Solve the System of Equations x^2+y^2=25 x^2=5-y

$x^{2} + y^{2} = 25$ $x^{2} = 5 - y$

Step 1

Solve for $x$ in $x^{2} = 5 - y$ .

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

$x = \pm \sqrt{5 - y}$

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

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.

$x = \sqrt{5 - y}$

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

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

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

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

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

$x = \sqrt{5 - y}$

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

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

$x = \sqrt{5 - y}$

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

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

$x = \sqrt{5 - y}$

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

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

Step 2

Solve the system $x = \sqrt{5 - y}, x^{2} + y^{2} = 25$ .

Replace all occurrences of $x$ with $\sqrt{5 - y}$ in each equation.

Replace all occurrences of $x$ in $x^{2} + y^{2} = 25$ with $\sqrt{5 - y}$ .

${(\sqrt{5 - y})}^{2} + y^{2} = 25$

$x = \sqrt{5 - y}$

Simplify the left side.

Rewrite ${\sqrt{5 - y}}^{2}$ as $5 - y$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 - y}$ as ${(5 - y)}^{\frac{1}{2}}$ .

${({(5 - y)}^{\frac{1}{2}})}^{2} + y^{2} = 25$

$x = \sqrt{5 - y}$

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

${(5 - y)}^{\frac{1}{2} \cdot 2} + y^{2} = 25$

$x = \sqrt{5 - y}$

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

${(5 - y)}^{\frac{2}{2}} + y^{2} = 25$

$x = \sqrt{5 - y}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(5 - y)}^{\frac{2}{2}} + y^{2} = 25$

$x = \sqrt{5 - y}$

Rewrite the expression.

$(5 - y) + y^{2} = 25$

$x = \sqrt{5 - y}$

$(5 - y) + y^{2} = 25$

$x = \sqrt{5 - y}$

Simplify.

$5 - y + y^{2} = 25$

$x = \sqrt{5 - y}$

$5 - y + y^{2} = 25$

$x = \sqrt{5 - y}$

$5 - y + y^{2} = 25$

$x = \sqrt{5 - y}$

$5 - y + y^{2} = 25$

$x = \sqrt{5 - y}$

Solve for $y$ in $5 - y + y^{2} = 25$ .

Subtract $25$ from both sides of the equation.

$5 - y + y^{2} - 25 = 0$

$x = \sqrt{5 - y}$

Subtract $25$ from $5$ .

$- y + y^{2} - 20 = 0$

$x = \sqrt{5 - y}$

Factor the left side of the equation.

Let $u = y$ . Substitute $u$ for all occurrences of $y$ .

$- u + u^{2} - 20 = 0$

$x = \sqrt{5 - y}$

Factor $- u + u^{2} - 20$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 20$ and whose sum is $- 1$ .

$- 5, 4$

$x = \sqrt{5 - y}$

Write the factored form using these integers.

$(u - 5) (u + 4) = 0$

$x = \sqrt{5 - y}$

$(u - 5) (u + 4) = 0$

$x = \sqrt{5 - y}$

Replace all occurrences of $u$ with $y$ .

$(y - 5) (y + 4) = 0$

$x = \sqrt{5 - y}$

$(y - 5) (y + 4) = 0$

$x = \sqrt{5 - y}$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y - 5 = 0$

$y + 4 = 0$

$x = \sqrt{5 - y}$

Set $y - 5$ equal to $0$ and solve for $y$ .

Set $y - 5$ equal to $0$ .

$y - 5 = 0$

$x = \sqrt{5 - y}$

Add $5$ to both sides of the equation.

$y = 5$

$x = \sqrt{5 - y}$

$y = 5$

$x = \sqrt{5 - y}$

Set $y + 4$ equal to $0$ and solve for $y$ .

Set $y + 4$ equal to $0$ .

$y + 4 = 0$

$x = \sqrt{5 - y}$

Subtract $4$ from both sides of the equation.

$y = - 4$

$x = \sqrt{5 - y}$

$y = - 4$

$x = \sqrt{5 - y}$

The final solution is all the values that make $(y - 5) (y + 4) = 0$ true.

$y = 5, - 4$

$x = \sqrt{5 - y}$

$y = 5, - 4$

$x = \sqrt{5 - y}$

Replace all occurrences of $y$ with $5$ in each equation.

Replace all occurrences of $y$ in $x = \sqrt{5 - y}$ with $5$ .

$x = \sqrt{5 - (5)}$

$y = 5$

Simplify the right side.

Simplify $\sqrt{5 - (5)}$ .

Multiply $- 1$ by $5$ .

$x = \sqrt{5 - 5}$

$y = 5$

Subtract $5$ from $5$ .

$x = \sqrt{0}$

$y = 5$

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

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

$y = 5$

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

$x = 0$

$y = 5$

$x = 0$

$y = 5$

$x = 0$

$y = 5$

$x = 0$

$y = 5$

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

Replace all occurrences of $y$ in $x = \sqrt{5 - y}$ with $- 4$ .

$x = \sqrt{5 - (- 4)}$

$y = - 4$

Simplify the right side.

Simplify $\sqrt{5 - (- 4)}$ .

Multiply $- 1$ by $- 4$ .

$x = \sqrt{5 + 4}$

$y = - 4$

Add $5$ and $4$ .

