solution or solutions that are no real numbers. We will take a look at those in a future post!

, ,

This post is for parents (ok that is probably clear from the title). Maybe you are homeschooling and just about to start algebra or your child is in school and just starting algebra. Either way you will probably find yourself in a discussion about it since this is usually a child’s first attempts at math with abstract ideas and concepts. I tackled this a little bit before in my article “What is Algebra?” but this time I am going to focus on what I think are the two big ideas with algebra.

BIG IDEA #1: We may have an unknown number (a variable) but we do know something about it!

In day to day life, things come up and you don’t have all of the information you need to make a conclusion. When the information you need is some kind of number, well, that’s where algebra comes in.

It’s like playing 20 questions but you already know some of the answers. For example in the equation x+3=5, I don’t know what x is but I have a clue: when I add 3 to x I get 5. Algebra is all about taking the clues that are given to us and using them to figure out what the unknown value is! The clues may be in words (like word/story problems or in real life) or in the form of an equation like the one above. In fact, when the clues are in words, we may go ahead and make an equation about it since we have well laid out steps to solve equations.

BIG IDEA #2: The game: Figure out the uknown value. Like any game, there are rules to do it!

I don’t completely like the word “rules” here but it gets the idea across. As I’ve said before, math is all about patterns and since the same patterns of things we know about a given variable (the unknown we are trying to find) come up over and over again we can develop a set of steps to use when we see it. When you take a class in algebra, what you are learning to do is recognize which type of pattern is in front of you and then choose one of the methods that works for that pattern.

That’s not to say that there is no creativity in algebra. Sometimes the step by step “rules” for a certain pattern may be overkill. Meaning that it will get you the right answer but maybe there is a shortcut for some reason. As you are learning the patterns and steps you start noticing these too. This is very much like any video game – you can always complete the level by doing every little thing you are “supposed” to do but after you get used to it, you start seeing all of the shortcuts!

It is really important for kids who are just starting out in algebra to understand these big picture ideas. Not only will it help them see why they are doing what they are doing but it will help them start connecting algebra to other areas of math, science, and life. (and help answer that whole “when do we need math?” thing that pops up) Think about it: Would you enjoy learning something if you didn’t really get or never talked about what the ideas are behind it?

, ,

While expressions with many exponents and variables can look complicated, simplifying them is really just about applying the laws of exponents one by one. When working on problems like these, make sure that you are never combining two terms with different bases and not leaving any negative exponents (this varies but it is most common to make sure all exponents are positive).

In this video I will go over four different types of examples and show you another way of thinking about these types of problems!

http://www.mathbootcamps.com/media/4-examples-simplifying-exponents.flv
, ,

To use this solver, your quadratic equation should be written in the form ax^2+bx+c=0. Simply input the coefficients (a,b,c) into the space below! While it will show you the steps, be aware that wolfram alpha (the program behind this widget) likes to complete the square for just about everything which may not be the most straightforward way to get your answer.

(I will probably add cool wolfram alpha widgets here over time but if you are looking for one for your class of website head on over this way: http://www.wolframalpha.com/widgets/

, ,

The rules of exponents (sometimes called the laws of exponents) allow us to simplify expressions by using our understanding of what an exponent means in much the same way that combining like terms uses the idea of what multiplication means. In both cases, we want to eliminate any extra/unnecessary information and literally make whatever we have in front of us simpler (while making sure we didn’t change the meaning!) Why write 5^{2}4^{3}5^{8}4^{2} when you could write 5^{10}4^{5}?

As a brief reminder, an exponent tells you how many times to multiply a number, variable or expression times itself. For example:

x^4=x \cdot x\cdot x\cdot x

(x+5)^3=(x+5)(x+5)(x+5)

All of the rules require that the terms have the SAME BASE. The base is whatever is being taken to a power. Above, x is the base in the first example and x+5 is the base in the second example.

When multiplying two terms with the same base, add the exponents

x^{m}x^{n}=x^{m+n}

For example, x^{9}x^{3}=x^{9+3}=x^{12}.Why does this work?

If I have 3^43^2, this means to multiply 3^4 by 3^2. But, 3^4=3\cdot3\cdot3\cdot3 while 3^2=3\cdot3. Looking at the original example, 3^43^2 = (3\cdot3\cdot3\cdot3)(3\cdot3) and if you count the number of 3′s being multiplied, there are 6 – so really, this is a complicated way to write 3^6.

When dividing two terms with the same base, subtract the exponents

\dfrac{x^m}{x^n}=x^{m-n}

For example, \dfrac{y^7}{y^4}=y^{7-4}=y^3 and \dfrac{x^2}{x^5}=x^{2-5}=x^{-3}=\dfrac{1}{x^3}. Why does this work?

