Solve the System of Equations (x+3)^2+(y+2)^2=17 x+y=-2

${(x + 3)}^{2} + {(y + 2)}^{2} = 17$ $x + y = - 2$

Step 1

Subtract $y$ from both sides of the equation.

$x = - 2 - y$

${(x + 3)}^{2} + {(y + 2)}^{2} = 17$

Step 2

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

Replace all occurrences of $x$ in ${(x + 3)}^{2} + {(y + 2)}^{2} = 17$ with $- 2 - y$ .

${((- 2 - y) + 3)}^{2} + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Simplify the left side.

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

Simplify each term.

Add $- 2$ and $3$ .

${(- y + 1)}^{2} + {(y + 2)}^{2} = 17$

$x = - 2 - y$

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

$(- y + 1) (- y + 1) + {(y + 2)}^{2} = 17$

$x = - 2 - y$

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

Apply the distributive property.

$- y (- y + 1) + 1 (- y + 1) + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Apply the distributive property.

$- y (- y) - y \cdot 1 + 1 (- y + 1) + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Apply the distributive property.

$- y (- y) - y \cdot 1 + 1 (- y) + 1 \cdot 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

$- y (- y) - y \cdot 1 + 1 (- y) + 1 \cdot 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$- 1 \cdot (- 1 y \cdot y) - y \cdot 1 + 1 (- y) + 1 \cdot 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

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

Move $y$ .

$- 1 \cdot (- 1 (y \cdot y)) - y \cdot 1 + 1 (- y) + 1 \cdot 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Multiply $y$ by $y$ .

$- 1 \cdot (- 1 y^{2}) - y \cdot 1 + 1 (- y) + 1 \cdot 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

$- 1 \cdot (- 1 y^{2}) - y \cdot 1 + 1 (- y) + 1 \cdot 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Multiply $- 1$ by $- 1$ .

$1 y^{2} - y \cdot 1 + 1 (- y) + 1 \cdot 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

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

$y^{2} - y \cdot 1 + 1 (- y) + 1 \cdot 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Multiply $- 1$ by $1$ .

$y^{2} - y + 1 (- y) + 1 \cdot 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Multiply $- y$ by $1$ .

$y^{2} - y - y + 1 \cdot 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Multiply $1$ by $1$ .

$y^{2} - y - y + 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

$y^{2} - y - y + 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Subtract $y$ from $- y$ .

$y^{2} - 2 y + 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

$y^{2} - 2 y + 1 + {(y + 2)}^{2} = 17$

$x = - 2 - y$

Rewrite ${(y + 2)}^{2}$ as $(y + 2) (y + 2)$ .

$y^{2} - 2 y + 1 + (y + 2) (y + 2) = 17$

$x = - 2 - y$

Expand $(y + 2) (y + 2)$ using the FOIL Method.

Apply the distributive property.

$y^{2} - 2 y + 1 + y (y + 2) + 2 (y + 2) = 17$

$x = - 2 - y$

Apply the distributive property.

$y^{2} - 2 y + 1 + y \cdot y + y \cdot 2 + 2 (y + 2) = 17$

$x = - 2 - y$

Apply the distributive property.

$y^{2} - 2 y + 1 + y \cdot y + y \cdot 2 + 2 y + 2 \cdot 2 = 17$

$x = - 2 - y$

$y^{2} - 2 y + 1 + y \cdot y + y \cdot 2 + 2 y + 2 \cdot 2 = 17$

$x = - 2 - y$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$y^{2} - 2 y + 1 + y^{2} + y \cdot 2 + 2 y + 2 \cdot 2 = 17$

$x = - 2 - y$

Move $2$ to the left of $y$ .

$y^{2} - 2 y + 1 + y^{2} + 2 \cdot y + 2 y + 2 \cdot 2 = 17$

$x = - 2 - y$

Multiply $2$ by $2$ .

$y^{2} - 2 y + 1 + y^{2} + 2 y + 2 y + 4 = 17$

$x = - 2 - y$

$y^{2} - 2 y + 1 + y^{2} + 2 y + 2 y + 4 = 17$

$x = - 2 - y$

Add $2 y$ and $2 y$ .

$y^{2} - 2 y + 1 + y^{2} + 4 y + 4 = 17$

$x = - 2 - y$

$y^{2} - 2 y + 1 + y^{2} + 4 y + 4 = 17$

$x = - 2 - y$

$y^{2} - 2 y + 1 + y^{2} + 4 y + 4 = 17$

$x = - 2 - y$

Simplify by adding terms.

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

$2 y^{2} - 2 y + 1 + 4 y + 4 = 17$

$x = - 2 - y$

Add $- 2 y$ and $4 y$ .

