Solve the System of Equations x^2=2y^2+18 x^2-3y^2=9

$x^{2} = 2 y^{2} + 18$ $x^{2} - 3 y^{2} = 9$

Step 1

Solve for $x$ in $x^{2} = 2 y^{2} + 18$ .

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

$x = \pm \sqrt{2 y^{2} + 18}$

$x^{2} - 3 y^{2} = 9$

Factor $2$ out of $2 y^{2} + 18$ .

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

$x = \pm \sqrt{2 (y^{2}) + 18}$

$x^{2} - 3 y^{2} = 9$

Factor $2$ out of $18$ .

$x = \pm \sqrt{2 y^{2} + 2 \cdot 9}$

$x^{2} - 3 y^{2} = 9$

Factor $2$ out of $2 y^{2} + 2 \cdot 9$ .

$x = \pm \sqrt{2 (y^{2} + 9)}$

$x^{2} - 3 y^{2} = 9$

$x = \pm \sqrt{2 (y^{2} + 9)}$

$x^{2} - 3 y^{2} = 9$

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.

$x = \sqrt{2 (y^{2} + 9)}$

$x^{2} - 3 y^{2} = 9$

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

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

$x^{2} - 3 y^{2} = 9$

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

$x = \sqrt{2 (y^{2} + 9)}$

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

$x^{2} - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

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

$x^{2} - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

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

$x^{2} - 3 y^{2} = 9$

Step 2

Solve the system $x = \sqrt{2 (y^{2} + 9)}, x^{2} - 3 y^{2} = 9$ .

Replace all occurrences of $x$ with $\sqrt{2 (y^{2} + 9)}$ in each equation.

Replace all occurrences of $x$ in $x^{2} - 3 y^{2} = 9$ with $\sqrt{2 (y^{2} + 9)}$ .

${(\sqrt{2 (y^{2} + 9)})}^{2} - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

Simplify the left side.

Simplify ${(\sqrt{2 (y^{2} + 9)})}^{2} - 3 y^{2}$ .

Simplify each term.

Rewrite ${\sqrt{2 (y^{2} + 9)}}^{2}$ as $2 (y^{2} + 9)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 (y^{2} + 9)}$ as ${(2 (y^{2} + 9))}^{\frac{1}{2}}$ .

${({(2 (y^{2} + 9))}^{\frac{1}{2}})}^{2} - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

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

${(2 (y^{2} + 9))}^{\frac{1}{2} \cdot 2} - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

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

${(2 (y^{2} + 9))}^{\frac{2}{2}} - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 (y^{2} + 9))}^{\frac{2}{2}} - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

Rewrite the expression.

$(2 (y^{2} + 9)) - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

$(2 (y^{2} + 9)) - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

Simplify.

$2 (y^{2} + 9) - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

$2 (y^{2} + 9) - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

Apply the distributive property.

$2 y^{2} + 2 \cdot 9 - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

Multiply $2$ by $9$ .

$2 y^{2} + 18 - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

$2 y^{2} + 18 - 3 y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

Subtract $3 y^{2}$ from $2 y^{2}$ .

$- y^{2} + 18 = 9$

$x = \sqrt{2 (y^{2} + 9)}$

$- y^{2} + 18 = 9$

$x = \sqrt{2 (y^{2} + 9)}$

$- y^{2} + 18 = 9$

$x = \sqrt{2 (y^{2} + 9)}$

$- y^{2} + 18 = 9$

$x = \sqrt{2 (y^{2} + 9)}$

Solve for $y$ in $- y^{2} + 18 = 9$ .

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

Subtract $18$ from both sides of the equation.

$- y^{2} = 9 - 18$

$x = \sqrt{2 (y^{2} + 9)}$

Subtract $18$ from $9$ .

$- y^{2} = - 9$

$x = \sqrt{2 (y^{2} + 9)}$

$- y^{2} = - 9$

$x = \sqrt{2 (y^{2} + 9)}$

Divide each term in $- y^{2} = - 9$ by $- 1$ and simplify.

Divide each term in $- y^{2} = - 9$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{- 9}{- 1}$

$x = \sqrt{2 (y^{2} + 9)}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{- 9}{- 1}$

$x = \sqrt{2 (y^{2} + 9)}$

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

$y^{2} = \frac{- 9}{- 1}$

$x = \sqrt{2 (y^{2} + 9)}$

$y^{2} = \frac{- 9}{- 1}$

$x = \sqrt{2 (y^{2} + 9)}$

Simplify the right side.

Divide $- 9$ by $- 1$ .

