Be the Professor: Algebra Skills

Hello, I’m Professor Small and I’m going to teach you how to use algebra. Algebra is fairly simple, used on a daily basis in everyday situations and, in my opinion, kinda fun. At its very core, algebra is all about finding “x.” Or whatever value you have to find. You do this by getting x by itself on one side of the equation. Or, you figure out what you have to take out to get “x = ...” You do this by doing the opposite operation seen in the equation, and doing it to both sides. You do it to both sides to keep everything balanced. For example, 5x = 10. In this problem you should see that in order to get x alone, you have to “lose” the 5. You would do this by dividing each side by 5, resulting in x = 2. This is the same for multiplication, addiction, subtraction, exponents, etc. The best way to understand algebra, again in my opinion, is to do examples upon examples upon examples.

1) x + 7 = 20

So, you want to get x by itself, what do you do? You subtract 7 from each side and get x = 13

2) 9x + 50 = 95

Again, wanting to get x alone. First, you’d want to get anything that doesn’t have x attached to it on one side. So, you’d take away 50 from each side and get 9x = 45. Next, you’d simply divide each side by 9. So, x = 5.

So, lets kick it up a notch. Now we’re going to factor. Factoring is when you solve x for 0. Or, what x could equal in order for the expression to be zero. You’ll also learn that there are different strategies that one can use to factor an expression. Here’s the equation for what we’re looking for: (z - q)(z + v) = 0. At first this won’t mean anything, but it will turn into an equation. Here’s how you multiply this out, you multiply the first value in the first parentheses (z in the above example) by the first value in the next parentheses (k). Then you multiply the first value (z) again with the next value in the next parentheses (v). You continue like this until there’s no more values in the second parentheses. Then you start with the next value in the first parentheses (q) and repeat. So, in the end, the above equation would be z^2 + zv - zk - qv. Every expression has an “a,” “b,” and “c” term, sometimes even more then that. This just means where they fall in line in the expression. The “a” term is found by multiplying the first two values of the parentheses. So, going back to that example above, the “a” term would be z^2. The “b” term is found by adding or subtracting the two values that will have an x in it. That means that the “b” term for the above example would be (zv - zk). Finally, the “c” term is the result of the final two values multiplied together. -qv for the above example.

ex1) 2x^2 - 9x + 10

In this example the “a” term is 2, the “b” term is -9 and the “c” term is 10. So, 2 is an easy “a” term because the only possible things it can be is a 2 and a 1. So the factored expression will start off to look like (2x + Q)(x + W). Next, look at the “c” term and think of all the multiples of that number and figure out which ones would work, so that the “b” term works out as well. With 10, the choices are 10 and 1 or 5 and 2. Wrong. The order definitely matters here. So, the choices are actually 10 and 1, 1 and 10, 5 and 2 or 2 and 5. And don’t forget about negative/positive. 10 is positive while 9 is negative. This means that the two values that are going to multiplied together to get 10 have to be negative. Since, they’re going to be added together to get -9. Here are the four choices we have at the moment:

      (2x - 10)(x - 1),          (2x - 1)(x - 10),    (2x - 5)(x - 2),      (2x - 2)(x - 5).

Then, we just multiply them out to see what you get.

   2x^2 - 2x - 10x + 10    2x^2 - 20x - 2x + 10    2x^2 - 4x - 5x + 10     2x^2 - 10x - 2x +10

2x^2 -12x + 10 2x^2 -22x + 10     2x^2 - 9x +10       2x^2 - 12x + 10

So, it looks like the answer is (2x - 5)(x - 2). Eventually, after a lot of practice, you will be able to see what gets multiplied by what in a moments notice.

ex2) 25x^4 - 100 = 0

Right away you should notice something about all four numerals in the example. They’re all perfect squares. This makes your job a lot easier when you factor. You could just subtract 100 from each side, then divide by 25, etc., but there’s a simpler way. Since all the numbers are perfect squares, they can be achieved by multiplying their square root together twice. 5 is the square root of 25, 10 is the square root of 100 and x^2 is the square root of x^4. So, 5x^2 times 5x^2 would equal 25x^4. This is factoring (as said earlier). The breakdown of an expression so that its easier to find x=0.  Or what times what equals the expression. Since there isn’t a “b” term and 100 is negative, we would multiply 10 and -10 to get -100 and to cancel out a “b” term. Here’s what it looks like written out, solved:

(5x^2 - 10)(5x^2 + 10) = 0

This is called “difference of squares.”

It is acceptable to leave it like this if the instructions ask for you to just factor. However, if the instructions ask you to solve for x, then you have get x alone on one side of the equation. So, 5x^2 = 10. x^2 = 2. x = \2.

ex3) a^2 + 16a + 64 = 0

This is another special case. This is called “perfect squares.” It is called that because the square root of the “c” term is also the “b” term added together twice. So, it would look like this (a + 8)(a + 8). Very simple.

ex4) 25x^4y^2 - 50x^5y^3 + 125x^6y^4

Another fun thing you can do when you’re factoring expressions is you can take out like terms in order to make the expression more easily readable. To start with this example, lets forget about the x and y exponents. So, right not the expression is 

25xy - 50xy + 125xy. Each of these terms are divisible by the same number, 25. You can rewrite this expression like this: 25 (1xy - 2xy + 5xy). Now, lets add back the x and y exponents: 25 (1x^4y^2 - 2x^5y^3 + 5x^6y^4). Just like the before, the exponents have common factors, x^4 and y^2. In other words, each term has at least x^4 and y^2. Put this in front with the 25 and you get: 25x^4y^2 (1 - 2xy + 5x^2y^2). You can do this because of PEMDAS.

This is obviously the base level, algebra is the foundation to almost every level of math including Pre-Calculus, Calculus, and Statistics. Without a good understanding of algebra, you will be more likely to be lost in future levels. That being said, algebra can also be found in everyday life and, therefore, you can get practice all of the time. Here are a few examples:

1) You have $800 to spend on a TV, a remote and movies all together. The TV costs $380, the remote costs $20 and the movies cost $10 each. How many movies can you get?

You’d make x = number of movies

$800 = $380 + $20 + $10x

$800 = $400 + $10x

$400 = $10x

x = 40 movies

2) You’re driving from Washington to New York, which is 228 miles. Lets say gas is $3.50 per gallon. How much money do you need for gas?

Let x = number of gallons

228miles = $3.50x

x = ~$65.14

3) Four oranges and five apples cost $3.56. Three oranges and four apples cost $2.76. What’s the cost of an orange and the cost of an apple?

Let x = oranges and y = apples

4x + 5y = $3.56 and 3x + 4y = $2.76

What you want to do here is combine the two problems. You solve for x (or y) in one of the problems, and you then plug it in the other. This way you have like variables. Here’s what I mean:

4x + 5y = $3.56

4x = $3.56 - 5y

x = ($3.56 - 5y)/4

So, you have x, now plug it into the other expression:

3 (($3.56 -5y)/4) + 4y = $2.76

3($3.56 - 5y) +16y = $11.04       (multiply everything by 4 to get rid of the fraction)

$10.68 - 15y + 16y = $11.04

y = $0.36

Now, plug that y value into either expression for x:

4x + 5($0.36) = $3.56

4x + $1.80 = $3.56

4x = $1.76

x = $0.44

So, oranges cost 44 cents each, while apples cost 36 cents