Solve by Substitution x^2+y^2=25 , x^2-y^2=7

$x^{2} + y^{2} = 25$ , $x^{2} - y^{2} = 7$

Step 1

Solve for $x$ in $x^{2} - y^{2} = 7$ .

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

$x^{2} = 7 + y^{2}$

$x^{2} + y^{2} = 25$

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

$x = \pm \sqrt{7 + y^{2}}$

$x^{2} + y^{2} = 25$

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.

$x = \sqrt{7 + y^{2}}$

$x^{2} + y^{2} = 25$

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

$x = - \sqrt{7 + y^{2}}$

$x^{2} + y^{2} = 25$

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

$x = \sqrt{7 + y^{2}}$

$x = - \sqrt{7 + y^{2}}$

$x^{2} + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

$x = - \sqrt{7 + y^{2}}$

$x^{2} + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

$x = - \sqrt{7 + y^{2}}$

$x^{2} + y^{2} = 25$

Step 2

Solve the system $x = \sqrt{7 + y^{2}}, x^{2} + y^{2} = 25$ .

Replace all occurrences of $x$ with $\sqrt{7 + y^{2}}$ in each equation.

Replace all occurrences of $x$ in $x^{2} + y^{2} = 25$ with $\sqrt{7 + y^{2}}$ .

${(\sqrt{7 + y^{2}})}^{2} + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

Simplify the left side.

Simplify ${(\sqrt{7 + y^{2}})}^{2} + y^{2}$ .

Rewrite ${\sqrt{7 + y^{2}}}^{2}$ as $7 + y^{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7 + y^{2}}$ as ${(7 + y^{2})}^{\frac{1}{2}}$ .

${({(7 + y^{2})}^{\frac{1}{2}})}^{2} + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

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

${(7 + y^{2})}^{\frac{1}{2} \cdot 2} + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

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

${(7 + y^{2})}^{\frac{2}{2}} + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(7 + y^{2})}^{\frac{2}{2}} + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

Rewrite the expression.

$(7 + y^{2}) + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

$(7 + y^{2}) + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

Simplify.

$7 + y^{2} + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

$7 + y^{2} + y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

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

$7 + 2 y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

$7 + 2 y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

$7 + 2 y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

$7 + 2 y^{2} = 25$

$x = \sqrt{7 + y^{2}}$

Solve for $y$ in $7 + 2 y^{2} = 25$ .

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

Subtract $7$ from both sides of the equation.

$2 y^{2} = 25 - 7$

$x = \sqrt{7 + y^{2}}$

Subtract $7$ from $25$ .

$2 y^{2} = 18$

$x = \sqrt{7 + y^{2}}$

$2 y^{2} = 18$

$x = \sqrt{7 + y^{2}}$

Divide each term in $2 y^{2} = 18$ by $2$ and simplify.

Divide each term in $2 y^{2} = 18$ by $2$ .

$\frac{2 y^{2}}{2} = \frac{18}{2}$

$x = \sqrt{7 + y^{2}}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y^{2}}{2} = \frac{18}{2}$

$x = \sqrt{7 + y^{2}}$

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

$y^{2} = \frac{18}{2}$

$x = \sqrt{7 + y^{2}}$

$y^{2} = \frac{18}{2}$

$x = \sqrt{7 + y^{2}}$

$y^{2} = \frac{18}{2}$

$x = \sqrt{7 + y^{2}}$

Simplify the right side.

Divide $18$ by $2$ .

$y^{2} = 9$

$x = \sqrt{7 + y^{2}}$

$y^{2} = 9$

$x = \sqrt{7 + y^{2}}$

$y^{2} = 9$

$x = \sqrt{7 + y^{2}}$

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

$y = \pm \sqrt{9}$

$x = \sqrt{7 + y^{2}}$

Simplify $\pm \sqrt{9}$ .

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

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

$x = \sqrt{7 + y^{2}}$

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

$y = \pm 3$

$x = \sqrt{7 + y^{2}}$

$y = \pm 3$

$x = \sqrt{7 + y^{2}}$

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.

$y = 3$

$x = \sqrt{7 + y^{2}}$

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

$y = - 3$

$x = \sqrt{7 + y^{2}}$

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

$y = 3, - 3$

$x = \sqrt{7 + y^{2}}$

$y = 3, - 3$

$x = \sqrt{7 + y^{2}}$

$y = 3, - 3$

$x = \sqrt{7 + y^{2}}$

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

Replace all occurrences of $y$ in $x = \sqrt{7 + y^{2}}$ with $3$ .

$x = \sqrt{7 + {(3)}^{2}}$

$y = 3$

Simplify the right side.

Simplify $\sqrt{7 + {(3)}^{2}}$ .

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

$x = \sqrt{7 + 9}$

$y = 3$

Add $7$ and $9$ .

$x = \sqrt{16}$

$y = 3$

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

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

$y = 3$

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

$x = 4$

$y = 3$

$x = 4$

$y = 3$

$x = 4$

$y = 3$

$x = 4$

$y = 3$

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

Replace all occurrences of $y$ in $x = \sqrt{7 + y^{2}}$ with $- 3$ .

$x = \sqrt{7 + {(- 3)}^{2}}$

$y = - 3$

Simplify the right side.

Simplify $\sqrt{7 + {(- 3)}^{2}}$ .

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

$x = \sqrt{7 + 9}$

$y = - 3$

Add $7$ and $9$ .

$x = \sqrt{16}$

$y = - 3$

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

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

$y = - 3$

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

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

Step 3

Solve the system $x = - \sqrt{7 + y^{2}}, x^{2} + y^{2} = 25$ .

Replace all occurrences of $x$ with $- \sqrt{7 + y^{2}}$ in each equation.

