Solve the System of Inequalities x^2+y^2<=49 x^2-4y^2>16

$x^{2} + y^{2} \leq 49$ $x^{2} - 4 y^{2} > 16$

Solve $x^{2} + y^{2} \leq 49$ for $y$ .

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

$y^{2} \leq 49 - x^{2}$

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

$\sqrt{y^{2}} \leq \sqrt{49 - x^{2}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | \leq \sqrt{49 - x^{2}}$

Simplify the right side.

Simplify $\sqrt{49 - x^{2}}$ .

Rewrite $49$ as $7^{2}$ .

$| y | \leq \sqrt{7^{2} - x^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 7$ and $b = x$ .

$| y | \leq \sqrt{(7 + x) (7 - x)}$

Write $| y | \leq \sqrt{(7 + x) (7 - x)}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

In the piece where $y$ is non-negative, remove the absolute value.

$y \leq \sqrt{(7 + x) (7 - x)}$

Find the domain of $y \leq \sqrt{(7 + x) (7 - x)}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \leq \sqrt{(7 + x) (7 - x)}$ .

Set the radicand in $\sqrt{(7 + x) (7 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(7 + x) (7 - x) \geq 0$

Solve for $y$ .

Simplify $(7 + x) (7 - x)$ .

Expand $(7 + x) (7 - x)$ using the FOIL Method.

Apply the distributive property.

$7 (7 - x) + x (7 - x) \geq 0$

Apply the distributive property.

$7 \cdot 7 + 7 (- x) + x (7 - x) \geq 0$

Apply the distributive property.

$7 \cdot 7 + 7 (- x) + x \cdot 7 + x (- x) \geq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $7$ by $7$ .

$49 + 7 (- x) + x \cdot 7 + x (- x) \geq 0$

Multiply $- 1$ by $7$ .

$49 - 7 x + x \cdot 7 + x (- x) \geq 0$

Move $7$ to the left of $x$ .

$49 - 7 x + 7 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$49 - 7 x + 7 x - x \cdot x \geq 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$49 - 7 x + 7 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$49 - 7 x + 7 x - x^{2} \geq 0$

Add $- 7 x$ and $7 x$ .

$49 + 0 - x^{2} \geq 0$

Add $49$ and $0$ .

$49 - x^{2} \geq 0$

Subtract $49$ from both sides of the inequality.

$- x^{2} \geq - 49$

Divide each term in $- x^{2} \geq - 49$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 49$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 49}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 49}{- 1}$

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

$x^{2} \leq \frac{- 49}{- 1}$

Simplify the right side.

Divide $- 49$ by $- 1$ .

$x^{2} \leq 49$

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

$\sqrt{x^{2}} \leq \sqrt{49}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{49}$

Simplify the right side.

Simplify $\sqrt{49}$ .

Rewrite $49$ as $7^{2}$ .

$| x | \leq \sqrt{7^{2}}$

Pull terms out from under the radical.

$| x | \leq | 7 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $7$ is $7$ .

$| x | \leq 7$

Write $| x | \leq 7$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 7$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 7$

Write as a piecewise.

${\begin{cases} x \leq 7 & x \geq 0 \\ - x \leq 7 & x < 0 \end{cases}$

Find the intersection of $x \leq 7$ and $x \geq 0$ .

$0 \leq x \leq 7$

Solve $- x \leq 7$ when $x < 0$ .

Divide each term in $- x \leq 7$ by $- 1$ and simplify.

Divide each term in $- x \leq 7$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{7}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{7}{- 1}$

Simplify the right side.

Divide $7$ by $- 1$ .

$x \geq - 7$

Find the intersection of $x \geq - 7$ and $x < 0$ .

$- 7 \leq x < 0$

Find the union of the solutions.

$- 7 \leq x \leq 7$

The domain is all values of $y$ that make the expression defined.

$[- 7, 7]$

Find the intersection of $y \geq 0$ and $[- 7, 7]$ .

$0 \leq y \leq 7$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

$- y \leq \sqrt{(7 + x) (7 - x)}$

Find the domain of $- y \leq \sqrt{(7 + x) (7 - x)}$ and find the intersection with $y < 0$ .

