Simplifying Complex Fractions, final step

In summary, the conversation involves a student seeking help with their homework, specifically with simplifying rational expressions. The student has tried multiple methods but has not been able to reach the given solution. They receive hints from another user who points out the need for factoring and provides step-by-step guidance for two specific problems. The conversation ends with the student understanding the importance of factoring and thanking the others for their help.
  • #1
DS2C

Homework Statement



Please see attachment.

Homework Equations



I don't know how to get the final product on the ones with the question marks (textbook answers written next to them). I've gotten to the last step (except for # 29 but don't mind that one, I haven't exhausted all ideas). I've tried a bunch of ideas but none give me the given solution. I've also looked back at the prior steps but found no mistakes.
Any hints?

The Attempt at a Solution


[/B]
Please see attachment.
 

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  • #2
Re uploaded for the proper orientation.
 

Attachments

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  • #3
If I can trust your steps, looking at #21, next to finish would be
(y-2+1)/(y^2-2y+1)

(y-1)/(y^2-2y+1)

(y-1)/((y-1)(y-1))

1/(y-1)
 
  • #4
symbolipoint said:
If I can trust your steps, looking at #21, next to finish would be
(y-2+1)/(y^2-2y+1)

(y-1)/(y^2-2y+1)

(y-1)/((y-1)(y-1))

1/(y-1)

On this step, did you just factor the expression? That may explain why I couldn't get it. We've been jumping around in the book and haven't even gone over factoring yet.
 
  • #5
#23 using just regular keyboard text,

(4y-8)/(4(6y-12))

(y-2)/(6y-12), and not simplifiable further from here
 
  • #6
symbolipoint said:
#23 using just regular keyboard text,

(4y-8)/(4(6y-12))

(y-2)/(6y-12), and not simplifiable further from here
It can be simplified further --
##\frac{4(y - 2)}{4 \cdot 6(y - 2)} = \frac 1 6##, provided that ##y \ne 2##.

@DS2C, you are misusing the ##\Rightarrow## symbol in the work you posted. The expressions you're working with are equal to one another, so the proper connect is =.
The "implication" symbol, ##\Rightarrow## should be used one one statement being true implies that the following statement is true.

For example, ##x = 2 \Rightarrow x^2 = 4##. Notice that x = 2 and ##x^2 = 4## are equations (a type of statement). A statement can be true or false, depending on the value of a variable. OTOH ##\frac {4y - 8}{8y - 16}## is an expression, which has a numeric value.
 
  • #7
DS2C said:
On this step, did you just factor the expression?
Yes, that's what he did, factoring ##y^2 - 2y + 1## to ##(y - 1)^2##.
DS2C said:
That may explain why I couldn't get it. We've been jumping around in the book and haven't even gone over factoring yet.
Factoring is at the heart of reducing rational expressions. Possibly your instructor is assuming that you've covered factoring in a previous course.
 
  • #8
For #29, ax + ab = a(x + b) and ##x^2 - b^2 = (x - b)(x + b)##. The original expression simplifies to ##\frac a {x + b}##, provided that ##x \ne b##. (If x = b, the original rational expression is undefined, but the final result doesn't have x - b in the denominator, so you need to stipulate that x can't be equal to b.
 
  • #9
Mark44 said:
It can be simplified further --
##\frac{4(y - 2)}{4 \cdot 6(y - 2)} = \frac 1 6##, provided that ##y \ne 2##.

@DS2C, you are misusing the ##\Rightarrow## symbol in the work you posted. The expressions you're working with are equal to one another, so the proper connect is =.
The "implication" symbol, ##\Rightarrow## should be used one one statement being true implies that the following statement is true.

For example, ##x = 2 \Rightarrow x^2 = 4##. Notice that x = 2 and ##x^2 = 4## are equations (a type of statement). A statement can be true or false, depending on the value of a variable. OTOH ##\frac {4y - 8}{8y - 16}## is an expression, which has a numeric value.
NOW I SEE. Denominator could be factored and then the next simplification.
 
  • #10
Thanks so much guys makes much more sense now that I've reviewed factoring.
Also @Mark44, thanks for pointing that misuse out. Honestly I didnt even know it was a real symbol. I made it up to show myself when a new step began rather than having a =. Ill just stick with the = symbol or something rather.
 

Related to Simplifying Complex Fractions, final step

1. What is the final step in simplifying complex fractions?

The final step in simplifying complex fractions is to find the greatest common factor (GCF) of the numerator and denominator and divide both by it.

2. Can complex fractions be simplified without a calculator?

Yes, complex fractions can be simplified without a calculator by using the method of finding the GCF and dividing both the numerator and denominator by it.

3. Is it necessary to simplify complex fractions?

Simplifying complex fractions is not always necessary, but it can make computations easier and help in finding the simplest form of a fraction.

4. What is the difference between simplifying and reducing complex fractions?

Simplifying complex fractions involves finding the GCF and dividing both the numerator and denominator by it, while reducing complex fractions involves converting them into equivalent fractions with smaller numbers in the numerator and denominator.

5. What should be done if the numerator and denominator of a complex fraction have no common factors?

If the numerator and denominator of a complex fraction have no common factors, then the fraction is already in its simplest form and cannot be simplified further.

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