Rearrange an Algebraic Equation

I would like to rearrange the equation:
X = (Y-A)/(Y-B) to look like this
B-A = ((X-1)*(Y-A))/X

I'm able to isolate B-A on the left hand side of the equation, but I
can't seem to get the right hand side to the form above.

I've tried the following steps:
1. XY-XB = Y-A
2. XY-XB-Y = -A
3. XY-Y = XB-A
4. XY-Y = X(B-(A/X))
5. (XY-Y)/X = B-(A/X)
6. B = (XY-Y)/X + (A/X) = (XY-Y+A)/X
7. B-A = ((XY-Y+A)/X)-A

and this is where I get stuck. I feel like I'm close but I can't
quite get it in the form shown above. What am I missing?
Thanks.

If the equation you're trying to prove is indeed true, given the original equation, I would do the following: solve for B and call that Equation 1; then solve for A and call that Equation 2; subtract the latter from the former, and simplify.

[EDIT]: This approach doesn't work. See next post.

Ok, the equation works. I went backwards from the equation you're trying to prove to the equation you started with. What I would do is look at the process used to do that reverse proof, and then reverse it again. All the steps are reversible. In other words, show that by assuming

$\displaystyle B-A=\frac{(X-1)(Y-A)}{X},$

you can derive

$\displaystyle X=\frac{Y-A}{Y-B}.$

Then simply reverse your proof to get your desired result.

Quote:

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Originally Posted by triviaguy

I would like to rearrange the equation:
X = (Y-A)/(Y-B) to look like this
B-A = ((X-1)*(Y-A))/X

I'm able to isolate B-A on the left hand side of the equation, but I
can't seem to get the right hand side to the form above.

I've tried the following steps:
1. XY-XB = Y-A
2. XY-XB-Y = -A
3. XY-Y = XB-A
4. XY-Y = X(B-(A/X))
5. (XY-Y)/X = B-(A/X)
6. B = (XY-Y)/X + (A/X) = (XY-Y+A)/X
7. B-A = ((XY-Y+A)/X)-A

and this is where I get stuck. I feel like I'm close but I can't
quite get it in the form shown above. What am I missing?
Thanks.

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All your calculations are correct. You only stopped 2 steps in front of the finish. I'll take your last line:

$\displaystyle B-A = \left(\dfrac{XY-Y+A}{X}\right) - A$

$\displaystyle B-A = \dfrac{XY-Y+A-AX}{X}$

$\displaystyle B-A = \dfrac{Y(X-1)-A(X-1)}{X}$

$\displaystyle B-A = \dfrac{(Y-A)(X-1)}{X}$

Thanks for the input Ackbeet, I believe I've come up with the correct procedure to rearrange the original equation as follows:

X = (Y-A)/(Y-B)

1. XY-XB = Y-A
2. XY-Y = XB-A
3. Y(X-1) = XB-A
4. Y(X-1)-XA = XB-A-XA
5. Y(X-1)-XA+A = XB-XA
6. (Y(X-1)-XA+A)/X = B-A
7. (Y(X-1)-A(X-1))/X = B-A
8. ((X-1)*(Y-A))/X = B-A

Thanks earboth, I found the solution as well. It's funny how you sometimes can't see something when it's right in front of you. Thanks again to all who responded.