Solve the System of Equations 2x^2+8y^2=16 x^2+y^2=5

$2 x^{2} + 8 y^{2} = 16$ $x^{2} + y^{2} = 5$

Step 1

Solve for $x$ in $x^{2} + y^{2} = 5$ .

Subtract $y^{2}$ from both sides of the equation.

$x^{2} = 5 - y^{2}$

$2 x^{2} + 8 y^{2} = 16$

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

$x = \pm \sqrt{5 - y^{2}}$

$2 x^{2} + 8 y^{2} = 16$

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.

$x = \sqrt{5 - y^{2}}$

$2 x^{2} + 8 y^{2} = 16$

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

$x = - \sqrt{5 - y^{2}}$

$2 x^{2} + 8 y^{2} = 16$

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

$x = \sqrt{5 - y^{2}}$

$x = - \sqrt{5 - y^{2}}$

$2 x^{2} + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

$x = - \sqrt{5 - y^{2}}$

$2 x^{2} + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

$x = - \sqrt{5 - y^{2}}$

$2 x^{2} + 8 y^{2} = 16$

Step 2

Solve the system $x = \sqrt{5 - y^{2}}, 2 x^{2} + 8 y^{2} = 16$ .

Replace all occurrences of $x$ with $\sqrt{5 - y^{2}}$ in each equation.

Replace all occurrences of $x$ in $2 x^{2} + 8 y^{2} = 16$ with $\sqrt{5 - y^{2}}$ .

$2 {(\sqrt{5 - y^{2}})}^{2} + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

Simplify the left side.

Simplify $2 {(\sqrt{5 - y^{2}})}^{2} + 8 y^{2}$ .

Simplify each term.

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

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

$2 {({(5 - y^{2})}^{\frac{1}{2}})}^{2} + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

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

$2 {(5 - y^{2})}^{\frac{1}{2} \cdot 2} + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

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

$2 {(5 - y^{2})}^{\frac{2}{2}} + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$2 {(5 - y^{2})}^{\frac{2}{2}} + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

Rewrite the expression.

$2 (5 - y^{2}) + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

$2 (5 - y^{2}) + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

Simplify.

$2 (5 - y^{2}) + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

$2 (5 - y^{2}) + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

Apply the distributive property.

$2 \cdot 5 + 2 (- y^{2}) + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

Multiply $2$ by $5$ .

$10 + 2 (- y^{2}) + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

Multiply $- 1$ by $2$ .

$10 - 2 y^{2} + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

$10 - 2 y^{2} + 8 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

Add $- 2 y^{2}$ and $8 y^{2}$ .

$10 + 6 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

$10 + 6 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

$10 + 6 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

$10 + 6 y^{2} = 16$

$x = \sqrt{5 - y^{2}}$

Solve for $y$ in $10 + 6 y^{2} = 16$ .

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

Subtract $10$ from both sides of the equation.

$6 y^{2} = 16 - 10$

$x = \sqrt{5 - y^{2}}$

Subtract $10$ from $16$ .

$6 y^{2} = 6$

$x = \sqrt{5 - y^{2}}$

$6 y^{2} = 6$

$x = \sqrt{5 - y^{2}}$

Divide each term in $6 y^{2} = 6$ by $6$ and simplify.

Divide each term in $6 y^{2} = 6$ by $6$ .

$\frac{6 y^{2}}{6} = \frac{6}{6}$

$x = \sqrt{5 - y^{2}}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 y^{2}}{6} = \frac{6}{6}$

$x = \sqrt{5 - y^{2}}$

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

$y^{2} = \frac{6}{6}$

$x = \sqrt{5 - y^{2}}$

$y^{2} = \frac{6}{6}$

$x = \sqrt{5 - y^{2}}$

$y^{2} = \frac{6}{6}$

$x = \sqrt{5 - y^{2}}$

Simplify the right side.

Divide $6$ by $6$ .

$y^{2} = 1$

$x = \sqrt{5 - y^{2}}$

$y^{2} = 1$

$x = \sqrt{5 - y^{2}}$

$y^{2} = 1$

$x = \sqrt{5 - y^{2}}$

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

$y = \pm \sqrt{1}$

$x = \sqrt{5 - y^{2}}$

Any root of $1$ is $1$ .

$y = \pm 1$

$x = \sqrt{5 - y^{2}}$

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.

