Solve the System of Inequalities x^2+y^2<=13 5x^2-y^2>=11

$x^{2} + y^{2} \leq 13$ $5 x^{2} - y^{2} \geq 11$

Step 1

Solve $x^{2} + y^{2} \leq 13$ for $y$ .

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

$y^{2} \leq 13 - x^{2}$

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

$\sqrt{y^{2}} \leq \sqrt{13 - x^{2}}$

Simplify the left side.

Pull terms out from under the radical.

$| y | \leq \sqrt{13 - x^{2}}$

Write $| y | \leq \sqrt{13 - x^{2}}$ 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.

$y \leq \sqrt{13 - x^{2}}$

Find the domain of $y \leq \sqrt{13 - x^{2}}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \leq \sqrt{13 - x^{2}}$ .

Set the radicand in $\sqrt{13 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$13 - x^{2} \geq 0$

Solve for $y$ .

Subtract $13$ from both sides of the inequality.

$- x^{2} \geq - 13$

Divide each term in $- x^{2} \geq - 13$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 13$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 13}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 13}{- 1}$

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

$x^{2} \leq \frac{- 13}{- 1}$

Simplify the right side.

Divide $- 13$ by $- 1$ .

$x^{2} \leq 13$

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

$\sqrt{x^{2}} \leq \sqrt{13}$

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{13}$

Write $| x | \leq \sqrt{13}$ 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.

$x \leq \sqrt{13}$

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 $- 1$ .

$- x \leq \sqrt{13}$

Write as a piecewise.

${\begin{cases} x \leq \sqrt{13} & x \geq 0 \\ - x \leq \sqrt{13} & x < 0 \end{cases}$

Find the intersection of $x \leq \sqrt{13}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{13}$

Solve $- x \leq \sqrt{13}$ when $x < 0$ .

Divide each term in $- x \leq \sqrt{13}$ by $- 1$ and simplify.

Divide each term in $- x \leq \sqrt{13}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{13}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{13}}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{13}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{13}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{13}$

Rewrite $- 1 \cdot \sqrt{13}$ as $- \sqrt{13}$ .

$x \geq - \sqrt{13}$

Find the intersection of $x \geq - \sqrt{13}$ and $x < 0$ .

$- \sqrt{13} \leq x < 0$

Find the union of the solutions.

$- \sqrt{13} \leq x \leq \sqrt{13}$

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

$[- \sqrt{13}, \sqrt{13}]$

Find the intersection of $y \geq 0$ and $[- \sqrt{13}, \sqrt{13}]$ .

$0 \leq y \leq \sqrt{13}$

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 $- 1$ .

$- y \leq \sqrt{13 - x^{2}}$

Find the domain of $- y \leq \sqrt{13 - x^{2}}$ and find the intersection with $y < 0$ .

Find the domain of $- y \leq \sqrt{13 - x^{2}}$ .

Set the radicand in $\sqrt{13 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$13 - x^{2} \geq 0$

Solve for $y$ .

Subtract $13$ from both sides of the inequality.

$- x^{2} \geq - 13$

Divide each term in $- x^{2} \geq - 13$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 13$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 13}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 13}{- 1}$

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

$x^{2} \leq \frac{- 13}{- 1}$

Simplify the right side.

Divide $- 13$ by $- 1$ .

$x^{2} \leq 13$

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

$\sqrt{x^{2}} \leq \sqrt{13}$

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{13}$

Write $| x | \leq \sqrt{13}$ 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.

$x \leq \sqrt{13}$

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 $- 1$ .

$- x \leq \sqrt{13}$

Write as a piecewise.

${\begin{cases} x \leq \sqrt{13} & x \geq 0 \\ - x \leq \sqrt{13} & x < 0 \end{cases}$

Find the intersection of $x \leq \sqrt{13}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{13}$

Solve $- x \leq \sqrt{13}$ when $x < 0$ .

Divide each term in $- x \leq \sqrt{13}$ by $- 1$ and simplify.

Divide each term in $- x \leq \sqrt{13}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{13}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{13}}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{13}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{13}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{13}$

Rewrite $- 1 \cdot \sqrt{13}$ as $- \sqrt{13}$ .

$x \geq - \sqrt{13}$

Find the intersection of $x \geq - \sqrt{13}$ and $x < 0$ .

$- \sqrt{13} \leq x < 0$

Find the union of the solutions.

$- \sqrt{13} \leq x \leq \sqrt{13}$

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

$[- \sqrt{13}, \sqrt{13}]$

Find the intersection of $y < 0$ and $[- \sqrt{13}, \sqrt{13}]$ .

