Solve by Substitution 2x^2-4y^2=-8 , 5x^2+3y^2=32

$2 x^{2} - 4 y^{2} = - 8$ , $5 x^{2} + 3 y^{2} = 32$

Step 1

Solve for $x$ in $2 x^{2} - 4 y^{2} = - 8$ .

Add $4 y^{2}$ to both sides of the equation.

$2 x^{2} = - 8 + 4 y^{2}$

$5 x^{2} + 3 y^{2} = 32$

Divide each term in $2 x^{2} = - 8 + 4 y^{2}$ by $2$ and simplify.

Divide each term in $2 x^{2} = - 8 + 4 y^{2}$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{- 8}{2} + \frac{4 y^{2}}{2}$

$5 x^{2} + 3 y^{2} = 32$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{- 8}{2} + \frac{4 y^{2}}{2}$

$5 x^{2} + 3 y^{2} = 32$

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

$x^{2} = \frac{- 8}{2} + \frac{4 y^{2}}{2}$

$5 x^{2} + 3 y^{2} = 32$

$x^{2} = \frac{- 8}{2} + \frac{4 y^{2}}{2}$

$5 x^{2} + 3 y^{2} = 32$

$x^{2} = \frac{- 8}{2} + \frac{4 y^{2}}{2}$

$5 x^{2} + 3 y^{2} = 32$

Simplify the right side.

Simplify each term.

Divide $- 8$ by $2$ .

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

$5 x^{2} + 3 y^{2} = 32$

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4 y^{2}$ .

$x^{2} = - 4 + \frac{2 (2 y^{2})}{2}$

$5 x^{2} + 3 y^{2} = 32$

Cancel the common factors.

Factor $2$ out of $2$ .

$x^{2} = - 4 + \frac{2 (2 y^{2})}{2 (1)}$

$5 x^{2} + 3 y^{2} = 32$

Cancel the common factor.

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

$5 x^{2} + 3 y^{2} = 32$

Rewrite the expression.

$x^{2} = - 4 + \frac{2 y^{2}}{1}$

$5 x^{2} + 3 y^{2} = 32$

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

$x^{2} = - 4 + 2 y^{2}$

$5 x^{2} + 3 y^{2} = 32$

$x^{2} = - 4 + 2 y^{2}$

$5 x^{2} + 3 y^{2} = 32$

$x^{2} = - 4 + 2 y^{2}$

$5 x^{2} + 3 y^{2} = 32$

$x^{2} = - 4 + 2 y^{2}$

$5 x^{2} + 3 y^{2} = 32$

$x^{2} = - 4 + 2 y^{2}$

$5 x^{2} + 3 y^{2} = 32$

$x^{2} = - 4 + 2 y^{2}$

$5 x^{2} + 3 y^{2} = 32$

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

$x = \pm \sqrt{- 4 + 2 y^{2}}$

$5 x^{2} + 3 y^{2} = 32$

Factor $2$ out of $- 4 + 2 y^{2}$ .

Factor $2$ out of $- 4$ .

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

$5 x^{2} + 3 y^{2} = 32$

Factor $2$ out of $2 \cdot - 2 + 2 y^{2}$ .

$x = \pm \sqrt{2 (- 2 + y^{2})}$

$5 x^{2} + 3 y^{2} = 32$

$x = \pm \sqrt{2 (- 2 + y^{2})}$

$5 x^{2} + 3 y^{2} = 32$

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.

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

$5 x^{2} + 3 y^{2} = 32$

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

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

$5 x^{2} + 3 y^{2} = 32$

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

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

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

$5 x^{2} + 3 y^{2} = 32$

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

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

$5 x^{2} + 3 y^{2} = 32$

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

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

$5 x^{2} + 3 y^{2} = 32$

Step 2

Solve the system $x = \sqrt{2 (- 2 + y^{2})}, 5 x^{2} + 3 y^{2} = 32$ .

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

Replace all occurrences of $x$ in $5 x^{2} + 3 y^{2} = 32$ with $\sqrt{2 (- 2 + y^{2})}$ .

$5 {(\sqrt{2 (- 2 + y^{2})})}^{2} + 3 y^{2} = 32$

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

Simplify the left side.

Simplify $5 {(\sqrt{2 (- 2 + y^{2})})}^{2} + 3 y^{2}$ .

Simplify each term.

Rewrite ${\sqrt{2 (- 2 + y^{2})}}^{2}$ as $2 (- 2 + y^{2})$ .

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

$5 {({(2 (- 2 + y^{2}))}^{\frac{1}{2}})}^{2} + 3 y^{2} = 32$

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

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

$5 {(2 (- 2 + y^{2}))}^{\frac{1}{2} \cdot 2} + 3 y^{2} = 32$

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

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

$5 {(2 (- 2 + y^{2}))}^{\frac{2}{2}} + 3 y^{2} = 32$

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

Cancel the common factor of $2$ .

Cancel the common factor.

