Un-simplifying a fraction..

[MarkSA]

This is rather embarrassing - I feel as if I should know this. But I seem to have forgotten how to do it.

I have the given fraction:
(y-2)/(y+3)

How do I "unsimplify" it, for lack of a better term?
I'm trying to convert it back to this: 1 - 5/(y+3), which is equivalent to the above fraction.

 

[A_G]

You can write y - 2 as (y + 3) - 5

So you have ((y + 3) - 5)/(y + 3) [caution: not using the parenthesis could mean totally different!]

Now compare this with (a - b)/a = a/a - b/a = 1 - b/a

In your case a = y + 3 and b = 5

I am sure you can finish it on from here.

 

[mmm4444bot]

I like A_G's steps, too, but my mind took a weird route.

(I'm the whack job around here.)

It's easy, of course, to initially decompose to the following because this is just a reverse of the last step to combining the following.

y/(y + 3) - 2/(y + 3)

Now, since I know in advance that we want 1 - 5/(y + 3), I recognize that I can get a 1 by adding 3 to the numerator of y/(y + 3).

I have to do this by adding 3/(y + 3) to that term -- we need a common denominator, right?

But if I add something, I have to take it away, too. (Just like what often happens during the steps of completing the square.) Thus, the following.

3/(y + 3) + y/(y + 3) - 2/(y + 3) - 3/(y + 3)

(y + 3)/(y + 3) - (2 + 3)/(y + 3)

1 - 5/(y + 3)

Okay, Denis. I'm watching for the inevitable.

 

[Denis]

Since z = 26, then (y-2)/(y+3) = (25-2)/(25+3) = 23/28 = 1 - 5/28

 

[soroban]

Hello, MarkSA!

\(\displaystyle \text{Given: }\:\frac{y-2}{y+3}\)

\(\displaystyle \text{How do I "unsimplify" it to: }\:1 - \frac{5}{y+3}\)

You can always try Long Division . . .

. . \(\displaystyle \begin{array}{cccccc}& & & & 1 \\ & & -- & -- & -- \\ y+3 & ) & y & - & 2 \\ & & y & + & 3 \\ & & -- & -- & -- \\ & & & - & 5 \end{array}\)


\(\displaystyle \text{Therefore: }\;\frac{y-2}{y+3} \;=\;1 - \frac{5}{y+3}\)

 

[MarkSA]

Thanks everyone