Just as I did before, I’m going to use the basic idea of an exponent to see this rule in action. Take for instance, \dfrac{5^4}{5^2}. This can be rewritten by writing out the mulitplication \dfrac{5\cdot5\cdot5\cdot5}{5\cdot5}. At this point, I can cancel any factors that are shared by the numerator and the denominator and I am left with \dfrac{5\cdot5}{1}=5^2. The “rule” is essentially doing this work for us by subtracting. If I had used the rule here I would have \dfrac{5^4}{5^2}=5^{4-2}=5^2. While I think this is useful, I have always found that it can be more useful to use the idea of cancelling factors directly. As with a lot of math “problems” though, it depends on the situation.

When taking a “power to a power”, multiply the exponents

(x^m)^n=x^{mn}.

For example (x^4)^4=x^{16}. Why does this work?

Alright, let’s use another example say – (8^2)^3. If an exponent tells me how many times to multiply something by itself, then this means to multiply 8^2 by itself 3 times: 8^28^28^2. Now, we have three terms with exponents multiplied that all have the same base. This is the first rule of exponents so I know to add them: 8^28^28^2 = 8^6. This is the same thing I would have found if I had multiplied the 2 and the 3.

Now that we have those down, a few things to keep in mind when you start using them:

, ,

Understanding what you have when you have a polynomial is one thing, but performing operations like addition, subtraction, multiplication, or even division is another. Today, we will just talk about a special case that comes up when you multiply two specific types of polynomials – two binomials.

Remember, a binomial is a polynomial with two terms that are either added or subtracted from eachother like 3x+2 or even something more complicated like 5xy-3x^2. When multiplying any two polynomials, the ultimate goal is to make sure every term in the first polynomial has been multiplied by every term in the second polynomial. With two binomials, this process can be simplifed into an easy to remember process that you have probably heard of: FOIL. What does FOIL stand for? First, outer, inner, last.

First: multiply the first terms of each binomial

Outer: multiply the “outer” terms in each binomial

Inner: multiply the “inner” terms in each binomial

Last: multiply the “last” terms in each binomial

If you think about it, this does in fact make sure every term is multiplied by every other term. Take the first term for instance. Once you have FOILed, it has been multiplied by the first term of the second binomial and the last (by the “outer” step). This is true for every term!

Once this process is complete, it is very likely you have some cleaning up to do in the sense of combining like terms, so you should make sure to simplify your answer once you are done. Here is how it works with an example:

(3x+1)(2x-3)

First: The first two terms are the 3x and the 2x. Multiplying these will give (3x)(2x) = 6x^2.

Outer: The “outer terms” are the 3x and the -3. Multiplying these terms will give (3x)(-3) = -9x.

Inner: The inner terms are the 1 and the 2x. Multiplying these terms will give (1)(2x) = 2x.

Last: The last terms are the 1 and the -3. Multiplying these terms will give (1)(-3) = -3.

The final answer would be the terms we got in each step above added together. The steps all together would look like this:

(3x+1)(2x-3)
= (3x)(2x)+(3x)(-3)+(1)(2x)+(1)(-3) (using the steps listed above)
= 6x^2-9x+2x-3
= 6x^2-7x-3 (after combining like terms)

This is also the same method you would use to find the square of any binomial like (x+y)^2. If you think about what an exponent means, you can rewrite this as (x+y)(x+y) and the use FOIL to determine that it is x^2+2xy+y^2. As you can see, this is definitely not x^2+y^2.

A last but important point: This method will NOT work if one of the parentheses contains more than two terms (in other words if either is not a binomial). For instance, to find (2x-1)(x^2+5x-1), you would need a different method.

Also, if you have more than one binomial, FOIL may not work by itself – you may need to use FOIL on the first two binomials and then some other method depending on the results of that. This is a very nice method but make sure you apply it only when it makes sense!

, ,

In math, there are some words that are practically foundations to everything else. Just like any topic, math has its own vocabulary and one of the first words you should get comfortable with in algebra is the word “polynomial”.

Polynomials are expressions involving terms which may have exponents that are multiplied, added, or subtracted from eachother. For example, each of the expressions below would qualify as a polynomial:

5x^2

2x^{4}-5x+11

\dfrac{2}{3}z^5+2xyz-xy

In each case above, notice that the exponents were POSITIVE INTEGERS (positive whole numbers). If an exponent is negative, then this implies you are dividing by that term (based on the definition of a negative exponent) and polynomials can not have any division involving the variables. A polynomial CAN have fractions involving just the numbers in front of the variables (the coefficients).

With that in mind, the following are NOT polynomials:

x^{-2}+x-1 (negative exponent)

xy^{\frac{1}{2}}+2 (fractional exponent)

There are a few special types of polynomials which get their own names. (Remember, we are thinking of a term as something like 3xy or 5):

  • Monomial A monomial is a polynomial with one term: 7x^5
  • Binomial A binomial is a polynomial with two terms being added or subtracted: x+5
  • Trinomial A trinomial is a polynomial with three terms being added or subtracted: x^2+2x-1
,