Solve by Substitution x^2+y^2=9 , x^2+(y-1)^2=4

$x^{2} + y^{2} = 9$ , $x^{2} + {(y - 1)}^{2} = 4$

Step 1

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

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

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

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

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

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

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

Simplify $\pm \sqrt{9 - y^{2}}$ .

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

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = y$ .

$x = \pm \sqrt{(3 + y) (3 - y)}$

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

$x = \pm \sqrt{(3 + y) (3 - y)}$

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

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.

$x = \sqrt{(3 + y) (3 - y)}$

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

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

$x = - \sqrt{(3 + y) (3 - y)}$

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

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

$x = \sqrt{(3 + y) (3 - y)}$

$x = - \sqrt{(3 + y) (3 - y)}$

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

$x = \sqrt{(3 + y) (3 - y)}$

$x = - \sqrt{(3 + y) (3 - y)}$

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

$x = \sqrt{(3 + y) (3 - y)}$

$x = - \sqrt{(3 + y) (3 - y)}$

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

Step 2

Solve the system $x = \sqrt{(3 + y) (3 - y)}, x^{2} + {(y - 1)}^{2} = 4$ .

Replace all occurrences of $x$ with $\sqrt{(3 + y) (3 - y)}$ in each equation.

Replace all occurrences of $x$ in $x^{2} + {(y - 1)}^{2} = 4$ with $\sqrt{(3 + y) (3 - y)}$ .

${(\sqrt{(3 + y) (3 - y)})}^{2} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Simplify the left side.

Simplify ${(\sqrt{(3 + y) (3 - y)})}^{2} + {(y - 1)}^{2}$ .

Simplify each term.

Rewrite ${\sqrt{(3 + y) (3 - y)}}^{2}$ as $(3 + y) (3 - y)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{(3 + y) (3 - y)}$ as ${((3 + y) (3 - y))}^{\frac{1}{2}}$ .

${({((3 + y) (3 - y))}^{\frac{1}{2}})}^{2} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

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

${((3 + y) (3 - y))}^{\frac{1}{2} \cdot 2} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

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

${((3 + y) (3 - y))}^{\frac{2}{2}} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Cancel the common factor of $2$ .

Cancel the common factor.

${((3 + y) (3 - y))}^{\frac{2}{2}} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Rewrite the expression.

$((3 + y) (3 - y)) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$((3 + y) (3 - y)) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Simplify.

$(3 + y) (3 - y) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$(3 + y) (3 - y) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Expand $(3 + y) (3 - y)$ using the FOIL Method.

Apply the distributive property.

$3 (3 - y) + y (3 - y) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Apply the distributive property.

$3 \cdot 3 + 3 (- y) + y (3 - y) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Apply the distributive property.

$3 \cdot 3 + 3 (- y) + y \cdot 3 + y (- y) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$3 \cdot 3 + 3 (- y) + y \cdot 3 + y (- y) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$9 + 3 (- y) + y \cdot 3 + y (- y) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Multiply $- 1$ by $3$ .

$9 - 3 y + y \cdot 3 + y (- y) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Move $3$ to the left of $y$ .

$9 - 3 y + 3 \cdot y + y (- y) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Rewrite using the commutative property of multiplication.

$9 - 3 y + 3 y - y \cdot y + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$9 - 3 y + 3 y - (y \cdot y) + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Multiply $y$ by $y$ .

$9 - 3 y + 3 y - y^{2} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$9 - 3 y + 3 y - y^{2} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$9 - 3 y + 3 y - y^{2} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Add $- 3 y$ and $3 y$ .

$9 + 0 - y^{2} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Add $9$ and $0$ .

$9 - y^{2} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$9 - y^{2} + {(y - 1)}^{2} = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Rewrite ${(y - 1)}^{2}$ as $(y - 1) (y - 1)$ .

$9 - y^{2} + (y - 1) (y - 1) = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Expand $(y - 1) (y - 1)$ using the FOIL Method.

Apply the distributive property.

$9 - y^{2} + y (y - 1) - 1 (y - 1) = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Apply the distributive property.

$9 - y^{2} + y \cdot y + y \cdot - 1 - 1 (y - 1) = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Apply the distributive property.

