Graph x^2-y^2<=8 x^2+y^2>=120

$x^{2} - y^{2} \leq 8$ $x^{2} + y^{2} \geq 120$

Solve for $y$ .

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

$- y^{2} \leq 8 - x^{2}$

Divide each term in $- y^{2} \leq 8 - x^{2}$ by $- 1$ and simplify.

Divide each term in $- y^{2} \leq 8 - x^{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y^{2}}{- 1} \geq \frac{8}{- 1} + \frac{- x^{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} \geq \frac{8}{- 1} + \frac{- x^{2}}{- 1}$

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

$y^{2} \geq \frac{8}{- 1} + \frac{- x^{2}}{- 1}$

Simplify the right side.

Simplify each term.

Divide $8$ by $- 1$ .

$y^{2} \geq - 8 + \frac{- x^{2}}{- 1}$

Dividing two negative values results in a positive value.

$y^{2} \geq - 8 + \frac{x^{2}}{1}$

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

$y^{2} \geq - 8 + x^{2}$

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

$\sqrt{y^{2}} \geq \sqrt{- 8 + x^{2}}$

Simplify the left side.

Pull terms out from under the radical.

$| y | \geq \sqrt{- 8 + x^{2}}$

Write $| y | \geq \sqrt{- 8 + x^{2}}$ 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.

$y \geq \sqrt{- 8 + x^{2}}$

Find the domain of $y \geq \sqrt{- 8 + x^{2}}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \geq \sqrt{- 8 + x^{2}}$ .

Set the radicand in $\sqrt{- 8 + x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$- 8 + x^{2} \geq 0$

Solve for $y$ .

Add $8$ to both sides of the inequality.

$x^{2} \geq 8$

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

$\sqrt{x^{2}} \geq \sqrt{8}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{8}$

Simplify the right side.

Simplify $\sqrt{8}$ .

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$| x | \geq \sqrt{4 (2)}$

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

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

Pull terms out from under the radical.

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

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

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

Write $| x | \geq 2 \sqrt{2}$ 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.

$x \geq 2 \sqrt{2}$

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 $- 1$ .

$- x \geq 2 \sqrt{2}$

Write as a piecewise.

${\begin{cases} x \geq 2 \sqrt{2} & x \geq 0 \\ - x \geq 2 \sqrt{2} & x < 0 \end{cases}$

Find the intersection of $x \geq 2 \sqrt{2}$ and $x \geq 0$ .

$x \geq 2 \sqrt{2}$

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

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

$\frac{- x}{- 1} \leq \frac{2 \sqrt{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{2 \sqrt{2}}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{2 \sqrt{2}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{2 \sqrt{2}}{- 1}$ .

$x \leq - 1 \cdot (2 \sqrt{2})$

Rewrite $- 1 \cdot (2 \sqrt{2})$ as $- (2 \sqrt{2})$ .

$x \leq - (2 \sqrt{2})$

Multiply $2$ by $- 1$ .

$x \leq - 2 \sqrt{2}$

Find the union of the solutions.

$x \leq - 2 \sqrt{2}$ or $x \geq 2 \sqrt{2}$

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

$(- \infty, - 2 \sqrt{2}] \cup [2 \sqrt{2}, \infty)$

Find the intersection of $y \geq 0$ and $(- \infty, - 2 \sqrt{2}] \cup [2 \sqrt{2}, \infty)$ .

$y \geq 2 \sqrt{2}$

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 $- 1$ .

$- y \geq \sqrt{- 8 + x^{2}}$

Find the domain of $- y \geq \sqrt{- 8 + x^{2}}$ and find the intersection with $y < 0$ .

Find the domain of $- y \geq \sqrt{- 8 + x^{2}}$ .

Set the radicand in $\sqrt{- 8 + x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$- 8 + x^{2} \geq 0$

Solve for $y$ .

Add $8$ to both sides of the inequality.

$x^{2} \geq 8$

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

$\sqrt{x^{2}} \geq \sqrt{8}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{8}$

Simplify the right side.

Simplify $\sqrt{8}$ .

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$| x | \geq \sqrt{4 (2)}$

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

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

Pull terms out from under the radical.

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

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

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

Write $| x | \geq 2 \sqrt{2}$ 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.

$x \geq 2 \sqrt{2}$

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 $- 1$ .

$- x \geq 2 \sqrt{2}$

Write as a piecewise.

${\begin{cases} x \geq 2 \sqrt{2} & x \geq 0 \\ - x \geq 2 \sqrt{2} & x < 0 \end{cases}$

Find the intersection of $x \geq 2 \sqrt{2}$ and $x \geq 0$ .

$x \geq 2 \sqrt{2}$

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

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

$\frac{- x}{- 1} \leq \frac{2 \sqrt{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{2 \sqrt{2}}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{2 \sqrt{2}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{2 \sqrt{2}}{- 1}$ .

$x \leq - 1 \cdot (2 \sqrt{2})$

Rewrite $- 1 \cdot (2 \sqrt{2})$ as $- (2 \sqrt{2})$ .

