# Solving for y

• Jul 16th 2010, 11:23 AM
Mike9182
Solving for y
How is y solved for in this equation?

$\displaystyle x=\frac{1-\sqrt{y}}{1+\sqrt{y}}$
• Jul 16th 2010, 11:28 AM
wonderboy1953
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 AM
Mike9182
Did you make a typo in that formula?
• Jul 16th 2010, 11:36 AM
wonderboy1953
Yes, should be multiplying both sides by $\displaystyle (1 + \sqrt{y})$
• Jul 16th 2010, 11:44 AM
1005
$\displaystyle x=\frac{1-\sqrt{y}}{1+\sqrt{y}}$
synthetically divide:
$\displaystyle x = \frac{2}{\sqrt{y}+1} - 1$
$\displaystyle x + 1= \frac{2}{\sqrt{y}+1}$
multiply by that denominator and divide by x+1:
$\displaystyle \sqrt{y} + 1 = \frac{2}{x + 1}$
subtract 1:
$\displaystyle \sqrt{y} = \frac{2}{x + 1} - 1$
$\displaystyle \sqrt{y} = \frac{-x + 1}{x+1}$
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 PM
Wilmer
Quote:

Originally Posted by Mike9182
$\displaystyle x=\frac{1-\sqrt{y}}{1+\sqrt{y}}$

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
• Jul 17th 2010, 03:40 AM
dhiab
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 AM
Wilmer
Quote:

Originally Posted by dhiab
Hello : 1005 and wilmer
you have one solution http://www.mathhelpforum.com/math-he...10db5142e3.png
but (1-x)/(1+x)>= 0

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