$- \sqrt{13} \leq y < 0$

Write as a piecewise.

${\begin{cases} y \leq \sqrt{13 - x^{2}} & 0 \leq y \leq \sqrt{13} \\ - y \leq \sqrt{13 - x^{2}} & - \sqrt{13} \leq y < 0 \end{cases}$

Find the intersection of $y \leq \sqrt{13 - x^{2}}$ and $0 \leq y \leq \sqrt{13}$ .

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

Solve $- y \leq \sqrt{13 - x^{2}}$ when $- \sqrt{13} \leq y < 0$ .

Divide each term in $- y \leq \sqrt{13 - x^{2}}$ by $- 1$ and simplify.

Divide each term in $- y \leq \sqrt{13 - x^{2}}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{\sqrt{13 - x^{2}}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{\sqrt{13 - x^{2}}}{- 1}$

Divide $y$ by $1$ .

$y \geq \frac{\sqrt{13 - x^{2}}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{13 - x^{2}}}{- 1}$ .

$y \geq - 1 \cdot \sqrt{13 - x^{2}}$

Rewrite $- 1 \cdot \sqrt{13 - x^{2}}$ as $- \sqrt{13 - x^{2}}$ .

$y \geq - \sqrt{13 - x^{2}}$

Find the intersection of $y \geq - \sqrt{13 - x^{2}}$ and $- \sqrt{13} \leq y < 0$ .

No solution

Find the union of the solutions.

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

Step 2

Solve $5 x^{2} - y^{2} \geq 11$ for $y$ .

Subtract $5 x^{2}$ from both sides of the inequality.

$- y^{2} \geq 11 - 5 x^{2}$

Divide each term in $- y^{2} \geq 11 - 5 x^{2}$ by $- 1$ and simplify.

Divide each term in $- y^{2} \geq 11 - 5 x^{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y^{2}}{- 1} \leq \frac{11}{- 1} + \frac{- 5 x^{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} \leq \frac{11}{- 1} + \frac{- 5 x^{2}}{- 1}$

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

$y^{2} \leq \frac{11}{- 1} + \frac{- 5 x^{2}}{- 1}$

Simplify the right side.

Simplify each term.

Divide $11$ by $- 1$ .

$y^{2} \leq - 11 + \frac{- 5 x^{2}}{- 1}$

Move the negative one from the denominator of $\frac{- 5 x^{2}}{- 1}$ .

$y^{2} \leq - 11 - 1 \cdot (- 5 x^{2})$

Rewrite $- 1 \cdot (- 5 x^{2})$ as $- (- 5 x^{2})$ .

$y^{2} \leq - 11 - (- 5 x^{2})$

Multiply $- 5$ by $- 1$ .

$y^{2} \leq - 11 + 5 x^{2}$

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

$\sqrt{y^{2}} \leq \sqrt{- 11 + 5 x^{2}}$

Simplify the left side.

Pull terms out from under the radical.

$| y | \leq \sqrt{- 11 + 5 x^{2}}$

Write $| y | \leq \sqrt{- 11 + 5 x^{2}}$ 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.

$y \leq \sqrt{- 11 + 5 x^{2}}$

Find the domain of $y \leq \sqrt{- 11 + 5 x^{2}}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \leq \sqrt{- 11 + 5 x^{2}}$ .

Set the radicand in $\sqrt{- 11 + 5 x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$- 11 + 5 x^{2} \geq 0$

Solve for $y$ .

Add $11$ to both sides of the inequality.

$5 x^{2} \geq 11$

Divide each term in $5 x^{2} \geq 11$ by $5$ and simplify.

Divide each term in $5 x^{2} \geq 11$ by $5$ .

$\frac{5 x^{2}}{5} \geq \frac{11}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x^{2}}{5} \geq \frac{11}{5}$

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

$x^{2} \geq \frac{11}{5}$

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

$\sqrt{x^{2}} \geq \sqrt{\frac{11}{5}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{\frac{11}{5}}$

Simplify the right side.

Simplify $\sqrt{\frac{11}{5}}$ .

Rewrite $\sqrt{\frac{11}{5}}$ as $\frac{\sqrt{11}}{\sqrt{5}}$ .