Find the domain of $- y \leq \sqrt{(7 + x) (7 - x)}$ .

Set the radicand in $\sqrt{(7 + x) (7 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(7 + x) (7 - x) \geq 0$

Solve for $y$ .

Simplify $(7 + x) (7 - x)$ .

Expand $(7 + x) (7 - x)$ using the FOIL Method.

Apply the distributive property.

$7 (7 - x) + x (7 - x) \geq 0$

Apply the distributive property.

$7 \cdot 7 + 7 (- x) + x (7 - x) \geq 0$

Apply the distributive property.

$7 \cdot 7 + 7 (- x) + x \cdot 7 + x (- x) \geq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $7$ by $7$ .

$49 + 7 (- x) + x \cdot 7 + x (- x) \geq 0$

Multiply $- 1$ by $7$ .

$49 - 7 x + x \cdot 7 + x (- x) \geq 0$

Move $7$ to the left of $x$ .

$49 - 7 x + 7 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$49 - 7 x + 7 x - x \cdot x \geq 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$49 - 7 x + 7 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$49 - 7 x + 7 x - x^{2} \geq 0$

Add $- 7 x$ and $7 x$ .

$49 + 0 - x^{2} \geq 0$

Add $49$ and $0$ .

$49 - x^{2} \geq 0$

Subtract $49$ from both sides of the inequality.

$- x^{2} \geq - 49$

Divide each term in $- x^{2} \geq - 49$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 49$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 49}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 49}{- 1}$

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

$x^{2} \leq \frac{- 49}{- 1}$

Simplify the right side.

Divide $- 49$ by $- 1$ .

$x^{2} \leq 49$

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

$\sqrt{x^{2}} \leq \sqrt{49}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{49}$

Simplify the right side.

Simplify $\sqrt{49}$ .

Rewrite $49$ as $7^{2}$ .

$| x | \leq \sqrt{7^{2}}$

Pull terms out from under the radical.

$| x | \leq | 7 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $7$ is $7$ .

$| x | \leq 7$

Write $| x | \leq 7$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 7$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 7$

Write as a piecewise.

${\begin{cases} x \leq 7 & x \geq 0 \\ - x \leq 7 & x < 0 \end{cases}$

Find the intersection of $x \leq 7$ and $x \geq 0$ .

$0 \leq x \leq 7$

Solve $- x \leq 7$ when $x < 0$ .

Divide each term in $- x \leq 7$ by $- 1$ and simplify.

Divide each term in $- x \leq 7$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{7}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{7}{- 1}$

Simplify the right side.

Divide $7$ by $- 1$ .

$x \geq - 7$

Find the intersection of $x \geq - 7$ and $x < 0$ .

$- 7 \leq x < 0$

Find the union of the solutions.

$- 7 \leq x \leq 7$

The domain is all values of $y$ that make the expression defined.

$[- 7, 7]$

Find the intersection of $y < 0$ and $[- 7, 7]$ .

$- 7 \leq y < 0$

Write as a piecewise.

${\begin{cases} y \leq \sqrt{(7 + x) (7 - x)} & 0 \leq y \leq 7 \\ - y \leq \sqrt{(7 + x) (7 - x)} & - 7 \leq y < 0 \end{cases}$

Find the intersection of $y \leq \sqrt{(7 + x) (7 - x)}$ and $0 \leq y \leq 7$ .

$0 \leq y \leq N o (Min i m u m)$

Solve $- y \leq \sqrt{(7 + x) (7 - x)}$ when $- 7 \leq y < 0$ .

Divide each term in $- y \leq \sqrt{(7 + x) (7 - x)}$ by $- 1$ and simplify.

Divide each term in $- y \leq \sqrt{(7 + x) (7 - x)}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{\sqrt{(7 + x) (7 - x)}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{\sqrt{(7 + x) (7 - x)}}{- 1}$

Divide $y$ by $1$ .

$y \geq \frac{\sqrt{(7 + x) (7 - x)}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{(7 + x) (7 - x)}}{- 1}$ .

$y \geq - 1 \cdot \sqrt{(7 + x) (7 - x)}$

Rewrite $- 1 \cdot \sqrt{(7 + x) (7 - x)}$ as $- \sqrt{(7 + x) (7 - x)}$ .