$x \leq - (2 \sqrt{2})$

Multiply $2$ by $- 1$ .

$x \leq - 2 \sqrt{2}$

Find the union of the solutions.

$x \leq - 2 \sqrt{2}$ or $x \geq 2 \sqrt{2}$

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

$(- \infty, - 2 \sqrt{2}] \cup [2 \sqrt{2}, \infty)$

Find the intersection of $y < 0$ and $(- \infty, - 2 \sqrt{2}] \cup [2 \sqrt{2}, \infty)$ .

$y \leq - 2 \sqrt{2}$

Write as a piecewise.

${\begin{cases} y \geq \sqrt{- 8 + x^{2}} & y \geq 2 \sqrt{2} \\ - y \geq \sqrt{- 8 + x^{2}} & y \leq - 2 \sqrt{2} \end{cases}$

Find the intersection of $y \geq \sqrt{- 8 + x^{2}}$ and $y \geq 2 \sqrt{2}$ .

$y \geq \sqrt{- 8 + x^{2}}$ and $y \geq 2 \sqrt{2}$

Solve $- y \geq \sqrt{- 8 + x^{2}}$ when $y \leq - 2 \sqrt{2}$ .

Divide each term in $- y \geq \sqrt{- 8 + x^{2}}$ by $- 1$ and simplify.

Divide each term in $- y \geq \sqrt{- 8 + x^{2}}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \leq \frac{\sqrt{- 8 + x^{2}}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \leq \frac{\sqrt{- 8 + x^{2}}}{- 1}$

Divide $y$ by $1$ .

$y \leq \frac{\sqrt{- 8 + x^{2}}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{- 8 + x^{2}}}{- 1}$ .

$y \leq - 1 \cdot \sqrt{- 8 + x^{2}}$

Rewrite $- 1 \cdot \sqrt{- 8 + x^{2}}$ as $- \sqrt{- 8 + x^{2}}$ .

$y \leq - \sqrt{- 8 + x^{2}}$

Find the intersection of $y \leq - \sqrt{- 8 + x^{2}}$ and $y \leq - 2 \sqrt{2}$ .

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

Find the union of the solutions.

$y \geq N o Max i m u m$ or $y \leq N o Min i m u m$

Solve for $y$ .

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

$y^{2} \geq 120 - x^{2}$

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

$\sqrt{y^{2}} \geq \sqrt{120 - x^{2}}$

Simplify the left side.

Pull terms out from under the radical.

$| y | \geq \sqrt{120 - x^{2}}$

Write $| y | \geq \sqrt{120 - x^{2}}$ 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.

$y \geq \sqrt{120 - x^{2}}$

Find the domain of $y \geq \sqrt{120 - x^{2}}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \geq \sqrt{120 - x^{2}}$ .

Set the radicand in $\sqrt{120 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$120 - x^{2} \geq 0$

Solve for $y$ .

Subtract $120$ from both sides of the inequality.

$- x^{2} \geq - 120$

Divide each term in $- x^{2} \geq - 120$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 120$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 120}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 120}{- 1}$

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

$x^{2} \leq \frac{- 120}{- 1}$

Simplify the right side.

Divide $- 120$ by $- 1$ .

$x^{2} \leq 120$

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

$\sqrt{x^{2}} \leq \sqrt{120}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{120}$

Simplify the right side.

Simplify $\sqrt{120}$ .

Rewrite $120$ as $2^{2} \cdot 30$ .

Factor $4$ out of $120$ .

$| x | \leq \sqrt{4 (30)}$

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

$| x | \leq \sqrt{2^{2} \cdot 30}$

Pull terms out from under the radical.

$| x | \leq | 2 | \sqrt{30}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| x | \leq 2 \sqrt{30}$

Write $| x | \leq 2 \sqrt{30}$ 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.

$x \leq 2 \sqrt{30}$

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 $- 1$ .

$- x \leq 2 \sqrt{30}$

Write as a piecewise.

${\begin{cases} x \leq 2 \sqrt{30} & x \geq 0 \\ - x \leq 2 \sqrt{30} & x < 0 \end{cases}$

Find the intersection of $x \leq 2 \sqrt{30}$ and $x \geq 0$ .

$0 \leq x \leq 2 \sqrt{30}$

Solve $- x \leq 2 \sqrt{30}$ when $x < 0$ .

Divide each term in $- x \leq 2 \sqrt{30}$ by $- 1$ and simplify.

Divide each term in $- x \leq 2 \sqrt{30}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{2 \sqrt{30}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{2 \sqrt{30}}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{2 \sqrt{30}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{2 \sqrt{30}}{- 1}$ .

$x \geq - 1 \cdot (2 \sqrt{30})$

Rewrite $- 1 \cdot (2 \sqrt{30})$ as $- (2 \sqrt{30})$ .

$x \geq - (2 \sqrt{30})$

Multiply $2$ by $- 1$ .

$x \geq - 2 \sqrt{30}$

Find the intersection of $x \geq - 2 \sqrt{30}$ and $x < 0$ .

$- 2 \sqrt{30} \leq x < 0$

Find the union of the solutions.

