Solving for y

**Mike9182** · July 16th 2010, 11:23 AM

How is y solved for in this equation?

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

**wonderboy1953** · July 16th 2010, 11:28 AM

Multiply both sides of the equation by $1/(1 + \sqrt{y})$ , then isolate $\sqrt{y}$ . The rest will follow.

**Mike9182** · July 16th 2010, 11:31 AM

Did you make a typo in that formula?

**wonderboy1953** · July 16th 2010, 11:36 AM

Yes, should be multiplying both sides by $(1 + \sqrt{y})$

**1005** · July 16th 2010, 11:44 AM

$x=\frac{1-\sqrt{y}}{1+\sqrt{y}}$
synthetically divide:
$x = \frac{2}{\sqrt{y}+1} - 1$
add 1:
$x + 1= \frac{2}{\sqrt{y}+1}$
multiply by that denominator and divide by x+1:
$\sqrt{y} + 1 = \frac{2}{x + 1}$
subtract 1:
$\sqrt{y} = \frac{2}{x + 1} - 1$
find common denominator and add:
$\sqrt{y} = \frac{-x + 1}{x+1}$
square both sides:
$y = \frac{(-x+1)^2}{(x+1)^2}$
factor out -1 from numerator:
$y = \frac{(-[x-1])^2}{(x+1)^2}$
distribute ^2:
$y = (-1)^2\frac{(x-1)^2}{(x+1)^2}$
-1*-1 = 1:
$y = \frac{(x-1)^2}{(x+1)^2}$

**Wilmer** · July 16th 2010, 12:39 PM

Originally Posted by Mike9182

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

Another way:

x(1 + SQRT[y]) = 1 - SQRT[y]

x + xSQRT[y] = 1 - SQRT[y]

xSQRT[y] + SQRT[y] = 1 - x

SQRT[y](x + 1) = -1(x - 1)

SQRT[y] / -1 = (x - 1) / (x + 1)

y = [(x - 1) / (x + 1)]^2

**dhiab** · July 17th 2010, 03:40 AM

Hello : 1005 and wilmer
you have one solution

but (1-x)/(1+x)>= 0

**Wilmer** · July 17th 2010, 07:01 AM

Originally Posted by dhiab

Hello : 1005 and wilmer
you have one solution

but (1-x)/(1+x)>= 0

Hmmm....isn't all that's required: x <> -1 ?

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