Solve the System of Equations (x+5)^2+y^2=90 x+3y=-5

${(x + 5)}^{2} + y^{2} = 90$ $x + 3 y = - 5$

Subtract $3 y$ from both sides of the equation.

$x = - 5 - 3 y$

${(x + 5)}^{2} + y^{2} = 90$

Replace all occurrences of $x$ with $- 5 - 3 y$ in each equation.

Replace all occurrences of $x$ in ${(x + 5)}^{2} + y^{2} = 90$ with $- 5 - 3 y$ .

${((- 5 - 3 y) + 5)}^{2} + y^{2} = 90$

$x = - 5 - 3 y$

Simplify the left side.

Simplify ${((- 5 - 3 y) + 5)}^{2} + y^{2}$ .

Combine the opposite terms in ${((- 5 - 3 y) + 5)}^{2} + y^{2}$ .

Add $- 5$ and $5$ .

${(- 3 y + 0)}^{2} + y^{2} = 90$

$x = - 5 - 3 y$

Add $- 3 y$ and $0$ .

${(- 3 y)}^{2} + y^{2} = 90$

$x = - 5 - 3 y$

${(- 3 y)}^{2} + y^{2} = 90$

$x = - 5 - 3 y$

Simplify each term.

Apply the product rule to $- 3 y$ .

${(- 3)}^{2} y^{2} + y^{2} = 90$

$x = - 5 - 3 y$

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

$9 y^{2} + y^{2} = 90$

$x = - 5 - 3 y$

$9 y^{2} + y^{2} = 90$

$x = - 5 - 3 y$

Add $9 y^{2}$ and $y^{2}$ .

$10 y^{2} = 90$

$x = - 5 - 3 y$

$10 y^{2} = 90$

$x = - 5 - 3 y$

$10 y^{2} = 90$

$x = - 5 - 3 y$

$10 y^{2} = 90$

$x = - 5 - 3 y$

Solve for $y$ in $10 y^{2} = 90$ .

Divide each term in $10 y^{2} = 90$ by $10$ and simplify.

Divide each term in $10 y^{2} = 90$ by $10$ .

$\frac{10 y^{2}}{10} = \frac{90}{10}$

$x = - 5 - 3 y$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 y^{2}}{10} = \frac{90}{10}$

$x = - 5 - 3 y$

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

$y^{2} = \frac{90}{10}$

$x = - 5 - 3 y$

$y^{2} = \frac{90}{10}$

$x = - 5 - 3 y$

$y^{2} = \frac{90}{10}$

$x = - 5 - 3 y$

Simplify the right side.

Divide $90$ by $10$ .

$y^{2} = 9$

$x = - 5 - 3 y$

$y^{2} = 9$

$x = - 5 - 3 y$

$y^{2} = 9$

$x = - 5 - 3 y$

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

$y = \pm \sqrt{9}$

$x = - 5 - 3 y$

Simplify $\pm \sqrt{9}$ .

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

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

$x = - 5 - 3 y$

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

$y = \pm 3$

$x = - 5 - 3 y$

$y = \pm 3$

$x = - 5 - 3 y$

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.

$y = 3$

$x = - 5 - 3 y$

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

$y = - 3$

$x = - 5 - 3 y$

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

$y = 3, - 3$

$x = - 5 - 3 y$

$y = 3, - 3$

$x = - 5 - 3 y$

$y = 3, - 3$

$x = - 5 - 3 y$

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

Replace all occurrences of $y$ in $x = - 5 - 3 y$ with $3$ .

$x = - 5 - 3 \cdot 3$

$y = 3$

Simplify the right side.

Simplify $- 5 - 3 \cdot 3$ .

Multiply $- 3$ by $3$ .

$x = - 5 - 9$

$y = 3$

Subtract $9$ from $- 5$ .

$x = - 14$

$y = 3$

$x = - 14$

$y = 3$

$x = - 14$

$y = 3$

$x = - 14$

$y = 3$

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

Replace all occurrences of $y$ in $x = - 5 - 3 y$ with $- 3$ .

$x = - 5 - 3 \cdot - 3$

$y = - 3$

Simplify the right side.

Simplify $- 5 - 3 \cdot - 3$ .

Multiply $- 3$ by $- 3$ .

$x = - 5 + 9$

$y = - 3$

Add $- 5$ and $9$ .

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

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

$(- 14, 3)$

$(4, - 3)$

The result can be shown in multiple forms.

Point Form:

$(- 14, 3), (4, - 3)$

Equation Form:

$x = - 14, y = 3$

$x = 4, y = - 3$

Solve the System of Equations (x+5)^2+y^2=90 x+3y=-5

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