Solve the System of Equations x^2+(y+1)^2=37 -x+y=-6

$x^{2} + {(y + 1)}^{2} = 37$ $- x + y = - 6$

Add $x$ to both sides of the equation.

$y = - 6 + x$

$x^{2} + {(y + 1)}^{2} = 37$

Replace all occurrences of $y$ with $- 6 + x$ in each equation.

Replace all occurrences of $y$ in $x^{2} + {(y + 1)}^{2} = 37$ with $- 6 + x$ .

$x^{2} + {((- 6 + x) + 1)}^{2} = 37$

$y = - 6 + x$

Simplify the left side.

Simplify $x^{2} + {((- 6 + x) + 1)}^{2}$ .

Simplify each term.

Add $- 6$ and $1$ .

$x^{2} + {(x - 5)}^{2} = 37$

$y = - 6 + x$

Rewrite ${(x - 5)}^{2}$ as $(x - 5) (x - 5)$ .

$x^{2} + (x - 5) (x - 5) = 37$

$y = - 6 + x$

Expand $(x - 5) (x - 5)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + x (x - 5) - 5 (x - 5) = 37$

$y = - 6 + x$

Apply the distributive property.

$x^{2} + x \cdot x + x \cdot - 5 - 5 (x - 5) = 37$

$y = - 6 + x$

Apply the distributive property.

$x^{2} + x \cdot x + x \cdot - 5 - 5 x - 5 \cdot - 5 = 37$

$y = - 6 + x$

$x^{2} + x \cdot x + x \cdot - 5 - 5 x - 5 \cdot - 5 = 37$

$y = - 6 + x$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x^{2} + x \cdot - 5 - 5 x - 5 \cdot - 5 = 37$

$y = - 6 + x$

Move $- 5$ to the left of $x$ .

$x^{2} + x^{2} - 5 \cdot x - 5 x - 5 \cdot - 5 = 37$

$y = - 6 + x$

Multiply $- 5$ by $- 5$ .

$x^{2} + x^{2} - 5 x - 5 x + 25 = 37$

$y = - 6 + x$

$x^{2} + x^{2} - 5 x - 5 x + 25 = 37$

$y = - 6 + x$

Subtract $5 x$ from $- 5 x$ .

$x^{2} + x^{2} - 10 x + 25 = 37$

$y = - 6 + x$

$x^{2} + x^{2} - 10 x + 25 = 37$

$y = - 6 + x$

$x^{2} + x^{2} - 10 x + 25 = 37$

$y = - 6 + x$

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

$2 x^{2} - 10 x + 25 = 37$

$y = - 6 + x$

$2 x^{2} - 10 x + 25 = 37$

$y = - 6 + x$

$2 x^{2} - 10 x + 25 = 37$

$y = - 6 + x$

$2 x^{2} - 10 x + 25 = 37$

$y = - 6 + x$

Solve for $x$ in $2 x^{2} - 10 x + 25 = 37$ .

Subtract $37$ from both sides of the equation.

$2 x^{2} - 10 x + 25 - 37 = 0$

$y = - 6 + x$

Subtract $37$ from $25$ .

$2 x^{2} - 10 x - 12 = 0$

$y = - 6 + x$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 10 x - 12$ .

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

$2 (x^{2}) - 10 x - 12 = 0$

$y = - 6 + x$

Factor $2$ out of $- 10 x$ .

$2 (x^{2}) + 2 (- 5 x) - 12 = 0$

$y = - 6 + x$

Factor $2$ out of $- 12$ .

$2 x^{2} + 2 (- 5 x) + 2 \cdot - 6 = 0$

$y = - 6 + x$

Factor $2$ out of $2 x^{2} + 2 (- 5 x)$ .

$2 (x^{2} - 5 x) + 2 \cdot - 6 = 0$

$y = - 6 + x$

Factor $2$ out of $2 (x^{2} - 5 x) + 2 \cdot - 6$ .

$2 (x^{2} - 5 x - 6) = 0$

$y = - 6 + x$

$2 (x^{2} - 5 x - 6) = 0$

$y = - 6 + x$

Factor.

Factor $x^{2} - 5 x - 6$ 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 $- 5$ .

$- 6, 1$

$y = - 6 + x$

Write the factored form using these integers.

$2 ((x - 6) (x + 1)) = 0$

$y = - 6 + x$

$2 ((x - 6) (x + 1)) = 0$

$y = - 6 + x$

Remove unnecessary parentheses.

$2 (x - 6) (x + 1) = 0$

$y = - 6 + x$

$2 (x - 6) (x + 1) = 0$

$y = - 6 + x$

$2 (x - 6) (x + 1) = 0$

$y = - 6 + x$

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

$x - 6 = 0$

$x + 1 = 0$

$y = - 6 + x$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

$y = - 6 + x$

Add $6$ to both sides of the equation.

$x = 6$

$y = - 6 + x$

$x = 6$

$y = - 6 + x$

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

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

$x + 1 = 0$

$y = - 6 + x$

Subtract $1$ from both sides of the equation.

$x = - 1$

$y = - 6 + x$

$x = - 1$

$y = - 6 + x$

The final solution is all the values that make $2 (x - 6) (x + 1) = 0$ true.

$x = 6, - 1$

$y = - 6 + x$

$x = 6, - 1$

$y = - 6 + x$

Replace all occurrences of $x$ with $6$ in each equation.

Replace all occurrences of $x$ in $y = - 6 + x$ with $6$ .

$y = - 6 + 6$

$x = 6$

Simplify $y = - 6 + 6$ .

Simplify the left side.

Remove parentheses.

$y = - 6 + 6$

$x = 6$

$y = - 6 + 6$

$x = 6$

Simplify the right side.

Add $- 6$ and $6$ .

$y = 0$

$x = 6$

$y = 0$

$x = 6$

$y = 0$

$x = 6$

$y = 0$

$x = 6$

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

Replace all occurrences of $x$ in $y = - 6 + x$ with $- 1$ .

$y = - 6 - 1$

$x = - 1$

Simplify $y = - 6 - 1$ .

Simplify the left side.

Remove parentheses.

$y = - 6 - 1$

$x = - 1$

$y = - 6 - 1$

$x = - 1$

Simplify the right side.

Subtract $1$ from $- 6$ .

$y = - 7$

$x = - 1$

$y = - 7$

$x = - 1$

$y = - 7$

$x = - 1$

$y = - 7$

$x = - 1$

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

$(6, 0)$

$(- 1, - 7)$

The result can be shown in multiple forms.

Point Form:

$(6, 0), (- 1, - 7)$

Equation Form:

$x = 6, y = 0$

$x = - 1, y = - 7$

Solve the System of Equations x^2+(y+1)^2=37 -x+y=-6

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