Solve the System of Equations x^2+y^2=16 x^2-y^2=20

$x^{2} + y^{2} = 16$ $x^{2} - y^{2} = 20$

Step 1

Solve for $x$ in $x^{2} - y^{2} = 20$ .

Add $y^{2}$ to both sides of the equation.

$x^{2} = 20 + y^{2}$

$x^{2} + y^{2} = 16$

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

$x = \pm \sqrt{20 + y^{2}}$

$x^{2} + y^{2} = 16$

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.

$x = \sqrt{20 + y^{2}}$

$x^{2} + y^{2} = 16$

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

$x = - \sqrt{20 + y^{2}}$

$x^{2} + y^{2} = 16$

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

$x = \sqrt{20 + y^{2}}$

$x = - \sqrt{20 + y^{2}}$

$x^{2} + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

$x = - \sqrt{20 + y^{2}}$

$x^{2} + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

$x = - \sqrt{20 + y^{2}}$

$x^{2} + y^{2} = 16$

Step 2

Solve the system $x = \sqrt{20 + y^{2}}, x^{2} + y^{2} = 16$ .

Replace all occurrences of $x$ with $\sqrt{20 + y^{2}}$ in each equation.

Replace all occurrences of $x$ in $x^{2} + y^{2} = 16$ with $\sqrt{20 + y^{2}}$ .

${(\sqrt{20 + y^{2}})}^{2} + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

Simplify the left side.

Simplify ${(\sqrt{20 + y^{2}})}^{2} + y^{2}$ .

Rewrite ${\sqrt{20 + y^{2}}}^{2}$ as $20 + y^{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{20 + y^{2}}$ as ${(20 + y^{2})}^{\frac{1}{2}}$ .

${({(20 + y^{2})}^{\frac{1}{2}})}^{2} + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

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

${(20 + y^{2})}^{\frac{1}{2} \cdot 2} + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

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

${(20 + y^{2})}^{\frac{2}{2}} + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(20 + y^{2})}^{\frac{2}{2}} + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

Rewrite the expression.

$(20 + y^{2}) + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

$(20 + y^{2}) + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

Simplify.

$20 + y^{2} + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

$20 + y^{2} + y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

Add $y^{2}$ and $y^{2}$ .

$20 + 2 y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

$20 + 2 y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

$20 + 2 y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

$20 + 2 y^{2} = 16$

$x = \sqrt{20 + y^{2}}$

Solve for $y$ in $20 + 2 y^{2} = 16$ .

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

Subtract $20$ from both sides of the equation.

$2 y^{2} = 16 - 20$

$x = \sqrt{20 + y^{2}}$

Subtract $20$ from $16$ .

$2 y^{2} = - 4$

$x = \sqrt{20 + y^{2}}$

$2 y^{2} = - 4$

$x = \sqrt{20 + y^{2}}$

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

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

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

$x = \sqrt{20 + y^{2}}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$x = \sqrt{20 + y^{2}}$

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

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

$x = \sqrt{20 + y^{2}}$

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

$x = \sqrt{20 + y^{2}}$

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

$x = \sqrt{20 + y^{2}}$

Simplify the right side.

Divide $- 4$ by $2$ .

$y^{2} = - 2$

$x = \sqrt{20 + y^{2}}$

$y^{2} = - 2$

$x = \sqrt{20 + y^{2}}$

$y^{2} = - 2$

$x = \sqrt{20 + y^{2}}$

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

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

$x = \sqrt{20 + y^{2}}$

Simplify $\pm \sqrt{- 2}$ .

Rewrite $- 2$ as $- 1 (2)$ .

$y = \pm \sqrt{- 1 \cdot 2}$

$x = \sqrt{20 + y^{2}}$

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

$y = \pm \sqrt{- 1} \cdot \sqrt{2}$

$x = \sqrt{20 + y^{2}}$

Rewrite $\sqrt{- 1}$ as $i$ .

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

$x = \sqrt{20 + y^{2}}$

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

$x = \sqrt{20 + y^{2}}$

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.

