Solve the System of Equations (x-2)^2+(y-3)^2=16 x+y-1=0

${(x - 2)}^{2} + {(y - 3)}^{2} = 16$ $x + y - 1 = 0$

Step 1

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

Subtract $y$ from both sides of the equation.

$x - 1 = - y$

${(x - 2)}^{2} + {(y - 3)}^{2} = 16$

Add $1$ to both sides of the equation.

$x = - y + 1$

${(x - 2)}^{2} + {(y - 3)}^{2} = 16$

$x = - y + 1$

${(x - 2)}^{2} + {(y - 3)}^{2} = 16$

Step 2

Replace all occurrences of $x$ with $- y + 1$ in each equation.

Replace all occurrences of $x$ in ${(x - 2)}^{2} + {(y - 3)}^{2} = 16$ with $- y + 1$ .

${((- y + 1) - 2)}^{2} + {(y - 3)}^{2} = 16$

$x = - y + 1$

Simplify the left side.

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

Simplify each term.

Subtract $2$ from $1$ .

${(- y - 1)}^{2} + {(y - 3)}^{2} = 16$

$x = - y + 1$

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

$(- y - 1) (- y - 1) + {(y - 3)}^{2} = 16$

$x = - y + 1$

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

Apply the distributive property.

$- y (- y - 1) - 1 (- y - 1) + {(y - 3)}^{2} = 16$

$x = - y + 1$

Apply the distributive property.

$- y (- y) - y \cdot - 1 - 1 (- y - 1) + {(y - 3)}^{2} = 16$

$x = - y + 1$

Apply the distributive property.

$- y (- y) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

$- y (- y) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$- 1 \cdot (- 1 y \cdot y) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

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

Move $y$ .

$- 1 \cdot (- 1 (y \cdot y)) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

Multiply $y$ by $y$ .

$- 1 \cdot (- 1 y^{2}) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

$- 1 \cdot (- 1 y^{2}) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

Multiply $- 1$ by $- 1$ .

$1 y^{2} - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

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

$y^{2} - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

Multiply $- y \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$y^{2} + 1 y - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

Multiply $y$ by $1$ .

$y^{2} + y - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

$y^{2} + y - 1 (- y) - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

Multiply $- 1 (- y)$ .

Multiply $- 1$ by $- 1$ .

$y^{2} + y + 1 y - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

Multiply $y$ by $1$ .

$y^{2} + y + y - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

$y^{2} + y + y - 1 \cdot - 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

Multiply $- 1$ by $- 1$ .

$y^{2} + y + y + 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

$y^{2} + y + y + 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

Add $y$ and $y$ .

$y^{2} + 2 y + 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

$y^{2} + 2 y + 1 + {(y - 3)}^{2} = 16$

$x = - y + 1$

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

$y^{2} + 2 y + 1 + (y - 3) (y - 3) = 16$

$x = - y + 1$

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

Apply the distributive property.

$y^{2} + 2 y + 1 + y (y - 3) - 3 (y - 3) = 16$

$x = - y + 1$

Apply the distributive property.

$y^{2} + 2 y + 1 + y \cdot y + y \cdot - 3 - 3 (y - 3) = 16$

$x = - y + 1$

Apply the distributive property.

$y^{2} + 2 y + 1 + y \cdot y + y \cdot - 3 - 3 y - 3 \cdot - 3 = 16$

$x = - y + 1$

$y^{2} + 2 y + 1 + y \cdot y + y \cdot - 3 - 3 y - 3 \cdot - 3 = 16$

$x = - y + 1$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$y^{2} + 2 y + 1 + y^{2} + y \cdot - 3 - 3 y - 3 \cdot - 3 = 16$

$x = - y + 1$

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

$y^{2} + 2 y + 1 + y^{2} - 3 \cdot y - 3 y - 3 \cdot - 3 = 16$

$x = - y + 1$

Multiply $- 3$ by $- 3$ .

$y^{2} + 2 y + 1 + y^{2} - 3 y - 3 y + 9 = 16$

$x = - y + 1$

$y^{2} + 2 y + 1 + y^{2} - 3 y - 3 y + 9 = 16$

$x = - y + 1$

Subtract $3 y$ from $- 3 y$ .