$5 {(2 (- 2 + y^{2}))}^{\frac{2}{2}} + 3 y^{2} = 32$

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

Rewrite the expression.

$5 (2 (- 2 + y^{2})) + 3 y^{2} = 32$

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

$5 (2 (- 2 + y^{2})) + 3 y^{2} = 32$

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

Simplify.

$5 (2 (- 2 + y^{2})) + 3 y^{2} = 32$

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

$5 (2 (- 2 + y^{2})) + 3 y^{2} = 32$

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

Apply the distributive property.

$5 (2 \cdot - 2 + 2 y^{2}) + 3 y^{2} = 32$

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

Multiply $2$ by $- 2$ .

$5 (- 4 + 2 y^{2}) + 3 y^{2} = 32$

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

Apply the distributive property.

$5 \cdot - 4 + 5 (2 y^{2}) + 3 y^{2} = 32$

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

Multiply $5$ by $- 4$ .

$- 20 + 5 (2 y^{2}) + 3 y^{2} = 32$

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

Multiply $2$ by $5$ .

$- 20 + 10 y^{2} + 3 y^{2} = 32$

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

$- 20 + 10 y^{2} + 3 y^{2} = 32$

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

Add $10 y^{2}$ and $3 y^{2}$ .

$- 20 + 13 y^{2} = 32$

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

$- 20 + 13 y^{2} = 32$

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

$- 20 + 13 y^{2} = 32$

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

$- 20 + 13 y^{2} = 32$

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

Solve for $y$ in $- 20 + 13 y^{2} = 32$ .

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

Add $20$ to both sides of the equation.

$13 y^{2} = 32 + 20$

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

Add $32$ and $20$ .

$13 y^{2} = 52$

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

$13 y^{2} = 52$

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

Divide each term in $13 y^{2} = 52$ by $13$ and simplify.

Divide each term in $13 y^{2} = 52$ by $13$ .

$\frac{13 y^{2}}{13} = \frac{52}{13}$

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

Simplify the left side.

Cancel the common factor of $13$ .

Cancel the common factor.

$\frac{13 y^{2}}{13} = \frac{52}{13}$

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

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

$y^{2} = \frac{52}{13}$

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

$y^{2} = \frac{52}{13}$

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

$y^{2} = \frac{52}{13}$

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

Simplify the right side.

Divide $52$ by $13$ .

$y^{2} = 4$

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

$y^{2} = 4$

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

$y^{2} = 4$

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

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

$y = \pm \sqrt{4}$

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

Simplify $\pm \sqrt{4}$ .

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

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

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

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

$y = \pm 2$

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

$y = \pm 2$

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

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.

$y = 2$

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

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

$y = - 2$

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

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

$y = 2, - 2$

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

$y = 2, - 2$

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

$y = 2, - 2$

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

Replace all occurrences of $y$ with $2$ in each equation.

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

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

$y = 2$

Simplify the right side.

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

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

$x = \sqrt{2 (- 2 + 4)}$

$y = 2$

Add $- 2$ and $4$ .

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

$y = 2$

Multiply $2$ by $2$ .

$x = \sqrt{4}$

$y = 2$

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

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

$y = 2$

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

$x = 2$

$y = 2$

$x = 2$

$y = 2$

$x = 2$

$y = 2$

$x = 2$

$y = 2$

Replace all occurrences of $y$ with $- 2$ in each equation.

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

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

$y = - 2$

Simplify the right side.

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

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

$x = \sqrt{2 (- 2 + 4)}$

$y = - 2$

Add $- 2$ and $4$ .

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

$y = - 2$

Multiply $2$ by $2$ .

$x = \sqrt{4}$

$y = - 2$

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

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

$y = - 2$

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

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

Step 3

Solve the system $x = - \sqrt{2 (- 2 + y^{2})}, 5 x^{2} + 3 y^{2} = 32$ .

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

Replace all occurrences of $x$ in $5 x^{2} + 3 y^{2} = 32$ with $- \sqrt{2 (- 2 + y^{2})}$ .

$5 {(- \sqrt{2 (- 2 + y^{2})})}^{2} + 3 y^{2} = 32$

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

Simplify the left side.

Simplify $5 {(- \sqrt{2 (- 2 + y^{2})})}^{2} + 3 y^{2}$ .

Simplify each term.

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

$5 ({(- 1)}^{2} {\sqrt{2 (- 2 + y^{2})}}^{2}) + 3 y^{2} = 32$

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

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

$5 (1 {\sqrt{2 (- 2 + y^{2})}}^{2}) + 3 y^{2} = 32$

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

Multiply ${\sqrt{2 (- 2 + y^{2})}}^{2}$ by $1$ .

$5 {\sqrt{2 (- 2 + y^{2})}}^{2} + 3 y^{2} = 32$

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

Rewrite ${\sqrt{2 (- 2 + y^{2})}}^{2}$ as $2 (- 2 + y^{2})$ .