Replace all occurrences of $x$ in $x^{2} + y^{2} = 25$ with $- \sqrt{7 + y^{2}}$ .

${(- \sqrt{7 + y^{2}})}^{2} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

Simplify the left side.

Simplify ${(- \sqrt{7 + y^{2}})}^{2} + y^{2}$ .

Simplify each term.

Apply the product rule to $- \sqrt{7 + y^{2}}$ .

${(- 1)}^{2} {\sqrt{7 + y^{2}}}^{2} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

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

$1 {\sqrt{7 + y^{2}}}^{2} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

Multiply ${\sqrt{7 + y^{2}}}^{2}$ by $1$ .

${\sqrt{7 + y^{2}}}^{2} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

Rewrite ${\sqrt{7 + y^{2}}}^{2}$ as $7 + y^{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7 + y^{2}}$ as ${(7 + y^{2})}^{\frac{1}{2}}$ .

${({(7 + y^{2})}^{\frac{1}{2}})}^{2} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

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

${(7 + y^{2})}^{\frac{1}{2} \cdot 2} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

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

${(7 + y^{2})}^{\frac{2}{2}} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(7 + y^{2})}^{\frac{2}{2}} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

Rewrite the expression.

$(7 + y^{2}) + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

$(7 + y^{2}) + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

Simplify.

$7 + y^{2} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

$7 + y^{2} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

$7 + y^{2} + y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

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

$7 + 2 y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

$7 + 2 y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

$7 + 2 y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

$7 + 2 y^{2} = 25$

$x = - \sqrt{7 + y^{2}}$

Solve for $y$ in $7 + 2 y^{2} = 25$ .

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

Subtract $7$ from both sides of the equation.

$2 y^{2} = 25 - 7$

$x = - \sqrt{7 + y^{2}}$

Subtract $7$ from $25$ .

$2 y^{2} = 18$

$x = - \sqrt{7 + y^{2}}$

$2 y^{2} = 18$

$x = - \sqrt{7 + y^{2}}$

Divide each term in $2 y^{2} = 18$ by $2$ and simplify.

Divide each term in $2 y^{2} = 18$ by $2$ .

$\frac{2 y^{2}}{2} = \frac{18}{2}$

$x = - \sqrt{7 + y^{2}}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y^{2}}{2} = \frac{18}{2}$

$x = - \sqrt{7 + y^{2}}$

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

$y^{2} = \frac{18}{2}$

$x = - \sqrt{7 + y^{2}}$

$y^{2} = \frac{18}{2}$

$x = - \sqrt{7 + y^{2}}$

$y^{2} = \frac{18}{2}$

$x = - \sqrt{7 + y^{2}}$

Simplify the right side.

Divide $18$ by $2$ .

$y^{2} = 9$

$x = - \sqrt{7 + y^{2}}$

$y^{2} = 9$

$x = - \sqrt{7 + y^{2}}$

$y^{2} = 9$

$x = - \sqrt{7 + y^{2}}$

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

$y = \pm \sqrt{9}$

$x = - \sqrt{7 + y^{2}}$

Simplify $\pm \sqrt{9}$ .

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

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

$x = - \sqrt{7 + y^{2}}$

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

$y = \pm 3$

$x = - \sqrt{7 + y^{2}}$

$y = \pm 3$

$x = - \sqrt{7 + y^{2}}$

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.

$y = 3$

$x = - \sqrt{7 + y^{2}}$

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

$y = - 3$

$x = - \sqrt{7 + y^{2}}$

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

$y = 3, - 3$

$x = - \sqrt{7 + y^{2}}$

$y = 3, - 3$

$x = - \sqrt{7 + y^{2}}$

$y = 3, - 3$

$x = - \sqrt{7 + y^{2}}$

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

Replace all occurrences of $y$ in $x = - \sqrt{7 + y^{2}}$ with $3$ .

$x = - \sqrt{7 + {(3)}^{2}}$

$y = 3$

Simplify the right side.

Simplify $- \sqrt{7 + {(3)}^{2}}$ .

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

$x = - \sqrt{7 + 9}$

$y = 3$

Add $7$ and $9$ .

$x = - \sqrt{16}$

$y = 3$

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

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

$y = 3$

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

$x = - 1 \cdot 4$

$y = 3$

Multiply $- 1$ by $4$ .

$x = - 4$

$y = 3$

$x = - 4$

$y = 3$

$x = - 4$

$y = 3$

$x = - 4$

$y = 3$

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

Replace all occurrences of $y$ in $x = - \sqrt{7 + y^{2}}$ with $- 3$ .

$x = - \sqrt{7 + {(- 3)}^{2}}$

$y = - 3$

Simplify the right side.

Simplify $- \sqrt{7 + {(- 3)}^{2}}$ .

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

$x = - \sqrt{7 + 9}$

$y = - 3$

Add $7$ and $9$ .

$x = - \sqrt{16}$

$y = - 3$

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

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

$y = - 3$

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

$x = - 1 \cdot 4$

$y = - 3$

Multiply $- 1$ by $4$ .

$x = - 4$

$y = - 3$

$x = - 4$

$y = - 3$

$x = - 4$

$y = - 3$

$x = - 4$

$y = - 3$

$x = - 4$

$y = - 3$

Step 4

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

$(4, 3)$

$(4, - 3)$

$(- 4, 3)$

$(- 4, - 3)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(4, 3), (4, - 3), (- 4, 3), (- 4, - 3)$

Equation Form:

$x = 4, y = 3$

$x = 4, y = - 3$

$x = - 4, y = 3$

$x = - 4, y = - 3$

Step 6

Solve by Substitution x^2+y^2=25 , x^2-y^2=7

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