Unsure how to work algebra problem

In summary, the equation can be solved by factoring out the numerator and denominator and then solving the equation. This process is quite simple to follow, and should be in precalc or interm. algebra.
  • #1
snoggerT
186
0
(s^2+4s+4)/(s^2+4s+5) = 1 - (1/(s^2+4s+5)



The Attempt at a Solution



I'm actually pretty far up into my upper level math classes (past calc 3), but I can't seem to figure out this one. How do you go from the left side to the right side of the equation?
 
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  • #2
Are you trying to solve the equation? If you are, you do not "go from the left side to the right side". You should eliminate the rational expressions by multiplying both sides of the the equation by s^2 + 4s + 5.
 
  • #3
I'm not trying to solve the equation. It's for a circuits problem and in order to solve the problem, you have to set things up so you can do fraction decomposition. The right side is just what the solutions manual went to from what is on the left side. I just can't figure out how they got to the right side from the left side. Here is the original equation:

F(s)= (s+2)^2/(s^2+4s+5).

they factored out the numerator and then went to what I have on the right side. I just can't figure out how they got to it.
 
  • #4
This is really simple actually, and this idea will give you the boost you need:
(note that the way I am expressing this is not official, I just want to give you the idea)

[tex]\frac{a+m}{a+m+n}[/tex]

[tex]= \frac{(a+m+n)-n}{a+m+n}[/tex]

[tex]= 1-\frac{n}{a+m+n}[/tex]

Notice how in the 2nd part there is an equal part of the numerator and the denominator, dealing with those is expressed as follows:

[tex]\frac{ax+by}{x} = \frac{ax}{x}+\frac{by}{x} = a+\frac{by}{x}[/tex]
 
  • #5
Are the numerator or denominator factorable? (I should try finding myself, but maybe your own effort might be just as helpful?).
...
I thought about it a bit more: the numerator is factorable, but the denominator seems NOT factorable. (x __ 1) and (x __ 5), not useful.
Are you allowed to use complex numbers?
 
  • #6
Mentallic, I saw your post late. Your method seems unfamiliar based on "intermediate algebra" content; I'm not sure if I ever saw that type of process in "PreCalculus" either.
 
  • #7
symbolipoint said:
Mentallic, I saw your post late. Your method seems unfamiliar based on "intermediate algebra" content; I'm not sure if I ever saw that type of process in "PreCalculus" either.

Well, that is a really common trick, especially in integration. But, since the OP is higher in math classes, that should fit him/her perfectly fine.
 
  • #8
symbolipoint said:
Are the numerator or denominator factorable? (I should try finding myself, but maybe your own effort might be just as helpful?).
...
I thought about it a bit more: the numerator is factorable, but the denominator seems NOT factorable. (x __ 1) and (x __ 5), not useful.
Are you allowed to use complex numbers?

Yes, the denominator eventually is factored out with complex numbers (this is for a circuits class), but they get to what I have on the right side before doing that step.

Mentallic, I'm not quite sure I follow what you posted for this particular problem.

edit: I think it's throwing me off because the problem I'm dealing with has 3 terms on the top and bottom.
 
  • #9
I've never been 'taught' this process. I just happened to stumble upon it when trying to find the asymptotes for functions with the variable in the denominator.
This process is quite simple to follow, I don't see why it shouldn't be in precalc or interm. algebra :smile:

Anyway, that's how the textbook would have jumped from the LHS to the RHS.
 
  • #10
snoggerT said:
Mentallic, I'm not quite sure I follow what you posted for this particular problem.

What mentallic, is saying is this
[tex]\frac{(s^2+4s+4)}{s^2+4s+5}=\frac{(s^2+4s+4+1)-1}{s^2+4s+5}=\frac{s^2+4s+5}{s^2+4s+5}-\frac{1}{s^2+4s+5}=1-\frac{1}{s^2+4s+5}[/tex]
 
  • #11
snoggerT said:
Yes, the denominator eventually is factored out with complex numbers (this is for a circuits class), but they get to what I have on the right side before doing that step.

Mentallic, I'm not quite sure I follow what you posted for this particular problem.

Hmm so you both aren't following what I posted?

SnoggerT you were trying to prove that the LHS is equal to the RHS correct? But if you mean calculus in university, my help may be a little too primitive since I'm still in high school.
 
  • #12
sutupidmath said:
What mentallic, is saying is this
[tex]\frac{(s^2+4s+4)}{s^2+4s+5}=\frac{(s^2+4s+4+1)-1}{s^2+4s+5}=\frac{s^2+4s+5}{s^2+4s+5}-\frac{1}{s^2+4s+5}=1-\frac{1}{s^2+4s+5}[/tex]

- Okay. I see now. I guess with it showing the n in the denominator and then moving it up to the numerator threw me off. Thanks.
 
  • #13
There is nothing wrong with Mentallic's method. It is short,nice, and easy!
 
  • #14
Mentallic said:
Hmm so you both aren't following what I posted?

SnoggerT you were trying to prove that the LHS is equal to the RHS correct? But if you mean calculus in university, my help may be a little too primitive since I'm still in high school.

- I guess proving it would be right. I just wasn't sure how they got to the right hand side (proving one side to the other wasn't part of the problem). I don't remember things like this because I hadn't seen algebra in 10+ years when I started back in school for another degree.
 
  • #15
sutupidmath said:
What mentallic, is saying is this
[tex]\frac{(s^2+4s+4)}{s^2+4s+5}=\frac{(s^2+4s+4+1)-1}{s^2+4s+5}=\frac{s^2+4s+5}{s^2+4s+5}-\frac{1}{s^2+4s+5}=1-\frac{1}{s^2+4s+5}[/tex]

Yes exactly, but I try to avoid just giving the solution away so that I can avoid the authorities and their harsh punishments :cry:
 
  • #16
Mentallic said:
Yes exactly, but I try to avoid just giving the solution away so that I can avoid the authorities and their harsh punishments :cry:

haha...I didn't really need the solution, I just needed to know what the 'n' really was in what you showed me. I was thinking way too much about it and overlooked that all I was doing was making the top and bottom equal for one part (to get the 1).
 
  • #17
snoggerT said:
haha...I didn't really need the solution, I just needed to know what the 'n' really was in what you showed me. I was thinking way too much about it and overlooked that all I was doing was making the top and bottom equal for one part (to get the 1).

I'm always willing to try explain something in more than 1 way :smile: Good luck with the rest of your problem!
 

1. How do I start solving an algebra problem?

The first step is to read the problem carefully and identify what the unknown variable is. Then, use the given information to set up an algebraic equation.

2. I am confused about the order of operations in algebra. Can you explain it?

The order of operations in algebra is PEMDAS, which stands for Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right). This means that any operations within parentheses should be done first, followed by exponents, then multiplication and division, and finally addition and subtraction.

3. I am stuck on a step while solving an algebra problem. What should I do?

If you are stuck on a step, try working backwards from the answer and see if you can identify where you went wrong. If you are still unsure, you can ask a teacher or tutor for help.

4. How can I check if my answer to an algebra problem is correct?

You can check your answer by plugging it back into the original equation and seeing if it satisfies the equation. Another method is to use a calculator to solve the equation and compare your answer to the calculator's answer.

5. I am struggling with understanding algebra concepts. What resources can I use to improve?

There are many resources available for learning and improving in algebra, such as textbooks, online tutorials, and educational websites. You can also seek help from a teacher, tutor, or study group for additional support.

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