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

$5 {({(2 (- 2 + y^{2}))}^{\frac{1}{2}})}^{2} + 3 y^{2} = 32$

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

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

$5 {(2 (- 2 + y^{2}))}^{\frac{1}{2} \cdot 2} + 3 y^{2} = 32$

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

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

$5 {(2 (- 2 + y^{2}))}^{\frac{2}{2}} + 3 y^{2} = 32$

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

Cancel the common factor of $2$ .

Cancel the common factor.

$5 {(2 (- 2 + y^{2}))}^{\frac{2}{2}} + 3 y^{2} = 32$

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

Rewrite the expression.

$5 (2 (- 2 + y^{2})) + 3 y^{2} = 32$

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

$5 (2 (- 2 + y^{2})) + 3 y^{2} = 32$

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

Simplify.

$5 (2 (- 2 + y^{2})) + 3 y^{2} = 32$

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

$5 (2 (- 2 + y^{2})) + 3 y^{2} = 32$

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

Apply the distributive property.

$5 (2 \cdot - 2 + 2 y^{2}) + 3 y^{2} = 32$

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

Multiply $2$ by $- 2$ .

$5 (- 4 + 2 y^{2}) + 3 y^{2} = 32$

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

Apply the distributive property.

$5 \cdot - 4 + 5 (2 y^{2}) + 3 y^{2} = 32$

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

Multiply $5$ by $- 4$ .

$- 20 + 5 (2 y^{2}) + 3 y^{2} = 32$

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

Multiply $2$ by $5$ .

$- 20 + 10 y^{2} + 3 y^{2} = 32$

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

$- 20 + 10 y^{2} + 3 y^{2} = 32$

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

Add $10 y^{2}$ and $3 y^{2}$ .

$- 20 + 13 y^{2} = 32$

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

$- 20 + 13 y^{2} = 32$

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

$- 20 + 13 y^{2} = 32$

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

$- 20 + 13 y^{2} = 32$

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

Solve for $y$ in $- 20 + 13 y^{2} = 32$ .

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

Add $20$ to both sides of the equation.

$13 y^{2} = 32 + 20$

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

Add $32$ and $20$ .

$13 y^{2} = 52$

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

$13 y^{2} = 52$

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

Divide each term in $13 y^{2} = 52$ by $13$ and simplify.

Divide each term in $13 y^{2} = 52$ by $13$ .

$\frac{13 y^{2}}{13} = \frac{52}{13}$

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

Simplify the left side.

Cancel the common factor of $13$ .

Cancel the common factor.

$\frac{13 y^{2}}{13} = \frac{52}{13}$

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

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

$y^{2} = \frac{52}{13}$

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

$y^{2} = \frac{52}{13}$

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

$y^{2} = \frac{52}{13}$

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

Simplify the right side.

Divide $52$ by $13$ .

$y^{2} = 4$

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

$y^{2} = 4$

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

$y^{2} = 4$

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

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

$y = \pm \sqrt{4}$

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

Simplify $\pm \sqrt{4}$ .

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

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

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

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

$y = \pm 2$

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

$y = \pm 2$

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

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.

$y = 2$

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

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

$y = - 2$

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

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

$y = 2, - 2$

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

$y = 2, - 2$

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

$y = 2, - 2$

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

Replace all occurrences of $y$ with $2$ in each equation.

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

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

$y = 2$

Simplify the right side.

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

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

$x = - \sqrt{2 (- 2 + 4)}$

$y = 2$

Add $- 2$ and $4$ .

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

$y = 2$

Multiply $2$ by $2$ .

$x = - \sqrt{4}$

$y = 2$

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

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

$y = 2$

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

$x = - 1 \cdot 2$

$y = 2$

Multiply $- 1$ by $2$ .

$x = - 2$

$y = 2$

$x = - 2$

$y = 2$

$x = - 2$

$y = 2$

$x = - 2$

$y = 2$

Replace all occurrences of $y$ with $- 2$ in each equation.

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

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

$y = - 2$

Simplify the right side.

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

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

$x = - \sqrt{2 (- 2 + 4)}$

$y = - 2$

Add $- 2$ and $4$ .

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

$y = - 2$

Multiply $2$ by $2$ .

$x = - \sqrt{4}$

$y = - 2$

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

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

$y = - 2$

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

$x = - 1 \cdot 2$

$y = - 2$

Multiply $- 1$ by $2$ .

$x = - 2$

$y = - 2$

$x = - 2$

$y = - 2$

$x = - 2$

$y = - 2$

$x = - 2$

$y = - 2$

$x = - 2$

$y = - 2$

Step 4

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

$(2, 2)$

$(2, - 2)$

$(- 2, 2)$

$(- 2, - 2)$

Step 5

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 2, y = 2$

$x = 2, y = - 2$

$x = - 2, y = 2$

$x = - 2, y = - 2$

Step 6

Solve by Substitution 2x^2-4y^2=-8 , 5x^2+3y^2=32

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