Adding and Subtracting Rational Expressions

 Today in class, we focused on adding and subtracting rational expressions. Here are a few rules that apply to solving these expressions:
-1) you must find a common denominator before adding or subtracting
-2) when trying to find a common denominator, you must do the same to the top as you did to the bottom.  
Applying these rules: here are some examples :
1/2 + 1/5: Common denominator is lowest number that terms in the denominator both go in to (in this case, the common denominator is 10). If the common denominator is 10, then in the first term, the 2 in the denominator of 1/2 must be multiplied by 5 to reach ten. Then, rule number 2 must be applied, and the top must also be multiplied by 5, making the fraction equal 5/10. With the second term, the 5 in the denominator must be multiplied by 2 to reach ten, meaning that the top must also be multiplied by 2, making the new fraction 2/10. Now, we are able to do the equation 5/10 + 2/10= 7/10

Sometimes, a common denominator cannot be found, so the terms in the denominators must be multiplied together to form the common denominator.

1/x - 1/x+4 = x+4-x/x(x+4) = 4/x(x+4)

Here, x and (x+4) did not have a common denominator, so they were multiplied together to form one.


There are also cases where the denominator of one fraction must be multiplied by a number to make it equal the other denominator.

x/x-1 + 1/1-x = x/x-1 + 1/-1(-1+x) = x/x-1 + -1/x-1 = x-1/x-1 = 1

Here, the denominator of the second fraction was multiplied by -1 so that it equaled the denominator of the first fraction.

Many examples and practice problems can be found on page 227

Here are some examples from page 227:

13. 1/2xy^4 + 1/x^3y^2 = x^2+2y^2/2x^3y^4

19. 1/4x^2 - 1/xy + 1/y^2 = y^2 - 4xy + 4x^2 / 4x^2y^2

Here is a video link that may help you better understand this: http://www.youtube.com/watch?v=FZdt73khrxA

Simplifying Rational Expressions

Hello class,

-For the past two days we have been learning and gathering information on Simplifying Rational Expressions. Simplifying Rational Expressions is basically finding the easiest form of the equation by simplifying it. And Rational expressions are usual in their simplest form, if it's numerator and denominator have no common polynomial factors. We've also been going over how to find the domain and zeros of certain rational functions.



First of all:

What is a rational #?
A rational number would be numbers such as .4=4/10=2/5. It can be written as fractions w/ integers in the numerator & denominator.

It would not be a number for example like: √5/2


Rational Expressions: ---> polynomial/polynomial or polynomial over polynomial


Ex:

x^3-4x/x^3-4x^2+4x = x(x^2-4)/x(x^2-4x+4) = x(x+2)(x-2)/x(x-2)(x-2) = x+2/x-2

Basically what you do is you simplify to the lowest possible terms possible. Then once you are done simplifying you cancel out the same values (that's what the crossed out black values stand for). And then you end with your final answer which will be in the simplest form.



Domain: Is the set of all x values that work in the expression.

Ex:

x^3-4x/x^3-4x^2+4x = x(x^2-4)/x(x^2-4x+4) = x(x+2)(x-2)/x(x-2)(x-2) = x+2/x-2

Domain: All Real #'s - {2}

Which basically means All numbers work in the place of x to make the equation set equal to 0 except for 2.




Zeros: All x value that makes the whole expression 0.

Equation:

x^3-4x/x^3-4x^2+4x = x(x^2-4)/x(x^2-4x+4) = x(x+2)(x-2)/x(x-2)(x-2) = x+2/x-2

x=-2--> zero  -2+2/-2-2 = 0/-4 = 0

Basically what you're trying to find is the number you can substitute the x with to make the whole fraction equivalent to 0.



Now here are some example problems:

Equation: (x-1)(x+1)^2/(x+1)(x^2-1)


Solving:

(x-1)(x+1)(x+1)/(x+1)(x-1)(x+1) = 1

Domain: All Real #'s

So basically what happened within this equation when you simplified was that you had all of the same common value as denominators and numerators. So therefore in result they canceled each other out and you end up with a result of 1.



