Solve by Substitution x^2+y^2=9 , x^2-y^2=1

$x^{2} + y^{2} = 9$ , $x^{2} - y^{2} = 1$

Step 1

Solve for $x$ in $x^{2} - y^{2} = 1$ .

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

$x^{2} = 1 + y^{2}$

$x^{2} + y^{2} = 9$

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

$x = \pm \sqrt{1 + y^{2}}$

$x^{2} + y^{2} = 9$

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.

$x = \sqrt{1 + y^{2}}$

$x^{2} + y^{2} = 9$

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

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

$x^{2} + y^{2} = 9$

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

$x = \sqrt{1 + y^{2}}$

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

$x^{2} + y^{2} = 9$

$x = \sqrt{1 + y^{2}}$

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

$x^{2} + y^{2} = 9$

$x = \sqrt{1 + y^{2}}$

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

$x^{2} + y^{2} = 9$

Step 2

Solve the system $x = \sqrt{1 + y^{2}}, x^{2} + y^{2} = 9$ .

Replace all occurrences of $x$ with $\sqrt{1 + y^{2}}$ in each equation.

Replace all occurrences of $x$ in $x^{2} + y^{2} = 9$ with $\sqrt{1 + y^{2}}$ .

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

$x = \sqrt{1 + y^{2}}$

Simplify the left side.

Simplify ${(\sqrt{1 + y^{2}})}^{2} + y^{2}$ .

Rewrite ${\sqrt{1 + y^{2}}}^{2}$ as $1 + y^{2}$ .

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

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

$x = \sqrt{1 + y^{2}}$

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

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

$x = \sqrt{1 + y^{2}}$

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

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

$x = \sqrt{1 + y^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

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

$x = \sqrt{1 + y^{2}}$

Rewrite the expression.

$(1 + y^{2}) + y^{2} = 9$

$x = \sqrt{1 + y^{2}}$

$(1 + y^{2}) + y^{2} = 9$

$x = \sqrt{1 + y^{2}}$

Simplify.

$1 + y^{2} + y^{2} = 9$

$x = \sqrt{1 + y^{2}}$

$1 + y^{2} + y^{2} = 9$

$x = \sqrt{1 + y^{2}}$

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

$1 + 2 y^{2} = 9$

$x = \sqrt{1 + y^{2}}$

$1 + 2 y^{2} = 9$

$x = \sqrt{1 + y^{2}}$

$1 + 2 y^{2} = 9$

$x = \sqrt{1 + y^{2}}$

$1 + 2 y^{2} = 9$

$x = \sqrt{1 + y^{2}}$

Solve for $y$ in $1 + 2 y^{2} = 9$ .

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

Subtract $1$ from both sides of the equation.

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

$x = \sqrt{1 + y^{2}}$

Subtract $1$ from $9$ .

$2 y^{2} = 8$

$x = \sqrt{1 + y^{2}}$

$2 y^{2} = 8$

$x = \sqrt{1 + y^{2}}$

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

Divide each term in $2 y^{2} = 8$ by $2$ .

$\frac{2 y^{2}}{2} = \frac{8}{2}$

$x = \sqrt{1 + y^{2}}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y^{2}}{2} = \frac{8}{2}$

$x = \sqrt{1 + y^{2}}$

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

$y^{2} = \frac{8}{2}$

$x = \sqrt{1 + y^{2}}$

$y^{2} = \frac{8}{2}$

$x = \sqrt{1 + y^{2}}$

$y^{2} = \frac{8}{2}$

$x = \sqrt{1 + y^{2}}$

Simplify the right side.

Divide $8$ by $2$ .

$y^{2} = 4$

$x = \sqrt{1 + y^{2}}$

$y^{2} = 4$

$x = \sqrt{1 + y^{2}}$

$y^{2} = 4$

$x = \sqrt{1 + y^{2}}$

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

$y = \pm \sqrt{4}$

$x = \sqrt{1 + y^{2}}$

Simplify $\pm \sqrt{4}$ .

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

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

$x = \sqrt{1 + y^{2}}$

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

$y = \pm 2$

$x = \sqrt{1 + y^{2}}$

$y = \pm 2$

$x = \sqrt{1 + y^{2}}$

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.

$y = 2$

$x = \sqrt{1 + y^{2}}$

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

$y = - 2$

$x = \sqrt{1 + y^{2}}$

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

$y = 2, - 2$

$x = \sqrt{1 + y^{2}}$

$y = 2, - 2$

$x = \sqrt{1 + y^{2}}$

$y = 2, - 2$

$x = \sqrt{1 + y^{2}}$

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

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

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

$y = 2$

Simplify the right side.