$y = i \sqrt{2}$

$x = \sqrt{20 + y^{2}}$

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

$y = - i \sqrt{2}$

$x = \sqrt{20 + y^{2}}$

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

$y = i \sqrt{2}, - i \sqrt{2}$

$x = \sqrt{20 + y^{2}}$

$y = i \sqrt{2}, - i \sqrt{2}$

$x = \sqrt{20 + y^{2}}$

$y = i \sqrt{2}, - i \sqrt{2}$

$x = \sqrt{20 + y^{2}}$

Replace all occurrences of $y$ with $i \sqrt{2}$ in each equation.

Replace all occurrences of $y$ in $x = \sqrt{20 + y^{2}}$ with $i \sqrt{2}$ .

$x = \sqrt{20 + {(i \sqrt{2})}^{2}}$

$y = i \sqrt{2}$

Simplify the right side.

Simplify $\sqrt{20 + {(i \sqrt{2})}^{2}}$ .

Apply the product rule to $i \sqrt{2}$ .

$x = \sqrt{20 + i^{2} {\sqrt{2}}^{2}}$

$y = i \sqrt{2}$

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

$x = \sqrt{20 - 1 {\sqrt{2}}^{2}}$

$y = i \sqrt{2}$

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

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

$x = \sqrt{20 - 1 {(2^{\frac{1}{2}})}^{2}}$

$y = i \sqrt{2}$

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

$x = \sqrt{20 - 1 \cdot 2^{\frac{1}{2} \cdot 2}}$

$y = i \sqrt{2}$

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

$x = \sqrt{20 - 1 \cdot 2^{\frac{2}{2}}}$

$y = i \sqrt{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = \sqrt{20 - 1 \cdot 2^{\frac{2}{2}}}$

$y = i \sqrt{2}$

Rewrite the expression.

$x = \sqrt{20 - 1 \cdot 2}$

$y = i \sqrt{2}$

$x = \sqrt{20 - 1 \cdot 2}$

$y = i \sqrt{2}$

Evaluate the exponent.

$x = \sqrt{20 - 1 \cdot 2}$

$y = i \sqrt{2}$

$x = \sqrt{20 - 1 \cdot 2}$

$y = i \sqrt{2}$

Simplify the expression.

Multiply $- 1$ by $2$ .

$x = \sqrt{20 - 2}$

$y = i \sqrt{2}$

Subtract $2$ from $20$ .

$x = \sqrt{18}$

$y = i \sqrt{2}$

$x = \sqrt{18}$

$y = i \sqrt{2}$

Rewrite $18$ as $3^{2} \cdot 2$ .

Factor $9$ out of $18$ .

$x = \sqrt{9 (2)}$

$y = i \sqrt{2}$

Rewrite $9$ as $3^{2}$ .

$x = \sqrt{3^{2} \cdot 2}$

$y = i \sqrt{2}$

$x = \sqrt{3^{2} \cdot 2}$

$y = i \sqrt{2}$

Pull terms out from under the radical.

$x = 3 \sqrt{2}$

$y = i \sqrt{2}$

$x = 3 \sqrt{2}$

$y = i \sqrt{2}$

$x = 3 \sqrt{2}$

$y = i \sqrt{2}$

$x = 3 \sqrt{2}$

$y = i \sqrt{2}$

Replace all occurrences of $y$ with $- i \sqrt{2}$ in each equation.

Replace all occurrences of $y$ in $x = \sqrt{20 + y^{2}}$ with $- i \sqrt{2}$ .

$x = \sqrt{20 + {(- i \sqrt{2})}^{2}}$

$y = - i \sqrt{2}$

Simplify the right side.

Simplify $\sqrt{20 + {(- i \sqrt{2})}^{2}}$ .

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- i \sqrt{2}$ .

$x = \sqrt{20 + {(- i)}^{2} {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

Apply the product rule to $- i$ .

$x = \sqrt{20 + {(- 1)}^{2} i^{2} {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

$x = \sqrt{20 + {(- 1)}^{2} i^{2} {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

Raise $- 1$ to the power of $2$ .

$x = \sqrt{20 + 1 i^{2} {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

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

$x = \sqrt{20 + i^{2} {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

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

$x = \sqrt{20 - 1 {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

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

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

$x = \sqrt{20 - 1 {(2^{\frac{1}{2}})}^{2}}$

$y = - i \sqrt{2}$

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

$x = \sqrt{20 - 1 \cdot 2^{\frac{1}{2} \cdot 2}}$

$y = - i \sqrt{2}$

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

$x = \sqrt{20 - 1 \cdot 2^{\frac{2}{2}}}$

$y = - i \sqrt{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = \sqrt{20 - 1 \cdot 2^{\frac{2}{2}}}$

$y = - i \sqrt{2}$

Rewrite the expression.

