How is y solved for in this equation?

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

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- Jul 16th 2010, 11:23 AMMike9182Solving for y
How is y solved for in this equation?

$\displaystyle x=\frac{1-\sqrt{y}}{1+\sqrt{y}}$ - Jul 16th 2010, 11:28 AMwonderboy1953
Multiply both sides of the equation by $\displaystyle 1/(1 + \sqrt{y})$, then isolate $\displaystyle \sqrt{y}$. The rest will follow.

- Jul 16th 2010, 11:31 AMMike9182
Did you make a typo in that formula?

- Jul 16th 2010, 11:36 AMwonderboy1953
Yes, should be multiplying both sides by $\displaystyle (1 + \sqrt{y})$

- Jul 16th 2010, 11:44 AM1005
$\displaystyle x=\frac{1-\sqrt{y}}{1+\sqrt{y}}$

synthetically divide:

$\displaystyle x = \frac{2}{\sqrt{y}+1} - 1$

add 1:

$\displaystyle x + 1= \frac{2}{\sqrt{y}+1}$

multiply by that denominator and divide by x+1:

$\displaystyle \sqrt{y} + 1 = \frac{2}{x + 1}$

subtract 1:

$\displaystyle \sqrt{y} = \frac{2}{x + 1} - 1$

find common denominator and add:

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

square both sides:

$\displaystyle y = \frac{(-x+1)^2}{(x+1)^2}$

factor out -1 from numerator:

$\displaystyle y = \frac{(-[x-1])^2}{(x+1)^2}$

distribute ^2:

$\displaystyle y = (-1)^2\frac{(x-1)^2}{(x+1)^2}$

-1*-1 = 1:

$\displaystyle y = \frac{(x-1)^2}{(x+1)^2}$ - Jul 16th 2010, 12:39 PMWilmer
- Jul 17th 2010, 03:40 AMdhiab
Hello : 1005 and wilmer

you have one solution http://www.mathhelpforum.com/math-he...10db5142e3.png

but (1-x)/(1+x)>= 0 - Jul 17th 2010, 07:01 AMWilmer