$y \geq - \sqrt{(7 + x) (7 - x)}$

Find the intersection of $y \geq - \sqrt{(7 + x) (7 - x)}$ and $- 7 \leq y < 0$ .

No solution

Find the union of the solutions.

$0 \leq y \leq i N o Min m^{2} u$

Solve $x^{2} - 4 y^{2} > 16$ for $y$ .

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

$- 4 y^{2} > 16 - x^{2}$

Divide each term in $- 4 y^{2} > 16 - x^{2}$ by $- 4$ and simplify.

Divide each term in $- 4 y^{2} > 16 - x^{2}$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 4 y^{2}}{- 4} < \frac{16}{- 4} + \frac{- x^{2}}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 y^{2}}{- 4} < \frac{16}{- 4} + \frac{- x^{2}}{- 4}$

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

$y^{2} < \frac{16}{- 4} + \frac{- x^{2}}{- 4}$

Simplify the right side.

Simplify each term.

Divide $16$ by $- 4$ .

$y^{2} < - 4 + \frac{- x^{2}}{- 4}$

Dividing two negative values results in a positive value.

$y^{2} < - 4 + \frac{x^{2}}{4}$

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

$\sqrt{y^{2}} < \sqrt{- 4 + \frac{x^{2}}{4}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | < \sqrt{- 4 + \frac{x^{2}}{4}}$

Simplify the right side.

Simplify $\sqrt{- 4 + \frac{x^{2}}{4}}$ .

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

$| y | < \sqrt{- 4 + \frac{x^{2}}{2^{2}}}$

Rewrite $\frac{x^{2}}{2^{2}}$ as ${(\frac{x}{2})}^{2}$ .

$| y | < \sqrt{- 4 + {(\frac{x}{2})}^{2}}$

Simplify the expression.

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

$| y | < \sqrt{- 2^{2} + {(\frac{x}{2})}^{2}}$

Reorder $- 2^{2}$ and ${(\frac{x}{2})}^{2}$ .

$| y | < \sqrt{{(\frac{x}{2})}^{2} - 2^{2}}$

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

$| y | < \sqrt{(\frac{x}{2} + 2) (\frac{x}{2} - 2)}$

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$| y | < \sqrt{(\frac{x}{2} + 2 \cdot \frac{2}{2}) (\frac{x}{2} - 2)}$

Combine $2$ and $\frac{2}{2}$ .

$| y | < \sqrt{(\frac{x}{2} + \frac{2 \cdot 2}{2}) (\frac{x}{2} - 2)}$

Combine the numerators over the common denominator.

$| y | < \sqrt{\frac{x + 2 \cdot 2}{2} (\frac{x}{2} - 2)}$

Multiply $2$ by $2$ .

$| y | < \sqrt{\frac{x + 4}{2} (\frac{x}{2} - 2)}$

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$| y | < \sqrt{\frac{x + 4}{2} (\frac{x}{2} - 2 \cdot \frac{2}{2})}$

Combine $- 2$ and $\frac{2}{2}$ .

$| y | < \sqrt{\frac{x + 4}{2} (\frac{x}{2} + \frac{- 2 \cdot 2}{2})}$

Combine the numerators over the common denominator.

$| y | < \sqrt{\frac{x + 4}{2} \cdot \frac{x - 2 \cdot 2}{2}}$

Multiply $- 2$ by $2$ .

$| y | < \sqrt{\frac{x + 4}{2} \cdot \frac{x - 4}{2}}$

Multiply $\frac{x + 4}{2}$ by $\frac{x - 4}{2}$ .

$| y | < \sqrt{\frac{(x + 4) (x - 4)}{2 \cdot 2}}$

Multiply $2$ by $2$ .

$| y | < \sqrt{\frac{(x + 4) (x - 4)}{4}}$

Rewrite $\frac{(x + 4) (x - 4)}{4}$ as ${(\frac{1}{2})}^{2} ((x + 4) (x - 4))$ .

Factor the perfect power $1^{2}$ out of $(x + 4) (x - 4)$ .