$| x | \geq \frac{\sqrt{11}}{\sqrt{5}}$

Multiply $\frac{\sqrt{11}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$| x | \geq \frac{\sqrt{11}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{11}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{\sqrt{5} \sqrt{5}}$

Raise $\sqrt{5}$ to the power of $1$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{{\sqrt{5}}^{1} \sqrt{5}}$

Raise $\sqrt{5}$ to the power of $1$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{{\sqrt{5}}^{1} {\sqrt{5}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{{\sqrt{5}}^{1 + 1}}$

Add $1$ and $1$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{{\sqrt{5}}^{2}}$

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{{(5^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{5^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{5^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{5^{\frac{2}{2}}}$

Rewrite the expression.

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{5^{1}}$

Evaluate the exponent.

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{5}$

Simplify the numerator.

Combine using the product rule for radicals.

$| x | \geq \frac{\sqrt{11 \cdot 5}}{5}$

Multiply $11$ by $5$ .

$| x | \geq \frac{\sqrt{55}}{5}$

Write $| x | \geq \frac{\sqrt{55}}{5}$ 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.

$x \geq \frac{\sqrt{55}}{5}$

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 $- 1$ .

$- x \geq \frac{\sqrt{55}}{5}$

Write as a piecewise.

${\begin{cases} x \geq \frac{\sqrt{55}}{5} & x \geq 0 \\ - x \geq \frac{\sqrt{55}}{5} & x < 0 \end{cases}$

Find the intersection of $x \geq \frac{\sqrt{55}}{5}$ and $x \geq 0$ .

$x \geq \frac{\sqrt{55}}{5}$

Divide each term in $- x \geq \frac{\sqrt{55}}{5}$ by $- 1$ and simplify.

Divide each term in $- x \geq \frac{\sqrt{55}}{5}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{\frac{\sqrt{55}}{5}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{\frac{\sqrt{55}}{5}}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{\frac{\sqrt{55}}{5}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\frac{\sqrt{55}}{5}}{- 1}$ .

$x \leq - 1 \cdot \frac{\sqrt{55}}{5}$

Rewrite $- 1 \cdot \frac{\sqrt{55}}{5}$ as $- \frac{\sqrt{55}}{5}$ .

$x \leq - \frac{\sqrt{55}}{5}$

Find the union of the solutions.

$x \leq - \frac{\sqrt{55}}{5}$ or $x \geq \frac{\sqrt{55}}{5}$

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

$(- \infty, - \frac{\sqrt{55}}{5}] \cup [\frac{\sqrt{55}}{5}, \infty)$

Find the intersection of $y \geq 0$ and $(- \infty, - \frac{\sqrt{55}}{5}] \cup [\frac{\sqrt{55}}{5}, \infty)$ .

$y \geq \frac{\sqrt{55}}{5}$

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 $- 1$ .

$- y \leq \sqrt{- 11 + 5 x^{2}}$

Find the domain of $- y \leq \sqrt{- 11 + 5 x^{2}}$ and find the intersection with $y < 0$ .

Find the domain of $- y \leq \sqrt{- 11 + 5 x^{2}}$ .

Set the radicand in $\sqrt{- 11 + 5 x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$- 11 + 5 x^{2} \geq 0$

Solve for $y$ .

Add $11$ to both sides of the inequality.

$5 x^{2} \geq 11$

Divide each term in $5 x^{2} \geq 11$ by $5$ and simplify.

Divide each term in $5 x^{2} \geq 11$ by $5$ .

$\frac{5 x^{2}}{5} \geq \frac{11}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x^{2}}{5} \geq \frac{11}{5}$

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

$x^{2} \geq \frac{11}{5}$

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

$\sqrt{x^{2}} \geq \sqrt{\frac{11}{5}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{\frac{11}{5}}$

Simplify the right side.

Simplify $\sqrt{\frac{11}{5}}$ .

Rewrite $\sqrt{\frac{11}{5}}$ as $\frac{\sqrt{11}}{\sqrt{5}}$ .

$| x | \geq \frac{\sqrt{11}}{\sqrt{5}}$

Multiply $\frac{\sqrt{11}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$| x | \geq \frac{\sqrt{11}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{11}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{\sqrt{5} \sqrt{5}}$

Raise $\sqrt{5}$ to the power of $1$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{{\sqrt{5}}^{1} \sqrt{5}}$

Raise $\sqrt{5}$ to the power of $1$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{{\sqrt{5}}^{1} {\sqrt{5}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{{\sqrt{5}}^{1 + 1}}$

Add $1$ and $1$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{{\sqrt{5}}^{2}}$

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{{(5^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{5^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{5^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{5^{\frac{2}{2}}}$

Rewrite the expression.

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{5^{1}}$

Evaluate the exponent.

$| x | \geq \frac{\sqrt{11} \sqrt{5}}{5}$

Simplify the numerator.

Combine using the product rule for radicals.

