Solve the System of Equations 3x^2+y^2=129 2x^2+y^2=113

$3 x^{2} + y^{2} = 129$ $2 x^{2} + y^{2} = 113$

Step 1

Solve for $y$ in $2 x^{2} + y^{2} = 113$ .

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

$y^{2} = 113 - 2 x^{2}$

$3 x^{2} + y^{2} = 129$

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

$y = \pm \sqrt{113 - 2 x^{2}}$

$3 x^{2} + y^{2} = 129$

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.

$y = \sqrt{113 - 2 x^{2}}$

$3 x^{2} + y^{2} = 129$

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

$y = - \sqrt{113 - 2 x^{2}}$

$3 x^{2} + y^{2} = 129$

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

$y = \sqrt{113 - 2 x^{2}}$

$y = - \sqrt{113 - 2 x^{2}}$

$3 x^{2} + y^{2} = 129$

$y = \sqrt{113 - 2 x^{2}}$

$y = - \sqrt{113 - 2 x^{2}}$

$3 x^{2} + y^{2} = 129$

$y = \sqrt{113 - 2 x^{2}}$

$y = - \sqrt{113 - 2 x^{2}}$

$3 x^{2} + y^{2} = 129$

Step 2

Solve the system $y = \sqrt{113 - 2 x^{2}}, 3 x^{2} + y^{2} = 129$ .

Replace all occurrences of $y$ with $\sqrt{113 - 2 x^{2}}$ in each equation.

Replace all occurrences of $y$ in $3 x^{2} + y^{2} = 129$ with $\sqrt{113 - 2 x^{2}}$ .

$3 x^{2} + {(\sqrt{113 - 2 x^{2}})}^{2} = 129$

$y = \sqrt{113 - 2 x^{2}}$

Simplify the left side.

Simplify $3 x^{2} + {(\sqrt{113 - 2 x^{2}})}^{2}$ .

Rewrite ${\sqrt{113 - 2 x^{2}}}^{2}$ as $113 - 2 x^{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{113 - 2 x^{2}}$ as ${(113 - 2 x^{2})}^{\frac{1}{2}}$ .

$3 x^{2} + {({(113 - 2 x^{2})}^{\frac{1}{2}})}^{2} = 129$

$y = \sqrt{113 - 2 x^{2}}$

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

$3 x^{2} + {(113 - 2 x^{2})}^{\frac{1}{2} \cdot 2} = 129$

$y = \sqrt{113 - 2 x^{2}}$

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

$3 x^{2} + {(113 - 2 x^{2})}^{\frac{2}{2}} = 129$

$y = \sqrt{113 - 2 x^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$3 x^{2} + {(113 - 2 x^{2})}^{\frac{2}{2}} = 129$

$y = \sqrt{113 - 2 x^{2}}$

Rewrite the expression.

$3 x^{2} + 113 - 2 x^{2} = 129$

$y = \sqrt{113 - 2 x^{2}}$

$3 x^{2} + 113 - 2 x^{2} = 129$

$y = \sqrt{113 - 2 x^{2}}$

Simplify.

$3 x^{2} + 113 - 2 x^{2} = 129$

$y = \sqrt{113 - 2 x^{2}}$

$3 x^{2} + 113 - 2 x^{2} = 129$

$y = \sqrt{113 - 2 x^{2}}$

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

$x^{2} + 113 = 129$

$y = \sqrt{113 - 2 x^{2}}$

$x^{2} + 113 = 129$

$y = \sqrt{113 - 2 x^{2}}$

$x^{2} + 113 = 129$

$y = \sqrt{113 - 2 x^{2}}$

$x^{2} + 113 = 129$

$y = \sqrt{113 - 2 x^{2}}$

Solve for $x$ in $x^{2} + 113 = 129$ .

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

Subtract $113$ from both sides of the equation.

$x^{2} = 129 - 113$

$y = \sqrt{113 - 2 x^{2}}$

Subtract $113$ from $129$ .

$x^{2} = 16$

$y = \sqrt{113 - 2 x^{2}}$

$x^{2} = 16$

$y = \sqrt{113 - 2 x^{2}}$

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

$x = \pm \sqrt{16}$

$y = \sqrt{113 - 2 x^{2}}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

$y = \sqrt{113 - 2 x^{2}}$

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

$x = \pm 4$

$y = \sqrt{113 - 2 x^{2}}$

$x = \pm 4$

$y = \sqrt{113 - 2 x^{2}}$

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.