$x = \sqrt{20 - 1 \cdot 2}$

$y = - i \sqrt{2}$

$x = \sqrt{20 - 1 \cdot 2}$

$y = - i \sqrt{2}$

Evaluate the exponent.

$x = \sqrt{20 - 1 \cdot 2}$

$y = - i \sqrt{2}$

$x = \sqrt{20 - 1 \cdot 2}$

$y = - i \sqrt{2}$

Simplify the expression.

Multiply $- 1$ by $2$ .

$x = \sqrt{20 - 2}$

$y = - i \sqrt{2}$

Subtract $2$ from $20$ .

$x = \sqrt{18}$

$y = - i \sqrt{2}$

$x = \sqrt{18}$

$y = - i \sqrt{2}$

Rewrite $18$ as $3^{2} \cdot 2$ .

Factor $9$ out of $18$ .

$x = \sqrt{9 (2)}$

$y = - i \sqrt{2}$

Rewrite $9$ as $3^{2}$ .

$x = \sqrt{3^{2} \cdot 2}$

$y = - i \sqrt{2}$

$x = \sqrt{3^{2} \cdot 2}$

$y = - i \sqrt{2}$

Pull terms out from under the radical.

$x = 3 \sqrt{2}$

$y = - i \sqrt{2}$

$x = 3 \sqrt{2}$

$y = - i \sqrt{2}$

$x = 3 \sqrt{2}$

$y = - i \sqrt{2}$

$x = 3 \sqrt{2}$

$y = - i \sqrt{2}$

$x = 3 \sqrt{2}$

$y = - i \sqrt{2}$

Step 3

Solve the system $x = - \sqrt{20 + y^{2}}, x^{2} + y^{2} = 16$ .

Replace all occurrences of $x$ with $- \sqrt{20 + y^{2}}$ in each equation.

Replace all occurrences of $x$ in $x^{2} + y^{2} = 16$ with $- \sqrt{20 + y^{2}}$ .

${(- \sqrt{20 + y^{2}})}^{2} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

Simplify the left side.

Simplify ${(- \sqrt{20 + y^{2}})}^{2} + y^{2}$ .

Simplify each term.

Apply the product rule to $- \sqrt{20 + y^{2}}$ .

${(- 1)}^{2} {\sqrt{20 + y^{2}}}^{2} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

Raise $- 1$ to the power of $2$ .

$1 {\sqrt{20 + y^{2}}}^{2} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

Multiply ${\sqrt{20 + y^{2}}}^{2}$ by $1$ .

${\sqrt{20 + y^{2}}}^{2} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

Rewrite ${\sqrt{20 + y^{2}}}^{2}$ as $20 + y^{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{20 + y^{2}}$ as ${(20 + y^{2})}^{\frac{1}{2}}$ .

${({(20 + y^{2})}^{\frac{1}{2}})}^{2} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

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

${(20 + y^{2})}^{\frac{1}{2} \cdot 2} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

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

${(20 + y^{2})}^{\frac{2}{2}} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(20 + y^{2})}^{\frac{2}{2}} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

Rewrite the expression.

$(20 + y^{2}) + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

$(20 + y^{2}) + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

Simplify.

$20 + y^{2} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

$20 + y^{2} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

$20 + y^{2} + y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

Add $y^{2}$ and $y^{2}$ .

$20 + 2 y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

$20 + 2 y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

$20 + 2 y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

$20 + 2 y^{2} = 16$

$x = - \sqrt{20 + y^{2}}$

Solve for $y$ in $20 + 2 y^{2} = 16$ .

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

Subtract $20$ from both sides of the equation.

$2 y^{2} = 16 - 20$

$x = - \sqrt{20 + y^{2}}$

Subtract $20$ from $16$ .