$| y | < \sqrt{\frac{1^{2} ((x + 4) (x - 4))}{4}}$

Factor the perfect power $2^{2}$ out of $4$ .

$| y | < \sqrt{\frac{1^{2} ((x + 4) (x - 4))}{2^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} ((x + 4) (x - 4))}{2^{2} \cdot 1}$ .

$| y | < \sqrt{{(\frac{1}{2})}^{2} ((x + 4) (x - 4))}$

Pull terms out from under the radical.

$| y | < | \frac{1}{2} | \sqrt{(x + 4) (x - 4)}$

$\frac{1}{2}$ is approximately $0.5$ which is positive so remove the absolute value

$| y | < \frac{1}{2} \sqrt{(x + 4) (x - 4)}$

Combine $\frac{1}{2}$ and $\sqrt{(x + 4) (x - 4)}$ .

$| y | < \frac{\sqrt{(x + 4) (x - 4)}}{2}$

Write $| y | < \frac{\sqrt{(x + 4) (x - 4)}}{2}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

In the piece where $y$ is non-negative, remove the absolute value.

$y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$

Find the domain of $y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$ and find the intersection with $y \geq 0$ .

Find the domain of $y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$ .

Set the radicand in $\sqrt{(x + 4) (x - 4)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 4) (x - 4) \geq 0$

Solve for $y$ .

Simplify $(x + 4) (x - 4)$ .

Expand $(x + 4) (x - 4)$ using the FOIL Method.

Apply the distributive property.

$x (x - 4) + 4 (x - 4) \geq 0$

Apply the distributive property.

$x \cdot x + x \cdot - 4 + 4 (x - 4) \geq 0$

Apply the distributive property.

$x \cdot x + x \cdot - 4 + 4 x + 4 \cdot - 4 \geq 0$

Simplify terms.

Combine the opposite terms in $x \cdot x + x \cdot - 4 + 4 x + 4 \cdot - 4$ .

Reorder the factors in the terms $x \cdot - 4$ and $4 x$ .

$x \cdot x - 4 x + 4 x + 4 \cdot - 4 \geq 0$

Add $- 4 x$ and $4 x$ .

$x \cdot x + 0 + 4 \cdot - 4 \geq 0$

Add $x \cdot x$ and $0$ .

$x \cdot x + 4 \cdot - 4 \geq 0$

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + 4 \cdot - 4 \geq 0$

Multiply $4$ by $- 4$ .

$x^{2} - 16 \geq 0$

Add $16$ to both sides of the inequality.

$x^{2} \geq 16$

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

$\sqrt{x^{2}} \geq \sqrt{16}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{16}$

Simplify the right side.

Simplify $\sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$| x | \geq \sqrt{4^{2}}$

Pull terms out from under the radical.

$| x | \geq | 4 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$| x | \geq 4$

Write $| x | \geq 4$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \geq 4$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \geq 4$

Write as a piecewise.

${\begin{cases} x \geq 4 & x \geq 0 \\ - x \geq 4 & x < 0 \end{cases}$

Find the intersection of $x \geq 4$ and $x \geq 0$ .

$x \geq 4$

Divide each term in $- x \geq 4$ by $- 1$ and simplify.

Divide each term in $- x \geq 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{4}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{4}{- 1}$

Simplify the right side.

Divide $4$ by $- 1$ .

$x \leq - 4$

Find the union of the solutions.

$x \leq - 4$ or $x \geq 4$

The domain is all values of $y$ that make the expression defined.

$(- \infty, - 4] \cup [4, \infty)$

Find the intersection of $y \geq 0$ and $(- \infty, - 4] \cup [4, \infty)$ .

$y \geq 4$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

$- y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$

Find the domain of $- y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$ and find the intersection with $y < 0$ .

Find the domain of $- y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$ .

Set the radicand in $\sqrt{(x + 4) (x - 4)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 4) (x - 4) \geq 0$

Solve for $y$ .

Simplify $(x + 4) (x - 4)$ .

Expand $(x + 4) (x - 4)$ using the FOIL Method.

Apply the distributive property.

$x (x - 4) + 4 (x - 4) \geq 0$

Apply the distributive property.