$x = 4$

$y = \sqrt{113 - 2 x^{2}}$

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

$x = - 4$

$y = \sqrt{113 - 2 x^{2}}$

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

$x = 4, - 4$

$y = \sqrt{113 - 2 x^{2}}$

$x = 4, - 4$

$y = \sqrt{113 - 2 x^{2}}$

$x = 4, - 4$

$y = \sqrt{113 - 2 x^{2}}$

Replace all occurrences of $x$ with $4$ in each equation.

Replace all occurrences of $x$ in $y = \sqrt{113 - 2 x^{2}}$ with $4$ .

$y = \sqrt{113 - 2 {(4)}^{2}}$

$x = 4$

Simplify the right side.

Simplify $\sqrt{113 - 2 {(4)}^{2}}$ .

Raise $4$ to the power of $2$ .

$y = \sqrt{113 - 2 \cdot 16}$

$x = 4$

Multiply $- 2$ by $16$ .

$y = \sqrt{113 - 32}$

$x = 4$

Subtract $32$ from $113$ .

$y = \sqrt{81}$

$x = 4$

Rewrite $81$ as $9^{2}$ .

$y = \sqrt{9^{2}}$

$x = 4$

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

$y = 9$

$x = 4$

$y = 9$

$x = 4$

$y = 9$

$x = 4$

$y = 9$

$x = 4$

Replace all occurrences of $x$ with $- 4$ in each equation.

Replace all occurrences of $x$ in $y = \sqrt{113 - 2 x^{2}}$ with $- 4$ .

$y = \sqrt{113 - 2 {(- 4)}^{2}}$

$x = - 4$

Simplify the right side.

Simplify $\sqrt{113 - 2 {(- 4)}^{2}}$ .

Raise $- 4$ to the power of $2$ .

$y = \sqrt{113 - 2 \cdot 16}$

$x = - 4$

Multiply $- 2$ by $16$ .

$y = \sqrt{113 - 32}$

$x = - 4$

Subtract $32$ from $113$ .

$y = \sqrt{81}$

$x = - 4$

Rewrite $81$ as $9^{2}$ .

$y = \sqrt{9^{2}}$

$x = - 4$

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

$y = 9$

$x = - 4$

$y = 9$

$x = - 4$

$y = 9$

$x = - 4$

$y = 9$

$x = - 4$

$y = 9$

$x = - 4$

Step 3

Solve the system $y = - \sqrt{113 - 2 x^{2}}, 3 x^{2} + y^{2} = 129$ .

Replace all occurrences of $y$ with $- \sqrt{113 - 2 x^{2}}$ in each equation.

Replace all occurrences of $y$ in $3 x^{2} + y^{2} = 129$ with $- \sqrt{113 - 2 x^{2}}$ .

$3 x^{2} + {(- \sqrt{113 - 2 x^{2}})}^{2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

Simplify the left side.

Simplify $3 x^{2} + {(- \sqrt{113 - 2 x^{2}})}^{2}$ .

Simplify each term.

Apply the product rule to $- \sqrt{113 - 2 x^{2}}$ .

$3 x^{2} + {(- 1)}^{2} {\sqrt{113 - 2 x^{2}}}^{2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

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

$3 x^{2} + 1 {\sqrt{113 - 2 x^{2}}}^{2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

Multiply ${\sqrt{113 - 2 x^{2}}}^{2}$ by $1$ .

$3 x^{2} + {\sqrt{113 - 2 x^{2}}}^{2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

Rewrite ${\sqrt{113 - 2 x^{2}}}^{2}$ as $113 - 2 x^{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{113 - 2 x^{2}}$ as ${(113 - 2 x^{2})}^{\frac{1}{2}}$ .

$3 x^{2} + {({(113 - 2 x^{2})}^{\frac{1}{2}})}^{2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

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

$3 x^{2} + {(113 - 2 x^{2})}^{\frac{1}{2} \cdot 2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

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

$3 x^{2} + {(113 - 2 x^{2})}^{\frac{2}{2}} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$3 x^{2} + {(113 - 2 x^{2})}^{\frac{2}{2}} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

Rewrite the expression.

$3 x^{2} + 113 - 2 x^{2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

$3 x^{2} + 113 - 2 x^{2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

Simplify.