$y = 1$

$x = \sqrt{5 - y^{2}}$

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

$y = - 1$

$x = \sqrt{5 - y^{2}}$

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

$y = 1, - 1$

$x = \sqrt{5 - y^{2}}$

$y = 1, - 1$

$x = \sqrt{5 - y^{2}}$

$y = 1, - 1$

$x = \sqrt{5 - y^{2}}$

Replace all occurrences of $y$ with $1$ in each equation.

Replace all occurrences of $y$ in $x = \sqrt{5 - y^{2}}$ with $1$ .

$x = \sqrt{5 - {(1)}^{2}}$

$y = 1$

Simplify the right side.

Simplify $\sqrt{5 - {(1)}^{2}}$ .

One to any power is one.

$x = \sqrt{5 - 1 \cdot 1}$

$y = 1$

Multiply $- 1$ by $1$ .

$x = \sqrt{5 - 1}$

$y = 1$

Subtract $1$ from $5$ .

$x = \sqrt{4}$

$y = 1$

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

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

$y = 1$

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

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

Replace all occurrences of $y$ with $- 1$ in each equation.

Replace all occurrences of $y$ in $x = \sqrt{5 - y^{2}}$ with $- 1$ .

$x = \sqrt{5 - {(- 1)}^{2}}$

$y = - 1$

Simplify the right side.

Simplify $\sqrt{5 - {(- 1)}^{2}}$ .

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

Multiply $- 1$ by ${(- 1)}^{2}$ .

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

$x = \sqrt{5 + (- 1) {(- 1)}^{2}}$

$y = - 1$

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

$x = \sqrt{5 + {(- 1)}^{1 + 2}}$

$y = - 1$

$x = \sqrt{5 + {(- 1)}^{1 + 2}}$

$y = - 1$

Add $1$ and $2$ .

$x = \sqrt{5 + {(- 1)}^{3}}$

$y = - 1$

$x = \sqrt{5 + {(- 1)}^{3}}$

$y = - 1$

Raise $- 1$ to the power of $3$ .

$x = \sqrt{5 - 1}$

$y = - 1$

Subtract $1$ from $5$ .

$x = \sqrt{4}$

$y = - 1$

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

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

$y = - 1$

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

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

Step 3

Solve the system $x = - \sqrt{5 - y^{2}}, 2 x^{2} + 8 y^{2} = 16$ .

Replace all occurrences of $x$ with $- \sqrt{5 - y^{2}}$ in each equation.

Replace all occurrences of $x$ in $2 x^{2} + 8 y^{2} = 16$ with $- \sqrt{5 - y^{2}}$ .

$2 {(- \sqrt{5 - y^{2}})}^{2} + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

Simplify the left side.

Simplify $2 {(- \sqrt{5 - y^{2}})}^{2} + 8 y^{2}$ .

Simplify each term.

Apply the product rule to $- \sqrt{5 - y^{2}}$ .

$2 ({(- 1)}^{2} {\sqrt{5 - y^{2}}}^{2}) + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

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

$2 (1 {\sqrt{5 - y^{2}}}^{2}) + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

Multiply ${\sqrt{5 - y^{2}}}^{2}$ by $1$ .

$2 {\sqrt{5 - y^{2}}}^{2} + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

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

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

$2 {({(5 - y^{2})}^{\frac{1}{2}})}^{2} + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

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

$2 {(5 - y^{2})}^{\frac{1}{2} \cdot 2} + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

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

$2 {(5 - y^{2})}^{\frac{2}{2}} + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$2 {(5 - y^{2})}^{\frac{2}{2}} + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

Rewrite the expression.

$2 (5 - y^{2}) + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

$2 (5 - y^{2}) + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

Simplify.

$2 (5 - y^{2}) + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

$2 (5 - y^{2}) + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

Apply the distributive property.

$2 \cdot 5 + 2 (- y^{2}) + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

Multiply $2$ by $5$ .

$10 + 2 (- y^{2}) + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

Multiply $- 1$ by $2$ .

$10 - 2 y^{2} + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

$10 - 2 y^{2} + 8 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

Add $- 2 y^{2}$ and $8 y^{2}$ .

$10 + 6 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

$10 + 6 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

$10 + 6 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

$10 + 6 y^{2} = 16$

$x = - \sqrt{5 - y^{2}}$

Solve for $y$ in $10 + 6 y^{2} = 16$ .

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

Subtract $10$ from both sides of the equation.

$6 y^{2} = 16 - 10$

$x = - \sqrt{5 - y^{2}}$

Subtract $10$ from $16$ .