$2 y^{2} + 2 y + 1 + 4 = 17$

$x = - 2 - y$

Add $1$ and $4$ .

$2 y^{2} + 2 y + 5 = 17$

$x = - 2 - y$

$2 y^{2} + 2 y + 5 = 17$

$x = - 2 - y$

$2 y^{2} + 2 y + 5 = 17$

$x = - 2 - y$

$2 y^{2} + 2 y + 5 = 17$

$x = - 2 - y$

$2 y^{2} + 2 y + 5 = 17$

$x = - 2 - y$

Step 3

Solve for $y$ in $2 y^{2} + 2 y + 5 = 17$ .

Subtract $17$ from both sides of the equation.

$2 y^{2} + 2 y + 5 - 17 = 0$

$x = - 2 - y$

Subtract $17$ from $5$ .

$2 y^{2} + 2 y - 12 = 0$

$x = - 2 - y$

Factor the left side of the equation.

Factor $2$ out of $2 y^{2} + 2 y - 12$ .

Factor $2$ out of $2 y^{2}$ .

$2 (y^{2}) + 2 y - 12 = 0$

$x = - 2 - y$

Factor $2$ out of $2 y$ .

$2 (y^{2}) + 2 (y) - 12 = 0$

$x = - 2 - y$

Factor $2$ out of $- 12$ .

$2 (y^{2}) + 2 y + 2 \cdot - 6 = 0$

$x = - 2 - y$

Factor $2$ out of $2 (y^{2}) + 2 y$ .

$2 (y^{2} + y) + 2 \cdot - 6 = 0$

$x = - 2 - y$

Factor $2$ out of $2 (y^{2} + y) + 2 \cdot - 6$ .

$2 (y^{2} + y - 6) = 0$

$x = - 2 - y$

$2 (y^{2} + y - 6) = 0$

$x = - 2 - y$

Factor.

Factor $y^{2} + y - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

$x = - 2 - y$

Write the factored form using these integers.

$2 ((y - 2) (y + 3)) = 0$

$x = - 2 - y$

$2 ((y - 2) (y + 3)) = 0$

$x = - 2 - y$

Remove unnecessary parentheses.

$2 (y - 2) (y + 3) = 0$

$x = - 2 - y$

$2 (y - 2) (y + 3) = 0$

$x = - 2 - y$

$2 (y - 2) (y + 3) = 0$

$x = - 2 - y$

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

$y - 2 = 0$

$y + 3 = 0$

$x = - 2 - y$

Set $y - 2$ equal to $0$ and solve for $y$ .

Set $y - 2$ equal to $0$ .

$y - 2 = 0$

$x = - 2 - y$

Add $2$ to both sides of the equation.

$y = 2$

$x = - 2 - y$

$y = 2$

$x = - 2 - y$

Set $y + 3$ equal to $0$ and solve for $y$ .

Set $y + 3$ equal to $0$ .

$y + 3 = 0$

$x = - 2 - y$

Subtract $3$ from both sides of the equation.

$y = - 3$

$x = - 2 - y$

$y = - 3$

$x = - 2 - y$

The final solution is all the values that make $2 (y - 2) (y + 3) = 0$ true.

$y = 2, - 3$

$x = - 2 - y$

$y = 2, - 3$

$x = - 2 - y$

Step 4

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

Replace all occurrences of $y$ in $x = - 2 - y$ with $2$ .

$x = - 2 - (2)$

$y = 2$

Simplify the right side.

Simplify $- 2 - (2)$ .

Multiply $- 1$ by $2$ .

$x = - 2 - 2$

$y = 2$

Subtract $2$ from $- 2$ .

$x = - 4$

$y = 2$

$x = - 4$

$y = 2$

$x = - 4$

$y = 2$

$x = - 4$

$y = 2$

Step 5

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

Replace all occurrences of $y$ in $x = - 2 - y$ with $- 3$ .

$x = - 2 - (- 3)$

$y = - 3$

Simplify the right side.

Simplify $- 2 - (- 3)$ .

Multiply $- 1$ by $- 3$ .

$x = - 2 + 3$

$y = - 3$

Add $- 2$ and $3$ .

$x = 1$

$y = - 3$

$x = 1$

$y = - 3$

$x = 1$

$y = - 3$

$x = 1$

$y = - 3$

Step 6

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

$(- 4, 2)$

$(1, - 3)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 4, 2), (1, - 3)$

Equation Form:

$x = - 4, y = 2$

$x = 1, y = - 3$

Step 8

Solve the System of Equations (x+3)^2+(y+2)^2=17 x+y=-2

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