$2 y^{2} = - 4$

$x = - \sqrt{20 + y^{2}}$

$2 y^{2} = - 4$

$x = - \sqrt{20 + y^{2}}$

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

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

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

$x = - \sqrt{20 + y^{2}}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$x = - \sqrt{20 + y^{2}}$

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

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

$x = - \sqrt{20 + y^{2}}$

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

$x = - \sqrt{20 + y^{2}}$

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

$x = - \sqrt{20 + y^{2}}$

Simplify the right side.

Divide $- 4$ by $2$ .

$y^{2} = - 2$

$x = - \sqrt{20 + y^{2}}$

$y^{2} = - 2$

$x = - \sqrt{20 + y^{2}}$

$y^{2} = - 2$

$x = - \sqrt{20 + y^{2}}$

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

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

$x = - \sqrt{20 + y^{2}}$

Simplify $\pm \sqrt{- 2}$ .

Rewrite $- 2$ as $- 1 (2)$ .

$y = \pm \sqrt{- 1 \cdot 2}$

$x = - \sqrt{20 + y^{2}}$

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

$y = \pm \sqrt{- 1} \cdot \sqrt{2}$

$x = - \sqrt{20 + y^{2}}$

Rewrite $\sqrt{- 1}$ as $i$ .

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

$x = - \sqrt{20 + y^{2}}$

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

$x = - \sqrt{20 + y^{2}}$

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.

$y = i \sqrt{2}$

$x = - \sqrt{20 + y^{2}}$

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

$y = - i \sqrt{2}$

$x = - \sqrt{20 + y^{2}}$

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

$y = i \sqrt{2}, - i \sqrt{2}$

$x = - \sqrt{20 + y^{2}}$

$y = i \sqrt{2}, - i \sqrt{2}$

$x = - \sqrt{20 + y^{2}}$

$y = i \sqrt{2}, - i \sqrt{2}$

$x = - \sqrt{20 + y^{2}}$

Replace all occurrences of $y$ with $i \sqrt{2}$ in each equation.

Replace all occurrences of $y$ in $x = - \sqrt{20 + y^{2}}$ with $i \sqrt{2}$ .

$x = - \sqrt{20 + {(i \sqrt{2})}^{2}}$

$y = i \sqrt{2}$

Simplify the right side.

Simplify $- \sqrt{20 + {(i \sqrt{2})}^{2}}$ .

Apply the product rule to $i \sqrt{2}$ .

$x = - \sqrt{20 + i^{2} {\sqrt{2}}^{2}}$

$y = i \sqrt{2}$

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

$x = - \sqrt{20 - 1 {\sqrt{2}}^{2}}$

$y = i \sqrt{2}$

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

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

$x = - \sqrt{20 - 1 {(2^{\frac{1}{2}})}^{2}}$

$y = i \sqrt{2}$

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

$x = - \sqrt{20 - 1 \cdot 2^{\frac{1}{2} \cdot 2}}$

$y = i \sqrt{2}$

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

$x = - \sqrt{20 - 1 \cdot 2^{\frac{2}{2}}}$

$y = i \sqrt{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = - \sqrt{20 - 1 \cdot 2^{\frac{2}{2}}}$

$y = i \sqrt{2}$

Rewrite the expression.

$x = - \sqrt{20 - 1 \cdot 2}$

$y = i \sqrt{2}$

$x = - \sqrt{20 - 1 \cdot 2}$

$y = i \sqrt{2}$

Evaluate the exponent.

$x = - \sqrt{20 - 1 \cdot 2}$

$y = i \sqrt{2}$

$x = - \sqrt{20 - 1 \cdot 2}$

$y = i \sqrt{2}$

Simplify the expression.

Multiply $- 1$ by $2$ .

$x = - \sqrt{20 - 2}$

$y = i \sqrt{2}$

Subtract $2$ from $20$ .

$x = - \sqrt{18}$

$y = i \sqrt{2}$

$x = - \sqrt{18}$

$y = i \sqrt{2}$

Rewrite $18$ as $3^{2} \cdot 2$ .

Factor $9$ out of $18$ .

$x = - \sqrt{9 (2)}$

$y = i \sqrt{2}$

Rewrite $9$ as $3^{2}$ .

$x = - \sqrt{3^{2} \cdot 2}$

$y = i \sqrt{2}$

$x = - \sqrt{3^{2} \cdot 2}$

$y = i \sqrt{2}$

Pull terms out from under the radical.