$y^{2} + 2 y + 1 + y^{2} - 6 y + 9 = 16$

$x = - y + 1$

$y^{2} + 2 y + 1 + y^{2} - 6 y + 9 = 16$

$x = - y + 1$

$y^{2} + 2 y + 1 + y^{2} - 6 y + 9 = 16$

$x = - y + 1$

Simplify by adding terms.

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

$2 y^{2} + 2 y + 1 - 6 y + 9 = 16$

$x = - y + 1$

Subtract $6 y$ from $2 y$ .

$2 y^{2} - 4 y + 1 + 9 = 16$

$x = - y + 1$

Add $1$ and $9$ .

$2 y^{2} - 4 y + 10 = 16$

$x = - y + 1$

$2 y^{2} - 4 y + 10 = 16$

$x = - y + 1$

$2 y^{2} - 4 y + 10 = 16$

$x = - y + 1$

$2 y^{2} - 4 y + 10 = 16$

$x = - y + 1$

$2 y^{2} - 4 y + 10 = 16$

$x = - y + 1$

Step 3

Solve for $y$ in $2 y^{2} - 4 y + 10 = 16$ .

Subtract $16$ from both sides of the equation.

$2 y^{2} - 4 y + 10 - 16 = 0$

$x = - y + 1$

Subtract $16$ from $10$ .

$2 y^{2} - 4 y - 6 = 0$

$x = - y + 1$

Factor the left side of the equation.

Factor $2$ out of $2 y^{2} - 4 y - 6$ .

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

$2 (y^{2}) - 4 y - 6 = 0$

$x = - y + 1$

Factor $2$ out of $- 4 y$ .

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

$x = - y + 1$

Factor $2$ out of $- 6$ .

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

$x = - y + 1$

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

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

$x = - y + 1$

Factor $2$ out of $2 (y^{2} - 2 y) + 2 \cdot - 3$ .

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

$x = - y + 1$

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

$x = - y + 1$

Factor.

Factor $y^{2} - 2 y - 3$ 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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

$x = - y + 1$

Write the factored form using these integers.

$2 ((y - 3) (y + 1)) = 0$

$x = - y + 1$

$2 ((y - 3) (y + 1)) = 0$

$x = - y + 1$

Remove unnecessary parentheses.

$2 (y - 3) (y + 1) = 0$

$x = - y + 1$

$2 (y - 3) (y + 1) = 0$

$x = - y + 1$

$2 (y - 3) (y + 1) = 0$

$x = - y + 1$

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

$y - 3 = 0$

$y + 1 = 0$

$x = - y + 1$

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

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

$y - 3 = 0$

$x = - y + 1$

Add $3$ to both sides of the equation.

$y = 3$

$x = - y + 1$

$y = 3$

$x = - y + 1$

Set $y + 1$ equal to $0$ and solve for $y$ .

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

$x = - y + 1$

Subtract $1$ from both sides of the equation.

$y = - 1$

$x = - y + 1$

$y = - 1$

$x = - y + 1$

The final solution is all the values that make $2 (y - 3) (y + 1) = 0$ true.

$y = 3, - 1$

$x = - y + 1$

$y = 3, - 1$

$x = - y + 1$

Step 4

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

Replace all occurrences of $y$ in $x = - y + 1$ with $3$ .

$x = - (3) + 1$

$y = 3$

Simplify the right side.

Simplify $- (3) + 1$ .

Multiply $- 1$ by $3$ .

$x = - 3 + 1$

$y = 3$

Add $- 3$ and $1$ .

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

Step 5

Replace all occurrences of $y$ with $- 1$ in each equation.

Replace all occurrences of $y$ in $x = - y + 1$ with $- 1$ .

$x = - (- 1) + 1$

$y = - 1$

Simplify the right side.

Simplify $- (- 1) + 1$ .

Multiply $- 1$ by $- 1$ .

$x = 1 + 1$

$y = - 1$

Add $1$ and $1$ .

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

Step 6

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

$(- 2, 3)$

$(2, - 1)$

Step 7

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = - 2, y = 3$

$x = 2, y = - 1$

Step 8

Solve the System of Equations (x-2)^2+(y-3)^2=16 x+y-1=0

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