Ask Me Help Desk - Maths - making "y" the subject *cries*

Hi all, I'm new to this forum, but it seems cool! But the proof is in the pudding: can you answer this question?
It came up in a specimen exam paper that I was revising with and, try as I did, I couldn't solve it. HELP! :confused:
I need make "y" the subject.. :

x = root (( y - a) / ( y - b))

How do I do it? I honestly can't do it!
Thanks in advance for your help. :D

Hello there:

$x=\sqrt{\frac{y-a}{y-b}}$

$x^{2}=\frac{y-a}{y-b}$

Rewrite the right side as: $\frac{b-a}{y-b}+1$

$x^{2}-1=\frac{b-a}{y-b}$

$\frac{1}{x^{2}-1}=\frac{y-b}{b-a}$

$\frac{b-a}{x^{2}-1}=y-b$

$y=\frac{b-a}{x^{2}-1}+b=\frac{bx^{2}-a}{x^{2}-1}$

Does that help? Maybe seeing a problem like this worked out will help you with your algebra skills.

Quote:

Originally Posted by galactus

Hello there:

$x=\sqrt{\frac{y-a}{y-b}}$

$x^{2}=\frac{y-a}{y-b}$

Rewrite the right side as: $\frac{b-a}{y-b}+1$

$x^{2}-1=\frac{b-a}{y-b}$

$\frac{1}{x^{2}-1}=\frac{y-b}{b-a}$

$\frac{b-a}{x^{2}-1}=y-b$

$y=\frac{b-a}{x^{2}-1}+b=\frac{bx^{2}-a}{x^{2}-1}$

Does that help?. Maybe seeing a problem like this worked out will help you with your algebra skills.

Hmm thanks but I (feel really stupid - I can't even do a GCSE question... ) don't understand how you get from step two to the rewriting:

$x^{2}=\frac{y-a}{y-b}$ => $\frac{b-a}{y-b}+1$

?? And I also don't get the very last = :

$y=\frac{b-a}{x^{2}-1}+b=\frac{bx^{2}-a}{x^{2}-1}$

??

Thanks for the answer although I don't quite see how you get to those above. If it was really dumbed down then it would probably help. If possible!
Thanks.
p.s. Apparently my algebraic skills are in need of improvement :( :)

That's just long division.