Solve for y x^2-y^2=1

$x^{2} - y^{2} = 1$

Subtract $x^{2}$ from both sides of the equation.

$- y^{2} = 1 - x^{2}$

Multiply each term in $- y^{2} = 1 - x^{2}$ by $- 1$

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Multiply each term in $- y^{2} = 1 - x^{2}$ by $- 1$ .

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

Multiply $(- y^{2}) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

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

Multiply $y^{2}$ by $1$ .

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

Simplify each term.

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Multiply $- 1$ by $1$ .

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

Multiply $(- x^{2}) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$y^{2} = - 1 + 1 x^{2}$

Multiply $x^{2}$ by $1$ .

$y^{2} = - 1 + x^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{- 1 + x^{2}}$

The complete solution is the result of both the positive and negative portions of the solution.

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Simplify the right side of the equation.

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Simplify the expression.

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Rewrite $1$ as $1^{2}$ .

$y = \pm \sqrt{- 1^{2} + x^{2}}$

Reorder $- 1^{2}$ and $x^{2}$ .

$y = \pm \sqrt{x^{2} - 1^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$y = \pm \sqrt{(x + 1) (x - 1)}$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{(x + 1) (x - 1)}$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt{(x + 1) (x - 1)}$

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{(x + 1) (x - 1)}$

$y = - \sqrt{(x + 1) (x - 1)}$

$y = \sqrt{(x + 1) (x - 1)}$

$y = - \sqrt{(x + 1) (x - 1)}$

$y = \sqrt{(x + 1) (x - 1)}$

$y = - \sqrt{(x + 1) (x - 1)}$

Solve for y x^2-y^2=1

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