Equation: 3x^2/6x^2-9x


Solving:

x(3x)/3x(2x-3) = x(3x)/3x(2x-3) = x/2x-3

What happened with this problem was it was simplified to simplest form then once there we canceled out the 3x from both of them leaving us with an answer of x/2x-3.

Tips:

• Be Careful with your simplifying (formulas on page 172 and 173)
• Make sure to cancel out the same integers
• Don't always suspect certain equations are simplified completely if it seems a bit harder then most problems

-http://www.youtube.com/watch?v=B4bVlDgHF5I

-http://www.youtube.com/watch?v=7Uos1ED3KHI

-If you didn't understand this concept before I hope this helped with the gist of it! For more explanation on this go to: Pages 218-219 in your book.

Dividing Monomials and Negative Exponents

Dividing Monomials:                        More examples on page 205 of our textbook

Ex. 15x3/5x = 3x2

To divide these two monomials, first divide the coefficients just like usual. 15/5 = 3. For the exponents with the same base, you subtract the exponents when dividing so 3 (from x3) minus 1 (from x1) equals 2 so your variable is x2. This gives you an answer of 3x2.

Ex. 24a3c6/16a4c3 = 3c3/2a

First divide the coefficients, they both divide by 8 and so we are left with 3 on the top and 2 on the bottom. There is an a3 on the top which cancels out with the a4 on the bottom but we are still left with one a on the bottom. There is a c6 on the top which cancels out the c3 on the bottom and we are left with c3 on the top. This gives us our final answer of 3c3/2a.

Video:
http://www.youtube.com/watch?v=Mn4WuvIGUgI

Negative Exponents: More examples on page 210 of our textbook
x2/x5 = x2-5 =x-3

If we follow our exponent rules and subtract the exponents when we divide then the answer we get is x-3

x2/x5 = (x)(x)/(x)(x)(x)(x)(x) = 1/x3

This is the same problem but solved slightly differently and with a different answer. From this we can assume that 

x-3 is equal to 1/x3

From this we can see that if you bring a negative exponent from the denominator to the numerator, and vice versa, that it then becomes a positive exponent.

Ex.

3a2x-2/axy-1 = 3a2y/ax3 = 3ay/x3

By bringing the bases with negative exponents from the numerator to the denominator, or from the denominator to the numerator we are left with all positive exponents. We moved the x-2 from the numerator to the denominator so it became 

x2

and we brought the y-1 from the denominator to the numerator making it just y. From here all we needed to do was cancel the a's and we got 3ay/x3 as our answer.

Video:

http://www.youtube.com/watch?v=c4aiYf3fzVQ

Tips:

• When Dividing subtract exponents
• When Multiplying add exponents
• When raising a monomial to a power multiply exponents
• When raising monomial don't forget to raise the coefficient

Test Review

We have been covering different kinds of factoring, finding the LCM, finding the GCF, word problems and setting equal to zero in class, here is a quick review for our test hope it helps. 

Factoring by grouping:

ax - bx + ay - by    =      x(a-b) + y(a-b)   =   (x+y)(a-b)

We use this when there is four terms. You group with two or sometimes even three terms and try to get a common factor which in this problem was (a-b). 

Difference of Two Squares:

x2 - 4 = (a+2)(a-2)

The formula for this is a2-b2 = (a+b)(a-b).

Trinomial Squares:

x2 + 2xy + y2 = (x+y)2

The formula for this is a2 + 2ab + b2 = (a+b)2 or a2 - 2ab + b2 = (a-b)2 depending if the middle term is positive or negative.