$3 x^{2} + 113 - 2 x^{2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

$3 x^{2} + 113 - 2 x^{2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

$3 x^{2} + 113 - 2 x^{2} = 129$

$y = - \sqrt{113 - 2 x^{2}}$

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

$x^{2} + 113 = 129$

$y = - \sqrt{113 - 2 x^{2}}$

$x^{2} + 113 = 129$

$y = - \sqrt{113 - 2 x^{2}}$

$x^{2} + 113 = 129$

$y = - \sqrt{113 - 2 x^{2}}$

$x^{2} + 113 = 129$

$y = - \sqrt{113 - 2 x^{2}}$

Solve for $x$ in $x^{2} + 113 = 129$ .

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

Subtract $113$ from both sides of the equation.

$x^{2} = 129 - 113$

$y = - \sqrt{113 - 2 x^{2}}$

Subtract $113$ from $129$ .

$x^{2} = 16$

$y = - \sqrt{113 - 2 x^{2}}$

$x^{2} = 16$

$y = - \sqrt{113 - 2 x^{2}}$

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

$x = \pm \sqrt{16}$

$y = - \sqrt{113 - 2 x^{2}}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

$y = - \sqrt{113 - 2 x^{2}}$

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

$x = \pm 4$

$y = - \sqrt{113 - 2 x^{2}}$

$x = \pm 4$

$y = - \sqrt{113 - 2 x^{2}}$

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.

$x = 4$

$y = - \sqrt{113 - 2 x^{2}}$

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

$x = - 4$

$y = - \sqrt{113 - 2 x^{2}}$

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

$x = 4, - 4$

$y = - \sqrt{113 - 2 x^{2}}$

$x = 4, - 4$

$y = - \sqrt{113 - 2 x^{2}}$

$x = 4, - 4$

$y = - \sqrt{113 - 2 x^{2}}$

Replace all occurrences of $x$ with $4$ in each equation.

Replace all occurrences of $x$ in $y = - \sqrt{113 - 2 x^{2}}$ with $4$ .

$y = - \sqrt{113 - 2 {(4)}^{2}}$

$x = 4$

Simplify the right side.

Simplify $- \sqrt{113 - 2 {(4)}^{2}}$ .

Raise $4$ to the power of $2$ .

$y = - \sqrt{113 - 2 \cdot 16}$

$x = 4$

Multiply $- 2$ by $16$ .

$y = - \sqrt{113 - 32}$

$x = 4$

Subtract $32$ from $113$ .

$y = - \sqrt{81}$

$x = 4$

Rewrite $81$ as $9^{2}$ .

$y = - \sqrt{9^{2}}$

$x = 4$

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

$y = - 1 \cdot 9$

$x = 4$

Multiply $- 1$ by $9$ .

$y = - 9$

$x = 4$

$y = - 9$

$x = 4$

$y = - 9$

$x = 4$

$y = - 9$

$x = 4$

Replace all occurrences of $x$ with $- 4$ in each equation.

Replace all occurrences of $x$ in $y = - \sqrt{113 - 2 x^{2}}$ with $- 4$ .

$y = - \sqrt{113 - 2 {(- 4)}^{2}}$

$x = - 4$

Simplify the right side.

Simplify $- \sqrt{113 - 2 {(- 4)}^{2}}$ .

Raise $- 4$ to the power of $2$ .

$y = - \sqrt{113 - 2 \cdot 16}$

$x = - 4$

Multiply $- 2$ by $16$ .

$y = - \sqrt{113 - 32}$

$x = - 4$

Subtract $32$ from $113$ .

$y = - \sqrt{81}$

$x = - 4$

Rewrite $81$ as $9^{2}$ .

$y = - \sqrt{9^{2}}$

$x = - 4$

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

$y = - 1 \cdot 9$

$x = - 4$

Multiply $- 1$ by $9$ .

$y = - 9$

$x = - 4$

$y = - 9$

$x = - 4$

$y = - 9$

$x = - 4$

$y = - 9$

$x = - 4$

$y = - 9$

$x = - 4$

Step 4

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

$(4, 9)$

$(- 4, 9)$

$(4, - 9)$

$(- 4, - 9)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(4, 9), (- 4, 9), (4, - 9), (- 4, - 9)$

Equation Form:

$x = 4, y = 9$

$x = - 4, y = 9$

$x = 4, y = - 9$

$x = - 4, y = - 9$

Step 6

Solve the System of Equations 3x^2+y^2=129 2x^2+y^2=113

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