$9 - y^{2} + y \cdot y + y \cdot - 1 - 1 y - 1 \cdot - 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$9 - y^{2} + y \cdot y + y \cdot - 1 - 1 y - 1 \cdot - 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$9 - y^{2} + y^{2} + y \cdot - 1 - 1 y - 1 \cdot - 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Move $- 1$ to the left of $y$ .

$9 - y^{2} + y^{2} - 1 \cdot y - 1 y - 1 \cdot - 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Rewrite $- 1 y$ as $- y$ .

$9 - y^{2} + y^{2} - y - 1 y - 1 \cdot - 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Rewrite $- 1 y$ as $- y$ .

$9 - y^{2} + y^{2} - y - y - 1 \cdot - 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Multiply $- 1$ by $- 1$ .

$9 - y^{2} + y^{2} - y - y + 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$9 - y^{2} + y^{2} - y - y + 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Subtract $y$ from $- y$ .

$9 - y^{2} + y^{2} - 2 y + 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$9 - y^{2} + y^{2} - 2 y + 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$9 - y^{2} + y^{2} - 2 y + 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Simplify by adding terms.

Combine the opposite terms in $9 - y^{2} + y^{2} - 2 y + 1$ .

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

$9 + 0 - 2 y + 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Add $9$ and $0$ .

$9 - 2 y + 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$9 - 2 y + 1 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Add $9$ and $1$ .

$- 2 y + 10 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$- 2 y + 10 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$- 2 y + 10 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$- 2 y + 10 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

$- 2 y + 10 = 4$

$x = \sqrt{(3 + y) (3 - y)}$

Solve for $y$ in $- 2 y + 10 = 4$ .

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

Subtract $10$ from both sides of the equation.

$- 2 y = 4 - 10$

$x = \sqrt{(3 + y) (3 - y)}$

Subtract $10$ from $4$ .

$- 2 y = - 6$

$x = \sqrt{(3 + y) (3 - y)}$

$- 2 y = - 6$

$x = \sqrt{(3 + y) (3 - y)}$

Divide each term in $- 2 y = - 6$ by $- 2$ and simplify.

Divide each term in $- 2 y = - 6$ by $- 2$ .

$\frac{- 2 y}{- 2} = \frac{- 6}{- 2}$

$x = \sqrt{(3 + y) (3 - y)}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 y}{- 2} = \frac{- 6}{- 2}$

$x = \sqrt{(3 + y) (3 - y)}$

Divide $y$ by $1$ .

$y = \frac{- 6}{- 2}$

$x = \sqrt{(3 + y) (3 - y)}$

$y = \frac{- 6}{- 2}$

$x = \sqrt{(3 + y) (3 - y)}$

$y = \frac{- 6}{- 2}$

$x = \sqrt{(3 + y) (3 - y)}$

Simplify the right side.

Divide $- 6$ by $- 2$ .

$y = 3$

$x = \sqrt{(3 + y) (3 - y)}$

$y = 3$

$x = \sqrt{(3 + y) (3 - y)}$

$y = 3$

$x = \sqrt{(3 + y) (3 - y)}$

$y = 3$

$x = \sqrt{(3 + y) (3 - y)}$

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

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

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

$y = 3$

Simplify $x = \sqrt{(3 + 3) (3 - (3))}$ .

Simplify the left side.

Remove parentheses.

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

$y = 3$

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

$y = 3$

Simplify the right side.

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

Add $3$ and $3$ .

$x = \sqrt{6 (3 - (3))}$

$y = 3$

Multiply $- 1$ by $3$ .

$x = \sqrt{6 (3 - 3)}$

$y = 3$

Subtract $3$ from $3$ .

$x = \sqrt{6 \cdot 0}$

$y = 3$

Multiply $6$ by $0$ .

$x = \sqrt{0}$

$y = 3$

Rewrite $0$ as $0^{2}$ .

$x = \sqrt{0^{2}}$

$y = 3$

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

$x = 0$

$y = 3$

$x = 0$

$y = 3$

$x = 0$

$y = 3$

$x = 0$

$y = 3$

$x = 0$

$y = 3$

$x = 0$

$y = 3$

Step 3

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

$(0, 3)$

Step 4

The result can be shown in multiple forms.

Point Form:

$(0, 3)$

Equation Form:

$x = 0, y = 3$

Step 5

Solve by Substitution x^2+y^2=9 , x^2+(y-1)^2=4

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