$x = \sqrt{9}$

$y = - 4$

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

$x = \sqrt{3^{2}}$

$y = - 4$

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

$x = 3$

$y = - 4$

$x = 3$

$y = - 4$

$x = 3$

$y = - 4$

$x = 3$

$y = - 4$

$x = 3$

$y = - 4$

Step 3

Solve the system $x = - \sqrt{5 - y}, x^{2} + y^{2} = 25$ .

Replace all occurrences of $x$ with $- \sqrt{5 - y}$ in each equation.

Replace all occurrences of $x$ in $x^{2} + y^{2} = 25$ with $- \sqrt{5 - y}$ .

${(- \sqrt{5 - y})}^{2} + y^{2} = 25$

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

Simplify the left side.

Simplify each term.

Apply the product rule to $- \sqrt{5 - y}$ .

${(- 1)}^{2} {\sqrt{5 - y}}^{2} + y^{2} = 25$

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

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

$1 {\sqrt{5 - y}}^{2} + y^{2} = 25$

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

Multiply ${\sqrt{5 - y}}^{2}$ by $1$ .

${\sqrt{5 - y}}^{2} + y^{2} = 25$

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

Rewrite ${\sqrt{5 - y}}^{2}$ as $5 - y$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 - y}$ as ${(5 - y)}^{\frac{1}{2}}$ .

${({(5 - y)}^{\frac{1}{2}})}^{2} + y^{2} = 25$

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

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

${(5 - y)}^{\frac{1}{2} \cdot 2} + y^{2} = 25$

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

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

${(5 - y)}^{\frac{2}{2}} + y^{2} = 25$

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

Cancel the common factor of $2$ .

Cancel the common factor.

${(5 - y)}^{\frac{2}{2}} + y^{2} = 25$

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

Rewrite the expression.

$(5 - y) + y^{2} = 25$

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

$(5 - y) + y^{2} = 25$

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

Simplify.

$5 - y + y^{2} = 25$

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

$5 - y + y^{2} = 25$

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

$5 - y + y^{2} = 25$

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

$5 - y + y^{2} = 25$

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

$5 - y + y^{2} = 25$

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

Solve for $y$ in $5 - y + y^{2} = 25$ .

Subtract $25$ from both sides of the equation.

$5 - y + y^{2} - 25 = 0$

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

Subtract $25$ from $5$ .

$- y + y^{2} - 20 = 0$

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

Factor the left side of the equation.

Let $u = y$ . Substitute $u$ for all occurrences of $y$ .

$- u + u^{2} - 20 = 0$

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

Factor $- u + u^{2} - 20$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 20$ and whose sum is $- 1$ .

$- 5, 4$

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

Write the factored form using these integers.

$(u - 5) (u + 4) = 0$

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

$(u - 5) (u + 4) = 0$

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

Replace all occurrences of $u$ with $y$ .

$(y - 5) (y + 4) = 0$

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

$(y - 5) (y + 4) = 0$

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y - 5 = 0$

$y + 4 = 0$

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

Set $y - 5$ equal to $0$ and solve for $y$ .

Set $y - 5$ equal to $0$ .

$y - 5 = 0$

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

Add $5$ to both sides of the equation.

$y = 5$

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

$y = 5$

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

Set $y + 4$ equal to $0$ and solve for $y$ .

Set $y + 4$ equal to $0$ .

$y + 4 = 0$

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

Subtract $4$ from both sides of the equation.

$y = - 4$

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

$y = - 4$

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

The final solution is all the values that make $(y - 5) (y + 4) = 0$ true.

$y = 5, - 4$

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

$y = 5, - 4$

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

Replace all occurrences of $y$ with $5$ in each equation.

Replace all occurrences of $y$ in $x = - \sqrt{5 - y}$ with $5$ .

$x = - \sqrt{5 - (5)}$

$y = 5$

Simplify the right side.

Simplify $- \sqrt{5 - (5)}$ .

Multiply $- 1$ by $5$ .

$x = - \sqrt{5 - 5}$

$y = 5$

Subtract $5$ from $5$ .

$x = - \sqrt{0}$

$y = 5$

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

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

$y = 5$

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

$x = - 0$

$y = 5$

Multiply $- 1$ by $0$ .

$x = 0$

$y = 5$

$x = 0$

$y = 5$

$x = 0$

$y = 5$

$x = 0$

$y = 5$

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

Replace all occurrences of $y$ in $x = - \sqrt{5 - y}$ with $- 4$ .

$x = - \sqrt{5 - (- 4)}$

$y = - 4$

Simplify the right side.

Simplify $- \sqrt{5 - (- 4)}$ .

Multiply $- 1$ by $- 4$ .

$x = - \sqrt{5 + 4}$

$y = - 4$

Add $5$ and $4$ .

$x = - \sqrt{9}$

$y = - 4$

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

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

$y = - 4$

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

$x = - 1 \cdot 3$

$y = - 4$

Multiply $- 1$ by $3$ .

$x = - 3$

$y = - 4$

$x = - 3$

$y = - 4$

$x = - 3$

$y = - 4$

$x = - 3$

$y = - 4$

$x = - 3$

$y = - 4$

Step 4

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

$(0, 5)$

$(3, - 4)$

$(- 3, - 4)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(0, 5), (3, - 4), (- 3, - 4)$

Equation Form:

$x = 0, y = 5$

$x = 3, y = - 4$

$x = - 3, y = - 4$

Step 6

Solve the System of Equations x^2+y^2=25 x^2=5-y

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