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

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

$x = \sqrt{1 + 4}$

$y = 2$

Add $1$ and $4$ .

$x = \sqrt{5}$

$y = 2$

$x = \sqrt{5}$

$y = 2$

$x = \sqrt{5}$

$y = 2$

$x = \sqrt{5}$

$y = 2$

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

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

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

$y = - 2$

Simplify the right side.

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

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

$x = \sqrt{1 + 4}$

$y = - 2$

Add $1$ and $4$ .

$x = \sqrt{5}$

$y = - 2$

$x = \sqrt{5}$

$y = - 2$

$x = \sqrt{5}$

$y = - 2$

$x = \sqrt{5}$

$y = - 2$

$x = \sqrt{5}$

$y = - 2$

Step 3

Solve the system $x = - \sqrt{1 + y^{2}}, x^{2} + y^{2} = 9$ .

Replace all occurrences of $x$ with $- \sqrt{1 + y^{2}}$ in each equation.

Replace all occurrences of $x$ in $x^{2} + y^{2} = 9$ with $- \sqrt{1 + y^{2}}$ .

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

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

Simplify the left side.

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

Simplify each term.

Apply the product rule to $- \sqrt{1 + y^{2}}$ .

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

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

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

$1 {\sqrt{1 + y^{2}}}^{2} + y^{2} = 9$

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

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

${\sqrt{1 + y^{2}}}^{2} + y^{2} = 9$

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

Rewrite ${\sqrt{1 + y^{2}}}^{2}$ as $1 + y^{2}$ .

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

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

Rewrite the expression.

$(1 + y^{2}) + y^{2} = 9$

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

$(1 + y^{2}) + y^{2} = 9$

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

Simplify.

$1 + y^{2} + y^{2} = 9$

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

$1 + y^{2} + y^{2} = 9$

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

$1 + y^{2} + y^{2} = 9$

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

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

$1 + 2 y^{2} = 9$

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

$1 + 2 y^{2} = 9$

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

$1 + 2 y^{2} = 9$

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

$1 + 2 y^{2} = 9$

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

Solve for $y$ in $1 + 2 y^{2} = 9$ .

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

Subtract $1$ from both sides of the equation.

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

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

Subtract $1$ from $9$ .

$2 y^{2} = 8$

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

$2 y^{2} = 8$

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

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

Divide each term in $2 y^{2} = 8$ by $2$ .

$\frac{2 y^{2}}{2} = \frac{8}{2}$

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y^{2}}{2} = \frac{8}{2}$

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

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

$y^{2} = \frac{8}{2}$

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

$y^{2} = \frac{8}{2}$

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

$y^{2} = \frac{8}{2}$

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

Simplify the right side.

Divide $8$ by $2$ .

$y^{2} = 4$

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

$y^{2} = 4$

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

$y^{2} = 4$

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

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

$y = \pm \sqrt{4}$

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

Simplify $\pm \sqrt{4}$ .

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

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

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

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

$y = \pm 2$

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

$y = \pm 2$

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

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.

$y = 2$

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

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

$y = - 2$

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

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

$y = 2, - 2$

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

$y = 2, - 2$

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

$y = 2, - 2$

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

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

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

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

$y = 2$

Simplify the right side.

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

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

$x = - \sqrt{1 + 4}$

$y = 2$

Add $1$ and $4$ .

$x = - \sqrt{5}$

$y = 2$

$x = - \sqrt{5}$

$y = 2$

$x = - \sqrt{5}$

$y = 2$

$x = - \sqrt{5}$

$y = 2$

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

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

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

$y = - 2$

Simplify the right side.

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

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

$x = - \sqrt{1 + 4}$

$y = - 2$

Add $1$ and $4$ .

$x = - \sqrt{5}$

$y = - 2$

$x = - \sqrt{5}$

$y = - 2$

$x = - \sqrt{5}$

$y = - 2$

$x = - \sqrt{5}$

$y = - 2$

$x = - \sqrt{5}$

$y = - 2$

Step 4

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

$(\sqrt{5}, 2)$

$(\sqrt{5}, - 2)$

$(- \sqrt{5}, 2)$

$(- \sqrt{5}, - 2)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(\sqrt{5}, 2), (\sqrt{5}, - 2), (- \sqrt{5}, 2), (- \sqrt{5}, - 2)$

Equation Form:

$x = \sqrt{5}, y = 2$

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

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

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

Step 6

Solve by Substitution x^2+y^2=9 , x^2-y^2=1

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