Solve for y limit as x approaches 1 of (x^2-y^2)/(x+2)=-1

$lim_{x \to 1} \frac{x^{2} - y^{2}}{x + 2} = - 1$

Simplify $lim_{x \to 1} \frac{x^{2} - y^{2}}{x + 2}$ .

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $1$ .

$\frac{lim_{x \to 1} x^{2} - y^{2}}{lim_{x \to 1} x + 2} = - 1$

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{lim_{x \to 1} x^{2} - lim_{x \to 1} y^{2}}{lim_{x \to 1} x + 2} = - 1$

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

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

Evaluate the limit of $y^{2}$ which is constant as $x$ approaches $1$ .

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

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

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

Evaluate the limit of $2$ which is constant as $x$ approaches $1$ .

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

Simplify the numerator.

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

$\frac{(lim_{x \to 1} x + y) (lim_{x \to 1} x - y)}{lim_{x \to 1} x + 2} = - 1$

Simplify.

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{(1 + y) (lim_{x \to 1} x - y)}{lim_{x \to 1} x + 2} = - 1$

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{(1 + y) (1 - y)}{lim_{x \to 1} x + 2} = - 1$

Simplify the denominator.

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

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

Add $1$ and $2$ .

$\frac{(1 + y) (1 - y)}{3} = - 1$

Multiply both sides by $3$ .

$\frac{(1 + y) (1 - y)}{3} \cdot 3 = - 1 \cdot 3$

Simplify.

Simplify the left side.

Simplify $\frac{(1 + y) (1 - y)}{3} \cdot 3$ .

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{(1 + y) (1 - y)}{3} \cdot 3 = - 1 \cdot 3$

Rewrite the expression.

$(1 + y) (1 - y) = - 1 \cdot 3$

Expand $(1 + y) (1 - y)$ using the FOIL Method.

Apply the distributive property.

$1 (1 - y) + y (1 - y) = - 1 \cdot 3$

Apply the distributive property.

$1 \cdot 1 + 1 (- y) + y (1 - y) = - 1 \cdot 3$

Apply the distributive property.

$1 \cdot 1 + 1 (- y) + y \cdot 1 + y (- y) = - 1 \cdot 3$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$1 + 1 (- y) + y \cdot 1 + y (- y) = - 1 \cdot 3$

Multiply $- y$ by $1$ .

$1 - y + y \cdot 1 + y (- y) = - 1 \cdot 3$

Multiply $y$ by $1$ .

$1 - y + y + y (- y) = - 1 \cdot 3$

Rewrite using the commutative property of multiplication.

$1 - y + y - y \cdot y = - 1 \cdot 3$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$1 - y + y - (y \cdot y) = - 1 \cdot 3$

Multiply $y$ by $y$ .

$1 - y + y - y^{2} = - 1 \cdot 3$

Add $- y$ and $y$ .

$1 + 0 - y^{2} = - 1 \cdot 3$

Add $1$ and $0$ .

$1 - y^{2} = - 1 \cdot 3$

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

$- y^{2} + 1 = - 1 \cdot 3$

Simplify the right side.

Multiply $- 1$ by $3$ .

$- y^{2} + 1 = - 3$

Solve for $y$ .

Move all terms not containing $y$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$- y^{2} = - 3 - 1$

Subtract $1$ from $- 3$ .

$- y^{2} = - 4$

Divide each term in $- y^{2} = - 4$ by $- 1$ and simplify.

Divide each term in $- y^{2} = - 4$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{- 4}{- 1}$

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

$y^{2} = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$y^{2} = 4$

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

$y = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$y = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm 2$

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

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

$y = 2$

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

$y = - 2$

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

$y = 2, - 2$

Solve for y limit as x approaches 1 of (x^2-y^2)/(x+2)=-1

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