Find the Points of Intersection -x+y=-2 x^2+y^2=2

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

Step 1

Add $x$ to both sides of the equation.

$y = - 2 + x$

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

Step 2

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

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

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

$y = - 2 + x$

Simplify the left side.

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

Simplify each term.

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

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

$y = - 2 + x$

Expand $(- 2 + x) (- 2 + x)$ using the FOIL Method.

Apply the distributive property.

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

$y = - 2 + x$

Apply the distributive property.

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

$y = - 2 + x$

Apply the distributive property.

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

$y = - 2 + x$

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

$y = - 2 + x$

Simplify and combine like terms.

Simplify each term.

Multiply $- 2$ by $- 2$ .

$x^{2} + 4 - 2 x + x \cdot - 2 + x \cdot x = 2$

$y = - 2 + x$

Move $- 2$ to the left of $x$ .

$x^{2} + 4 - 2 x - 2 \cdot x + x \cdot x = 2$

$y = - 2 + x$

Multiply $x$ by $x$ .

$x^{2} + 4 - 2 x - 2 x + x^{2} = 2$

$y = - 2 + x$

$x^{2} + 4 - 2 x - 2 x + x^{2} = 2$

$y = - 2 + x$

Subtract $2 x$ from $- 2 x$ .

$x^{2} + 4 - 4 x + x^{2} = 2$

$y = - 2 + x$

$x^{2} + 4 - 4 x + x^{2} = 2$

$y = - 2 + x$

$x^{2} + 4 - 4 x + x^{2} = 2$

$y = - 2 + x$

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

$2 x^{2} + 4 - 4 x = 2$

$y = - 2 + x$

$2 x^{2} + 4 - 4 x = 2$

$y = - 2 + x$

$2 x^{2} + 4 - 4 x = 2$

$y = - 2 + x$

$2 x^{2} + 4 - 4 x = 2$

$y = - 2 + x$

Step 3

Solve for $x$ in $2 x^{2} + 4 - 4 x = 2$ .

Subtract $2$ from both sides of the equation.

$2 x^{2} + 4 - 4 x - 2 = 0$

$y = - 2 + x$

Subtract $2$ from $4$ .

$2 x^{2} - 4 x + 2 = 0$

$y = - 2 + x$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 4 x + 2$ .

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

$2 (x^{2}) - 4 x + 2 = 0$

$y = - 2 + x$

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

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

$y = - 2 + x$

Factor $2$ out of $2$ .

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

$y = - 2 + x$

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

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

$y = - 2 + x$

Factor $2$ out of $2 (x^{2} - 2 x) + 2 (1)$ .

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

$y = - 2 + x$

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

$y = - 2 + x$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

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

$y = - 2 + x$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

$y = - 2 + x$

Rewrite the polynomial.

$2 (x^{2} - 2 \cdot x \cdot 1 + 1^{2}) = 0$

$y = - 2 + x$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

$2 {(x - 1)}^{2} = 0$

$y = - 2 + x$

$2 {(x - 1)}^{2} = 0$

$y = - 2 + x$

$2 {(x - 1)}^{2} = 0$

$y = - 2 + x$

Divide each term in $2 {(x - 1)}^{2} = 0$ by $2$ and simplify.

Divide each term in $2 {(x - 1)}^{2} = 0$ by $2$ .

$\frac{2 {(x - 1)}^{2}}{2} = \frac{0}{2}$

$y = - 2 + x$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 {(x - 1)}^{2}}{2} = \frac{0}{2}$

$y = - 2 + x$

Divide ${(x - 1)}^{2}$ by $1$ .

${(x - 1)}^{2} = \frac{0}{2}$

$y = - 2 + x$

${(x - 1)}^{2} = \frac{0}{2}$

$y = - 2 + x$

${(x - 1)}^{2} = \frac{0}{2}$

$y = - 2 + x$

Simplify the right side.

Divide $0$ by $2$ .

${(x - 1)}^{2} = 0$

$y = - 2 + x$

${(x - 1)}^{2} = 0$

$y = - 2 + x$

${(x - 1)}^{2} = 0$

$y = - 2 + x$

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

$y = - 2 + x$

Add $1$ to both sides of the equation.

$x = 1$

$y = - 2 + x$

$x = 1$

$y = - 2 + x$

Step 4

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

Replace all occurrences of $x$ in $y = - 2 + x$ with $1$ .

$y = - 2 + 1$

$x = 1$

Simplify $y = - 2 + 1$ .

Simplify the left side.

Remove parentheses.

$y = - 2 + 1$

$x = 1$

$y = - 2 + 1$

$x = 1$

Simplify the right side.

Add $- 2$ and $1$ .

$y = - 1$

$x = 1$

$y = - 1$

$x = 1$

$y = - 1$

$x = 1$

$y = - 1$

$x = 1$

Step 5

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

$(1, - 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(1, - 1)$

Equation Form:

$x = 1, y = - 1$

Step 7

Find the Points of Intersection -x+y=-2 x^2+y^2=2

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