$6 y^{2} = 6$

$x = - \sqrt{5 - y^{2}}$

$6 y^{2} = 6$

$x = - \sqrt{5 - y^{2}}$

Divide each term in $6 y^{2} = 6$ by $6$ and simplify.

Divide each term in $6 y^{2} = 6$ by $6$ .

$\frac{6 y^{2}}{6} = \frac{6}{6}$

$x = - \sqrt{5 - y^{2}}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 y^{2}}{6} = \frac{6}{6}$

$x = - \sqrt{5 - y^{2}}$

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

$y^{2} = \frac{6}{6}$

$x = - \sqrt{5 - y^{2}}$

$y^{2} = \frac{6}{6}$

$x = - \sqrt{5 - y^{2}}$

$y^{2} = \frac{6}{6}$

$x = - \sqrt{5 - y^{2}}$

Simplify the right side.

Divide $6$ by $6$ .

$y^{2} = 1$

$x = - \sqrt{5 - y^{2}}$

$y^{2} = 1$

$x = - \sqrt{5 - y^{2}}$

$y^{2} = 1$

$x = - \sqrt{5 - y^{2}}$

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

$y = \pm \sqrt{1}$

$x = - \sqrt{5 - y^{2}}$

Any root of $1$ is $1$ .

$y = \pm 1$

$x = - \sqrt{5 - y^{2}}$

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.

$y = 1$

$x = - \sqrt{5 - y^{2}}$

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

$y = - 1$

$x = - \sqrt{5 - y^{2}}$

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

$y = 1, - 1$

$x = - \sqrt{5 - y^{2}}$

$y = 1, - 1$

$x = - \sqrt{5 - y^{2}}$

$y = 1, - 1$

$x = - \sqrt{5 - y^{2}}$

Replace all occurrences of $y$ with $1$ in each equation.

Replace all occurrences of $y$ in $x = - \sqrt{5 - y^{2}}$ with $1$ .

$x = - \sqrt{5 - {(1)}^{2}}$

$y = 1$

Simplify the right side.

Simplify $- \sqrt{5 - {(1)}^{2}}$ .

One to any power is one.

$x = - \sqrt{5 - 1 \cdot 1}$

$y = 1$

Multiply $- 1$ by $1$ .

$x = - \sqrt{5 - 1}$

$y = 1$

Subtract $1$ from $5$ .

$x = - \sqrt{4}$

$y = 1$

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

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

$y = 1$

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

$x = - 1 \cdot 2$

$y = 1$

Multiply $- 1$ by $2$ .

$x = - 2$

$y = 1$

$x = - 2$

$y = 1$

$x = - 2$

$y = 1$

$x = - 2$

$y = 1$

Replace all occurrences of $y$ with $- 1$ in each equation.

Replace all occurrences of $y$ in $x = - \sqrt{5 - y^{2}}$ with $- 1$ .

$x = - \sqrt{5 - {(- 1)}^{2}}$

$y = - 1$

Simplify the right side.

Simplify $- \sqrt{5 - {(- 1)}^{2}}$ .

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

Multiply $- 1$ by ${(- 1)}^{2}$ .

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

$x = - \sqrt{5 + (- 1) {(- 1)}^{2}}$

$y = - 1$

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

$x = - \sqrt{5 + {(- 1)}^{1 + 2}}$

$y = - 1$

$x = - \sqrt{5 + {(- 1)}^{1 + 2}}$

$y = - 1$

Add $1$ and $2$ .

$x = - \sqrt{5 + {(- 1)}^{3}}$

$y = - 1$

$x = - \sqrt{5 + {(- 1)}^{3}}$

$y = - 1$

Raise $- 1$ to the power of $3$ .

$x = - \sqrt{5 - 1}$

$y = - 1$

Subtract $1$ from $5$ .

$x = - \sqrt{4}$

$y = - 1$

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

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

$y = - 1$

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

$x = - 1 \cdot 2$

$y = - 1$

Multiply $- 1$ by $2$ .

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

Step 4

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(2, 1)$

$(2, - 1)$

$(- 2, 1)$

$(- 2, - 1)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(2, 1), (2, - 1), (- 2, 1), (- 2, - 1)$

Equation Form:

$x = 2, y = 1$

$x = 2, y = - 1$

$x = - 2, y = 1$

$x = - 2, y = - 1$

Step 6

Solve the System of Equations 2x^2+8y^2=16 x^2+y^2=5

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