$| x | \geq \frac{\sqrt{11 \cdot 5}}{5}$

Multiply $11$ by $5$ .

$| x | \geq \frac{\sqrt{55}}{5}$

Write $| x | \geq \frac{\sqrt{55}}{5}$ 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.

$x \geq \frac{\sqrt{55}}{5}$

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 $- 1$ .

$- x \geq \frac{\sqrt{55}}{5}$

Write as a piecewise.

${\begin{cases} x \geq \frac{\sqrt{55}}{5} & x \geq 0 \\ - x \geq \frac{\sqrt{55}}{5} & x < 0 \end{cases}$

Find the intersection of $x \geq \frac{\sqrt{55}}{5}$ and $x \geq 0$ .

$x \geq \frac{\sqrt{55}}{5}$

Divide each term in $- x \geq \frac{\sqrt{55}}{5}$ by $- 1$ and simplify.

Divide each term in $- x \geq \frac{\sqrt{55}}{5}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{\frac{\sqrt{55}}{5}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{\frac{\sqrt{55}}{5}}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{\frac{\sqrt{55}}{5}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\frac{\sqrt{55}}{5}}{- 1}$ .

$x \leq - 1 \cdot \frac{\sqrt{55}}{5}$

Rewrite $- 1 \cdot \frac{\sqrt{55}}{5}$ as $- \frac{\sqrt{55}}{5}$ .

$x \leq - \frac{\sqrt{55}}{5}$

Find the union of the solutions.

$x \leq - \frac{\sqrt{55}}{5}$ or $x \geq \frac{\sqrt{55}}{5}$

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

$(- \infty, - \frac{\sqrt{55}}{5}] \cup [\frac{\sqrt{55}}{5}, \infty)$

Find the intersection of $y < 0$ and $(- \infty, - \frac{\sqrt{55}}{5}] \cup [\frac{\sqrt{55}}{5}, \infty)$ .

$y \leq - \frac{\sqrt{55}}{5}$

Write as a piecewise.

${\begin{cases} y \leq \sqrt{- 11 + 5 x^{2}} & y \geq \frac{\sqrt{55}}{5} \\ - y \leq \sqrt{- 11 + 5 x^{2}} & y \leq - \frac{\sqrt{55}}{5} \end{cases}$

Find the intersection of $y \leq \sqrt{- 11 + 5 x^{2}}$ and $y \geq \frac{\sqrt{55}}{5}$ .

$\frac{\sqrt{55}}{5} \leq y \leq \sqrt{- 11 + 5 x^{2}}$

Solve $- y \leq \sqrt{- 11 + 5 x^{2}}$ when $y \leq - \frac{\sqrt{55}}{5}$ .

Divide each term in $- y \leq \sqrt{- 11 + 5 x^{2}}$ by $- 1$ and simplify.

Divide each term in $- y \leq \sqrt{- 11 + 5 x^{2}}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{\sqrt{- 11 + 5 x^{2}}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{\sqrt{- 11 + 5 x^{2}}}{- 1}$

Divide $y$ by $1$ .

$y \geq \frac{\sqrt{- 11 + 5 x^{2}}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{- 11 + 5 x^{2}}}{- 1}$ .

$y \geq - 1 \cdot \sqrt{- 11 + 5 x^{2}}$

Rewrite $- 1 \cdot \sqrt{- 11 + 5 x^{2}}$ as $- \sqrt{- 11 + 5 x^{2}}$ .

$y \geq - \sqrt{- 11 + 5 x^{2}}$

Find the intersection of $y \geq - \sqrt{- 11 + 5 x^{2}}$ and $y \leq - \frac{\sqrt{55}}{5}$ .

$- \sqrt{- 11 + 5 x^{2}} \leq y \leq - \frac{\sqrt{55}}{5}$

Find the union of the solutions.

$\frac{\sqrt{55}}{5} \leq y \leq \sqrt{- 11 + 5 x^{2}}$ or $- \sqrt{- 11 + 5 x^{2}} \leq y \leq - \frac{\sqrt{55}}{5}$

Step 3

Find the intersection of $0 \leq y \leq i N o Min m^{2} u$ and $\frac{\sqrt{55}}{5} \leq y \leq \sqrt{- 11 + 5 x^{2}} or - \sqrt{- 11 + 5 x^{2}} \leq y \leq - \frac{\sqrt{55}}{5}$ .

$\frac{\sqrt{55}}{5} \leq y \leq N o (Min i m u m)$

Step 4

Solve the System of Inequalities x^2+y^2<=13 5x^2-y^2>=11

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