1. ## Solving for x

Can someone help me solve for x? I tried many different ways, but I never got the answer so can someone show me the steps so I can understand what I did wrong?

1.
$y = 1 - \frac {1}{1-x}$
The answer should be $x = \frac {y}{y-1}$, but I don't know how to get that answer...

2.
$y = 1 + \frac {1}{1+ \frac {1}{1-x}}$
The same for this one, i don't know how to get the answer $x= \frac {2y -3}{y-2}$

2. I have

$y = 1 - \frac {1}{1-x}$

$\frac {1}{1-x}=1-y$

$1=(1-y)(1-x)$

$\frac {1}{1-y}=1-x$

$\frac {1}{1-y}-1=-x$

$-\frac {1}{1-y}+1=x$

$1-\frac {1}{1-y}=x$

$x=1-\frac {1}{1-y}$

$x=\frac{1-y}{1-y}-\frac {1}{1-y}$

$x=\frac {1-y-1}{1-y}$

$x=\frac {-y}{1-y}$

$x=\frac {y}{y-1}$

3. 2. $
y = 1 + \frac {1}{1+ \frac {1}{1-x}}
$

For the second one, is the next step:
$y-1 = \frac {1}{1+ \frac {1}{1-x}}$

then:
$(y-1)(1+ \frac {1}{1-x}) = \frac {1}{1+ \frac {1}{1-x}}(1+ \frac {1}{1-x})$
$(y-1)(1+ \frac {1}{1-x}) = 1$

I don't know if I am right, but this is where I'm stuck at...
can anyone help me with the next steps to get the answer $x= \frac {2y -3}{y-2}$?

4. Originally Posted by ninjuhtime
2. $
y = 1 + \frac {1}{1+ \frac {1}{1-x}}
$

For the second one, is the next step:
$y-1 = \frac {1}{1+ \frac {1}{1-x}}$
Yes!

then:
$(y-1)(1+ \frac {1}{1-x}) = \frac {1}{1+ \frac {1}{1-x}}(1+ \frac {1}{1-x})$
$(y-1)(1+ \frac {1}{1-x}) = 1$
You can do that but I think it is simpler first to clear the fraction on the right side by multiplying both numerator and denominator by 1-x:
$y-1= \frac {1(1-x)}{(1+ \frac {1}{1-x})(1-x)}= \frac{1-x}{(1-x)+ 1}$ $= \frac{1-x}{2-x}$

I don't know if I am right, but this is where I'm stuck at...
can anyone help me with the next steps to get the answer $x= \frac {2y -3}{y-2}$?
From $y- 1= \frac{1-x}{2-x}$, you can multiply both sides by 2- x to get (2-x)(y-1)= 1-x so 2y-2- (y-1)x= 1- x. Add x to both sides and subtract 2y-2 from both sides: x-(y-1)x= 1+2- 2y or (2-y)x= 3- 2y.

Finally, divide both sides by 2-y.