$x = - (3 \sqrt{2})$

$y = i \sqrt{2}$

Multiply $3$ by $- 1$ .

$x = - 3 \sqrt{2}$

$y = i \sqrt{2}$

$x = - 3 \sqrt{2}$

$y = i \sqrt{2}$

$x = - 3 \sqrt{2}$

$y = i \sqrt{2}$

$x = - 3 \sqrt{2}$

$y = i \sqrt{2}$

Replace all occurrences of $y$ with $- i \sqrt{2}$ in each equation.

Replace all occurrences of $y$ in $x = - \sqrt{20 + y^{2}}$ with $- i \sqrt{2}$ .

$x = - \sqrt{20 + {(- i \sqrt{2})}^{2}}$

$y = - i \sqrt{2}$

Simplify the right side.

Simplify $- \sqrt{20 + {(- i \sqrt{2})}^{2}}$ .

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- i \sqrt{2}$ .

$x = - \sqrt{20 + {(- i)}^{2} {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

Apply the product rule to $- i$ .

$x = - \sqrt{20 + {(- 1)}^{2} i^{2} {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

$x = - \sqrt{20 + {(- 1)}^{2} i^{2} {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

Raise $- 1$ to the power of $2$ .

$x = - \sqrt{20 + 1 i^{2} {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

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

$x = - \sqrt{20 + i^{2} {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

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

$x = - \sqrt{20 - 1 {\sqrt{2}}^{2}}$

$y = - i \sqrt{2}$

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

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

$x = - \sqrt{20 - 1 {(2^{\frac{1}{2}})}^{2}}$

$y = - i \sqrt{2}$

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

$x = - \sqrt{20 - 1 \cdot 2^{\frac{1}{2} \cdot 2}}$

$y = - i \sqrt{2}$

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

$x = - \sqrt{20 - 1 \cdot 2^{\frac{2}{2}}}$

$y = - i \sqrt{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = - \sqrt{20 - 1 \cdot 2^{\frac{2}{2}}}$

$y = - i \sqrt{2}$

Rewrite the expression.

$x = - \sqrt{20 - 1 \cdot 2}$

$y = - i \sqrt{2}$

$x = - \sqrt{20 - 1 \cdot 2}$

$y = - i \sqrt{2}$

Evaluate the exponent.

$x = - \sqrt{20 - 1 \cdot 2}$

$y = - i \sqrt{2}$

$x = - \sqrt{20 - 1 \cdot 2}$

$y = - i \sqrt{2}$

Simplify the expression.

Multiply $- 1$ by $2$ .

$x = - \sqrt{20 - 2}$

$y = - i \sqrt{2}$

Subtract $2$ from $20$ .

$x = - \sqrt{18}$

$y = - i \sqrt{2}$

$x = - \sqrt{18}$

$y = - i \sqrt{2}$

Rewrite $18$ as $3^{2} \cdot 2$ .

Factor $9$ out of $18$ .

$x = - \sqrt{9 (2)}$

$y = - i \sqrt{2}$

Rewrite $9$ as $3^{2}$ .

$x = - \sqrt{3^{2} \cdot 2}$

$y = - i \sqrt{2}$

$x = - \sqrt{3^{2} \cdot 2}$

$y = - i \sqrt{2}$

Pull terms out from under the radical.

$x = - (3 \sqrt{2})$

$y = - i \sqrt{2}$

Multiply $3$ by $- 1$ .

$x = - 3 \sqrt{2}$

$y = - i \sqrt{2}$

$x = - 3 \sqrt{2}$

$y = - i \sqrt{2}$

$x = - 3 \sqrt{2}$

$y = - i \sqrt{2}$

$x = - 3 \sqrt{2}$

$y = - i \sqrt{2}$

$x = - 3 \sqrt{2}$

$y = - i \sqrt{2}$

Step 4

List all of the solutions.

$x = 3 \sqrt{2}, y = i \sqrt{2}$

$x = 3 \sqrt{2}, y = - i \sqrt{2}$

$x = - 3 \sqrt{2}, y = i \sqrt{2}$

$x = - 3 \sqrt{2}, y = - i \sqrt{2}$

Step 5

Solve the System of Equations x^2+y^2=16 x^2-y^2=20

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