$x \cdot x + x \cdot - 4 + 4 (x - 4) \geq 0$

Apply the distributive property.

$x \cdot x + x \cdot - 4 + 4 x + 4 \cdot - 4 \geq 0$

Simplify terms.

Combine the opposite terms in $x \cdot x + x \cdot - 4 + 4 x + 4 \cdot - 4$ .

Reorder the factors in the terms $x \cdot - 4$ and $4 x$ .

$x \cdot x - 4 x + 4 x + 4 \cdot - 4 \geq 0$

Add $- 4 x$ and $4 x$ .

$x \cdot x + 0 + 4 \cdot - 4 \geq 0$

Add $x \cdot x$ and $0$ .

$x \cdot x + 4 \cdot - 4 \geq 0$

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + 4 \cdot - 4 \geq 0$

Multiply $4$ by $- 4$ .

$x^{2} - 16 \geq 0$

Add $16$ to both sides of the inequality.

$x^{2} \geq 16$

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

$\sqrt{x^{2}} \geq \sqrt{16}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{16}$

Simplify the right side.

Simplify $\sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$| x | \geq \sqrt{4^{2}}$

Pull terms out from under the radical.

$| x | \geq | 4 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$| x | \geq 4$

Write $| x | \geq 4$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \geq 4$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \geq 4$

Write as a piecewise.

${\begin{cases} x \geq 4 & x \geq 0 \\ - x \geq 4 & x < 0 \end{cases}$

Find the intersection of $x \geq 4$ and $x \geq 0$ .

$x \geq 4$

Divide each term in $- x \geq 4$ by $- 1$ and simplify.

Divide each term in $- x \geq 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{4}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{4}{- 1}$

Simplify the right side.

Divide $4$ by $- 1$ .

$x \leq - 4$

Find the union of the solutions.

$x \leq - 4$ or $x \geq 4$

The domain is all values of $y$ that make the expression defined.

$(- \infty, - 4] \cup [4, \infty)$

Find the intersection of $y < 0$ and $(- \infty, - 4] \cup [4, \infty)$ .

$y \leq - 4$

Write as a piecewise.

${\begin{cases} y < \frac{\sqrt{(x + 4) (x - 4)}}{2} & y \geq 4 \\ - y < \frac{\sqrt{(x + 4) (x - 4)}}{2} & y \leq - 4 \end{cases}$

Find the intersection of $y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$ and $y \geq 4$ .

$4 \leq y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$

Solve $- y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$ when $y \leq - 4$ .

Divide each term in $- y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$ by $- 1$ and simplify.

Divide each term in $- y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} > \frac{\frac{\sqrt{(x + 4) (x - 4)}}{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} > \frac{\frac{\sqrt{(x + 4) (x - 4)}}{2}}{- 1}$

Divide $y$ by $1$ .

$y > \frac{\frac{\sqrt{(x + 4) (x - 4)}}{2}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\frac{\sqrt{(x + 4) (x - 4)}}{2}}{- 1}$ .

$y > - 1 \cdot \frac{\sqrt{(x + 4) (x - 4)}}{2}$

Rewrite $- 1 \cdot \frac{\sqrt{(x + 4) (x - 4)}}{2}$ as $- \frac{\sqrt{(x + 4) (x - 4)}}{2}$ .

$y > - \frac{\sqrt{(x + 4) (x - 4)}}{2}$

Find the intersection of $y > - \frac{\sqrt{(x + 4) (x - 4)}}{2}$ and $y \leq - 4$ .

$- \frac{\sqrt{(x + 4) (x - 4)}}{2} < y \leq - 4$

Find the union of the solutions.

$4 \leq y < \frac{\sqrt{(x + 4) (x - 4)}}{2}$ or $- \frac{\sqrt{(x + 4) (x - 4)}}{2} < y \leq - 4$

Find the intersection of $0 \leq y \leq i N o Min m^{2} u$ and $4 \leq y < \frac{\sqrt{(x + 4) (x - 4)}}{2} or - \frac{\sqrt{(x + 4) (x - 4)}}{2} < y \leq - 4$ .

$4 \leq y < N o (Min i m u m)$

Solve the System of Inequalities x^2+y^2<=49 x^2-4y^2>16

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