$y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

$y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

$y^{2} = 9$

$x = \sqrt{2 (y^{2} + 9)}$

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

$y = \pm \sqrt{9}$

$x = \sqrt{2 (y^{2} + 9)}$

Simplify $\pm \sqrt{9}$ .

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

$y = \pm \sqrt{3^{2}}$

$x = \sqrt{2 (y^{2} + 9)}$

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

$y = \pm 3$

$x = \sqrt{2 (y^{2} + 9)}$

$y = \pm 3$

$x = \sqrt{2 (y^{2} + 9)}$

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.

$y = 3$

$x = \sqrt{2 (y^{2} + 9)}$

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

$y = - 3$

$x = \sqrt{2 (y^{2} + 9)}$

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

$y = 3, - 3$

$x = \sqrt{2 (y^{2} + 9)}$

$y = 3, - 3$

$x = \sqrt{2 (y^{2} + 9)}$

$y = 3, - 3$

$x = \sqrt{2 (y^{2} + 9)}$

Replace all occurrences of $y$ with $3$ in each equation.

Replace all occurrences of $y$ in $x = \sqrt{2 (y^{2} + 9)}$ with $3$ .

$x = \sqrt{2 ({(3)}^{2} + 9)}$

$y = 3$

Simplify the right side.

Simplify $\sqrt{2 ({(3)}^{2} + 9)}$ .

Raise $3$ to the power of $2$ .

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

$y = 3$

Add $9$ and $9$ .

$x = \sqrt{2 \cdot 18}$

$y = 3$

Multiply $2$ by $18$ .

$x = \sqrt{36}$

$y = 3$

Rewrite $36$ as $6^{2}$ .

$x = \sqrt{6^{2}}$

$y = 3$

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

$x = 6$

$y = 3$

$x = 6$

$y = 3$

$x = 6$

$y = 3$

$x = 6$

$y = 3$

Replace all occurrences of $y$ with $- 3$ in each equation.

Replace all occurrences of $y$ in $x = \sqrt{2 (y^{2} + 9)}$ with $- 3$ .

$x = \sqrt{2 ({(- 3)}^{2} + 9)}$

$y = - 3$

Simplify the right side.

Simplify $\sqrt{2 ({(- 3)}^{2} + 9)}$ .

Raise $- 3$ to the power of $2$ .

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

$y = - 3$

Add $9$ and $9$ .

$x = \sqrt{2 \cdot 18}$

$y = - 3$

Multiply $2$ by $18$ .

$x = \sqrt{36}$

$y = - 3$

Rewrite $36$ as $6^{2}$ .

$x = \sqrt{6^{2}}$

$y = - 3$

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

$x = 6$

$y = - 3$

$x = 6$

$y = - 3$

$x = 6$

$y = - 3$

$x = 6$

$y = - 3$

$x = 6$

$y = - 3$

Step 3

Solve the system $x = - \sqrt{2 (y^{2} + 9)}, x^{2} - 3 y^{2} = 9$ .

Replace all occurrences of $x$ with $- \sqrt{2 (y^{2} + 9)}$ in each equation.

Replace all occurrences of $x$ in $x^{2} - 3 y^{2} = 9$ with $- \sqrt{2 (y^{2} + 9)}$ .

${(- \sqrt{2 (y^{2} + 9)})}^{2} - 3 y^{2} = 9$

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

Simplify the left side.

Simplify ${(- \sqrt{2 (y^{2} + 9)})}^{2} - 3 y^{2}$ .

Simplify each term.

Apply the product rule to $- \sqrt{2 (y^{2} + 9)}$ .

${(- 1)}^{2} {\sqrt{2 (y^{2} + 9)}}^{2} - 3 y^{2} = 9$

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

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

$1 {\sqrt{2 (y^{2} + 9)}}^{2} - 3 y^{2} = 9$

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

Multiply ${\sqrt{2 (y^{2} + 9)}}^{2}$ by $1$ .

${\sqrt{2 (y^{2} + 9)}}^{2} - 3 y^{2} = 9$

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

Rewrite ${\sqrt{2 (y^{2} + 9)}}^{2}$ as $2 (y^{2} + 9)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 (y^{2} + 9)}$ as ${(2 (y^{2} + 9))}^{\frac{1}{2}}$ .

${({(2 (y^{2} + 9))}^{\frac{1}{2}})}^{2} - 3 y^{2} = 9$

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

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

${(2 (y^{2} + 9))}^{\frac{1}{2} \cdot 2} - 3 y^{2} = 9$

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

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

${(2 (y^{2} + 9))}^{\frac{2}{2}} - 3 y^{2} = 9$

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

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 (y^{2} + 9))}^{\frac{2}{2}} - 3 y^{2} = 9$

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

Rewrite the expression.

