Substraction: (y-2/y+3)-(y/2y+6)

[dkarolasz]

(y-2/y+3)-(y/2y+6)

My next step is

[y-2/y+3]-[y/2(y+6)]

then;

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

next:

2y-4-y/2(y+3)

=

y-4/2(y+3)

is that right...

 

[skeeter]

one of these days you'll learn the correct form for posting algebraic fractions ...

(y-2)/(y+3) - y/(2y+6) =

2(y-2)/[2(y+3)] - y/[2(y+3)] =

[2(y-2) - y]/[2(y+3)] =

(2y - 4 - y)/[2(y+3)] =

(y-4)/[2(y+3)]

 

[dkarolasz]

so I was right, thanks, I tried reading karls's notes, and fingering it out, but I can't

 

[tkhunny]

I tried reading karls's notes, and fingering it out, but I can't

What does that even mean? Try again until you figure it out!

 

[pka]

I tried reading karls's notes, and fingering it out, but I can't

What does that even mean? Try again until you figure it out!

I fear that it means that if one does not understands the very basic operations, then that means one has no idea how to use Karl’s notes.

 

[Mrspi]

(y-2/y+3)-(y/2y+6)

My next step is

[y-2/y+3]-[y/2(y+6)]

then;

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

next:

2y-4-y/2(y+3)

=

y-4/2(y+3)

is that right...

"Karl's Notes" aside, I think you mean this:

 y - 2        y
------   -  ------
 y + 3     2y + 6

To add or subtract two fractions, they must have the same denominator.  The best way to find the Least Common Denominator is to [b]factor[/b] the denominator of each fraction:

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

Do you see that the least common multiple (LCD) of the two fractions is 2(y + 3)?

Multiply both numerator and denominator of the first fraction by 2.  The second fraction already has 2(y + 3) as it's denominator:

  (y - 2)*2              y
 -------------  -  -----------
  (y + 3)*2       2(y + 3)

Now you've got two fractions with the same denominator.  Subtract the numerators, and put the result over the common denominator:

   2y - 4 - y
-----------------
     2(y + 3)

Simplify the numerator. See if you can reduce the resulting fraction.