Simplifying Complex Fractions, final step

[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

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Please see attachment.

 

[DS2C]

Re uploaded for the proper orientation.

 

[symbolipoint]

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)

 

[DS2C]

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.

 

[symbolipoint]

#23 using just regular keyboard text,

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

(y-2)/(6y-12), and not simplifiable further from here

 

[Mark44]

#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.

 

[Mark44]

On this step, did you just factor the expression?

Yes, that's what he did, factoring $y^2 - 2y + 1$ to $(y - 1)^2$.

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.

 

[Mark44]

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.

 

[symbolipoint]

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.

 

[DS2C]

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.