$- 2 \sqrt{30} \leq x \leq 2 \sqrt{30}$

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

$[- 2 \sqrt{30}, 2 \sqrt{30}]$

Find the intersection of $y \geq 0$ and $[- 2 \sqrt{30}, 2 \sqrt{30}]$ .

$0 \leq y \leq 2 \sqrt{30}$

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 $- 1$ .

$- y \geq \sqrt{120 - x^{2}}$

Find the domain of $- y \geq \sqrt{120 - x^{2}}$ and find the intersection with $y < 0$ .

Find the domain of $- y \geq \sqrt{120 - x^{2}}$ .

Set the radicand in $\sqrt{120 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$120 - x^{2} \geq 0$

Solve for $y$ .

Subtract $120$ from both sides of the inequality.

$- x^{2} \geq - 120$

Divide each term in $- x^{2} \geq - 120$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 120$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 120}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 120}{- 1}$

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

$x^{2} \leq \frac{- 120}{- 1}$

Simplify the right side.

Divide $- 120$ by $- 1$ .

$x^{2} \leq 120$

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

$\sqrt{x^{2}} \leq \sqrt{120}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{120}$

Simplify the right side.

Simplify $\sqrt{120}$ .

Rewrite $120$ as $2^{2} \cdot 30$ .

Factor $4$ out of $120$ .

$| x | \leq \sqrt{4 (30)}$

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

$| x | \leq \sqrt{2^{2} \cdot 30}$

Pull terms out from under the radical.

$| x | \leq | 2 | \sqrt{30}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| x | \leq 2 \sqrt{30}$

Write $| x | \leq 2 \sqrt{30}$ 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.

$x \leq 2 \sqrt{30}$

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 $- 1$ .

$- x \leq 2 \sqrt{30}$

Write as a piecewise.

${\begin{cases} x \leq 2 \sqrt{30} & x \geq 0 \\ - x \leq 2 \sqrt{30} & x < 0 \end{cases}$

Find the intersection of $x \leq 2 \sqrt{30}$ and $x \geq 0$ .

$0 \leq x \leq 2 \sqrt{30}$

Solve $- x \leq 2 \sqrt{30}$ when $x < 0$ .

Divide each term in $- x \leq 2 \sqrt{30}$ by $- 1$ and simplify.

Divide each term in $- x \leq 2 \sqrt{30}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{2 \sqrt{30}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{2 \sqrt{30}}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{2 \sqrt{30}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{2 \sqrt{30}}{- 1}$ .

$x \geq - 1 \cdot (2 \sqrt{30})$

Rewrite $- 1 \cdot (2 \sqrt{30})$ as $- (2 \sqrt{30})$ .

$x \geq - (2 \sqrt{30})$

Multiply $2$ by $- 1$ .

$x \geq - 2 \sqrt{30}$

Find the intersection of $x \geq - 2 \sqrt{30}$ and $x < 0$ .

$- 2 \sqrt{30} \leq x < 0$

Find the union of the solutions.

$- 2 \sqrt{30} \leq x \leq 2 \sqrt{30}$

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

$[- 2 \sqrt{30}, 2 \sqrt{30}]$

Find the intersection of $y < 0$ and $[- 2 \sqrt{30}, 2 \sqrt{30}]$ .

$- 2 \sqrt{30} \leq y < 0$

Write as a piecewise.

${\begin{cases} y \geq \sqrt{120 - x^{2}} & 0 \leq y \leq 2 \sqrt{30} \\ - y \geq \sqrt{120 - x^{2}} & - 2 \sqrt{30} \leq y < 0 \end{cases}$

Find the intersection of $y \geq \sqrt{120 - x^{2}}$ and $0 \leq y \leq 2 \sqrt{30}$ .

No solution

Solve $- y \geq \sqrt{120 - x^{2}}$ when $- 2 \sqrt{30} \leq y < 0$ .

Divide each term in $- y \geq \sqrt{120 - x^{2}}$ by $- 1$ and simplify.

Divide each term in $- y \geq \sqrt{120 - x^{2}}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \leq \frac{\sqrt{120 - x^{2}}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \leq \frac{\sqrt{120 - x^{2}}}{- 1}$

Divide $y$ by $1$ .

$y \leq \frac{\sqrt{120 - x^{2}}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{120 - x^{2}}}{- 1}$ .

$y \leq - 1 \cdot \sqrt{120 - x^{2}}$

Rewrite $- 1 \cdot \sqrt{120 - x^{2}}$ as $- \sqrt{120 - x^{2}}$ .

$y \leq - \sqrt{120 - x^{2}}$

Find the intersection of $y \leq - \sqrt{120 - x^{2}}$ and $- 2 \sqrt{30} \leq y < 0$ .

$- 2 \sqrt{30} \leq y < N o (Min i m u m)$

Find the union of the solutions.

$- 2 \sqrt{30} \leq y < i N o Min m^{2} u$

Plot each graph on the same coordinate system.

$x^{2} - y^{2} \leq 8$

$x^{2} + y^{2} \geq 120$

Graph x^2-y^2<=8 x^2+y^2>=120

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