Algebra II - Complex Fractions URGENT

[Destroyer2000]

Alright, I'm having trouble understanding even the basic concept of this...I don't know how to do it. Here's one of the trouble problems:

a-b
-------
a^-1 - b^-1

I don't know how to do this. And these I've never even seen (I was given the worksheet and told I would have a test tomorrow...stupid makeup work.)

x___1__2
-_+_-_=_-
9___6__3
(Disregard the '_'s)

'Solve each open sentence'?

EDIT: Since I can't seem to get the second problem to appear correctly, it is x over 9 plus 1 over 6 equals 2/3s. How do I do this?

Help, please.

[Dante]

Alright, I'm having trouble understanding even the basic concept of this...I don't know how to do it. Here's one of the trouble problems:

a-b
-------
a^-1 - b^-1

I don't know how to do this. And these I've never even seen (I was given the worksheet and told I would have a test tomorrow...stupid makeup work.)

x___1__2
-_+_-_=_-
9___6__3
(Disregard the '_'s)

'Solve each open sentence'?

I have no idea what these mean, what in the world is with the _ s if they are to be disregaured? Please rewrite this in standard format i.e. ((x-1+y)^z)/(g*x+4*z)=5

EDIT: Since I can't seem to get the second problem to appear correctly, it is x over 9 plus 1 over 6 equals 2/3s. How do I do this?

Help, please.
__________________

Ok, I think this bottem one states (x/9)+(1/6)=(2/3)

The first step is to isolate your x (That way you can figure out what it "equals", to do this multiply both sides by 9... simple enough,

x*(9/9)+(9/6)=((9*2)/3)

simplifying this, the 9/9 is 1 so it just vanishes, 9/6=3/2 when you divide the top and bottem by three, and (9*2)/3=18/3= 6/1=6 when you divide the top and bottem by 3.

x+3/2=6

Now to solve for x, just isolate it on the left side by subtracting 3/2s from both sides,

x=6-3/2

to simplify this, change the common denominator to two. 6=12/2

x=(12/2)-(3/2)=(12-3)/2=9/2

x=9/2

This is the answer if I gather the question correctly.

Pascal

EDIT: By the way this isn't complex, to be complex it requires the use of imaginary numbers a+b*i, where i is the square root of negative one.

[Destroyer2000]

I have no idea what these mean, what in the world is with the _ s if they are to be disregaured? Please rewrite this in standard format i.e. ((x-1+y)^z)/(g*x+4*z)=5

EDIT: Since I can't seem to get the second problem to appear correctly, it is x over 9 plus 1 over 6 equals 2/3s. How do I do this?

Help, please.
__________________

Ok, I think this bottem one states (x/9)+(1/6)=(2/3)

The first step is to isolate your x (That way you can figure out what it "equals", to do this multiply both sides by 9... simple enough,

x*(9/9)+(9/6)=((9*2)/3)

simplifying this, the 9/9 is 1 so it just vanishes, 9/6=3/2 when you divide the top and bottem by three, and (9*2)/3=18/3= 6/1=6 when you divide the top and bottem by 3.

x+3/2=6

Now to solve for x, just isolate it on the left side by subtracting 3/2s from both sides,

x=6-3/2

to simplify this, change the common denominator to two. 6=12/2

x=(12/2)-(3/2)=(12-3)/2=9/2

x=9/2

This is the answer if I gather the question correctly.

Pascal

EDIT: By the way this isn't complex, to be complex it requires the use of imaginary numbers a+b*i, where i is the square root of negative one.

THanks. The '_'s were there as spacers; they weren't in the original problem.

[everdred12a]

I know this problem has pretty much been solved, but I was just going to say that a little work could have been saved if you isolated the x first, like this:

(x/9)+(1/6)=(2/3)
Basically what I'm saying is do this first:
(x/9)=(2/3)-(1/6) = (x/9)=(1/2)
Then, follow as Pascal said, multiplying both sides by 9.

The only reason I suggested this is because it eliminates the problem of having to multiply three different terms by 9, and instead only multiplies two terms. I personally like to make my math as simple as possible (because I'm not so good at it... xD)