Sum and Difference of Cubes

z3 + y6 =  (z+y2)(z2- zy+b2) Sum of Cubes

z3 - y6 =  (z-y2)(z2+zy+b2) Difference of Cubes

The formula for the sum of cubes is a3+b3 = (a+b)(a2-ab+b2)

While the formula for the difference of cubes is a3-b3 = (a-b)(a2+ab+b2)

Tips for factoring:

• Always look for GCF (greatest common factor before starting)
• When you have finished make sure your answer is completely factored

Finding the LCM and GCF

To find the GCF make a factor tree for each of the numbers and then create a ven diagram. Put all the factors they all have in common in the middle. Multiply all the numbers in the middle and this is your GCF coefficient. Next look at the variables and take out the variables they all have in common. In this case they all only have a Y in common. To find the LCM you must multiply all of the numbers in the ven diagram, but make sure not to multiply the numbers in the middle more than once. This number is your coefficient. Now look at the variables and take the all the variables from all of the original numbers and if some of the numbers have the same variable, pick the one with the highest exponent.




Word problems/Setting Equal to zero
Find two numbers differing by 3 and whose product is 88

x2+3x=88

x2+3x-88=0

(x+11)(x-8)=0

x+11=0     x-8=0

x=-11        x=8

x=-11, 8

First factor the equation and then set both factored sets equal to zero and then solve.

Zero-Product Property and Consecutive Numbers

Hi Class!

-On wednesday we went over polynomial equations and the zero-product property. A polynomial is an equation that is equivalent to an equation with a polynomial as one side of the = sign, and 0 on the other side. Examples of this type of equation are x^2 + 4x + 4 = 0, and x^4 + x= = 0. To solve these types of equations, you must factor the polynomial into linear factors. If you cannot factor the polynomial, then it is prime. 


-Explanation of the zero-product property in book: p183-185


-Visual example: △ x □  = 0

          ~Either △ = 0, □ = 0, or □ and △ = 0
          ~This can also be shown as: ab = 0
                                   where a = 0 and or b = 0
                 
           *There can be more than one value for X

-The steps to solving polynomial equations with the zero-product property are:
          
          1. Write the equation with one side as zero (set the equation equal to zero)
          2. Factor the non-zero side of the equation
          3. Set both of the factors equal to zero and solve

-Example problem:

          1. One side as zero:  x^2 - 49 = 0
          2. Factor:  (x+2) (x-2) = 0
          3. Set factors equal to zero:  x+2 = 0, x-2 = 0
                                                       x = 2, -2 ---> {2, -2}

-Alternate ways to solve: In class we also learned another way to solve certain problems:

          ~9x^2 - 81 = 0: Instead of factoring, you could also solve for x normally.
            9x^2 = 81
            x^2 = 9
            x = 3, -3

                                *Do not forget that the square root can be positive AND negative

          ~Here is the same problem (^), but using the zero-product property:
            9x^2 - 81 = 0
            (3x-9) (3x+9) = 0
            3x-9 = 0, 3x+9 = 0
            3x = 9, 3x = -9
            x = 3, x= -3
            {3, -3}


-Another example of zero-product property:

          ~x^2 + 10x + 25 = 0
            (x-5) (x-5) = 0 
            x = 5  
            ^When the factor occurs twice (same factor more than once) it's called a 'Double Root'

- Consecutive Numbers: In class we also learned how to express consecutive integers, or consecutive even integers and consecutive odd integers.

          ~For example: 2 consecutive odd integers numbers that multiply to 99
                                    x (x+2) = 99
                                           *x represents the first integer and x+2 represents the second integer
                                   x^2 + 2x = 99
                                   x^2 + 2x - 99 = 0
                                  (x+11) (x-9) = 0
                                   x = -11, x = 9 {-11, 9)
                                            *-11 and 9 are the two possibilities for x NOT the answers to the problem.
                                             To find the answer  ---> plug -11 and 9 into x+2 (add two)

         ~Another example: 2 consecutive even integers  ---> the difference of who's squares is 68
                                         x^2 - (x+2)^2 = 68
                                         (x+2)^2 -x^2 - 68 = 0
                                         x^2 + 4x + 4 - x^2  - 68 = 0
                                         4x - 64 = 0
                                         4x = 64
                                          x = 16     *Now plug into original equation
                                                           (16 + 2)^2 - 16^2 = 68
                                                            {16,18} {-16,-18} 



- http://www.youtube.com/watch?v=1Iay8rFBQ6o


                                                           Hope this helped!
                                           Problems are on pages 185-191 in the book!