$(2 (y^{2} + 9)) - 3 y^{2} = 9$

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

$(2 (y^{2} + 9)) - 3 y^{2} = 9$

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

Simplify.

$2 (y^{2} + 9) - 3 y^{2} = 9$

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

$2 (y^{2} + 9) - 3 y^{2} = 9$

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

Apply the distributive property.

$2 y^{2} + 2 \cdot 9 - 3 y^{2} = 9$

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

Multiply $2$ by $9$ .

$2 y^{2} + 18 - 3 y^{2} = 9$

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

$2 y^{2} + 18 - 3 y^{2} = 9$

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

Subtract $3 y^{2}$ from $2 y^{2}$ .

$- y^{2} + 18 = 9$

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

$- y^{2} + 18 = 9$

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

$- y^{2} + 18 = 9$

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

$- y^{2} + 18 = 9$

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

Solve for $y$ in $- y^{2} + 18 = 9$ .

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

Subtract $18$ from both sides of the equation.

$- y^{2} = 9 - 18$

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

Subtract $18$ from $9$ .

$- y^{2} = - 9$

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

$- y^{2} = - 9$

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

Divide each term in $- y^{2} = - 9$ by $- 1$ and simplify.

Divide each term in $- y^{2} = - 9$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{- 9}{- 1}$

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

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{- 9}{- 1}$

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

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

$y^{2} = \frac{- 9}{- 1}$

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

$y^{2} = \frac{- 9}{- 1}$

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

Simplify the right side.

Divide $- 9$ by $- 1$ .

$y^{2} = 9$

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

$y^{2} = 9$

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

$y^{2} = 9$

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

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

$y = \pm \sqrt{9}$

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

Simplify $\pm \sqrt{9}$ .

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

$y = \pm \sqrt{3^{2}}$

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

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

$y = \pm 3$

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

$y = \pm 3$

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

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.

$y = 3$

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

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

$y = - 3$

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

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

$y = 3, - 3$

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

$y = 3, - 3$

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

$y = 3, - 3$

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

Replace all occurrences of $y$ with $3$ in each equation.

Replace all occurrences of $y$ in $x = - \sqrt{2 (y^{2} + 9)}$ with $3$ .

$x = - \sqrt{2 ({(3)}^{2} + 9)}$

$y = 3$

Simplify the right side.

Simplify $- \sqrt{2 ({(3)}^{2} + 9)}$ .

Raise $3$ to the power of $2$ .

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

$y = 3$

Add $9$ and $9$ .

$x = - \sqrt{2 \cdot 18}$

$y = 3$

Multiply $2$ by $18$ .

$x = - \sqrt{36}$

$y = 3$

Rewrite $36$ as $6^{2}$ .

$x = - \sqrt{6^{2}}$

$y = 3$

Multiply.

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

$x = - 1 \cdot 6$

$y = 3$

Multiply $- 1$ by $6$ .

$x = - 6$

$y = 3$

$x = - 6$

$y = 3$

$x = - 6$

$y = 3$

$x = - 6$

$y = 3$

$x = - 6$

$y = 3$

Replace all occurrences of $y$ with $- 3$ in each equation.

Replace all occurrences of $y$ in $x = - \sqrt{2 (y^{2} + 9)}$ with $- 3$ .

$x = - \sqrt{2 ({(- 3)}^{2} + 9)}$

$y = - 3$

Simplify the right side.

Simplify $- \sqrt{2 ({(- 3)}^{2} + 9)}$ .

Raise $- 3$ to the power of $2$ .

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

$y = - 3$

Add $9$ and $9$ .

$x = - \sqrt{2 \cdot 18}$

$y = - 3$

Multiply $2$ by $18$ .

$x = - \sqrt{36}$

$y = - 3$

Rewrite $36$ as $6^{2}$ .

$x = - \sqrt{6^{2}}$

$y = - 3$

Multiply.

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

$x = - 1 \cdot 6$

$y = - 3$

Multiply $- 1$ by $6$ .

$x = - 6$

$y = - 3$

$x = - 6$

$y = - 3$

$x = - 6$

$y = - 3$

$x = - 6$

$y = - 3$

$x = - 6$

$y = - 3$

$x = - 6$

$y = - 3$

Step 4

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

$(6, 3)$

$(6, - 3)$

$(- 6, 3)$

$(- 6, - 3)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(6, 3), (6, - 3), (- 6, 3), (- 6, - 3)$

Equation Form:

$x = 6, y = 3$

$x = 6, y = - 3$

$x = - 6, y = 3$

$x = - 6, y = - 3$

Step 6

Solve the System of Equations x^2=2y^2+18 x^2-3y^2=9

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