Find the Points of Intersection x=3-y^2 y=x-1

$x = 3 - y^{2}$ $y = x - 1$

Step 1

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

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

$x = 3 - {(x - 1)}^{2}$

$y = x - 1$

Simplify the right side.

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

Simplify each term.

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

$x = 3 - ((x - 1) (x - 1))$

$y = x - 1$

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

Apply the distributive property.

$x = 3 - (x (x - 1) - 1 (x - 1))$

$y = x - 1$

Apply the distributive property.

$x = 3 - (x \cdot x + x \cdot - 1 - 1 (x - 1))$

$y = x - 1$

Apply the distributive property.

$x = 3 - (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1)$

$y = x - 1$

$x = 3 - (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1)$

$y = x - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

$y = x - 1$

Move $- 1$ to the left of $x$ .

$x = 3 - (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1)$

$y = x - 1$

Rewrite $- 1 x$ as $- x$ .

$x = 3 - (x^{2} - x - 1 x - 1 \cdot - 1)$

$y = x - 1$

Rewrite $- 1 x$ as $- x$ .

$x = 3 - (x^{2} - x - x - 1 \cdot - 1)$

$y = x - 1$

Multiply $- 1$ by $- 1$ .

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

$y = x - 1$

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

$y = x - 1$

Subtract $x$ from $- x$ .

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

$y = x - 1$

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

$y = x - 1$

Apply the distributive property.

$x = 3 - x^{2} - (- 2 x) - 1 \cdot 1$

$y = x - 1$

Simplify.

Multiply $- 2$ by $- 1$ .

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

$y = x - 1$

Multiply $- 1$ by $1$ .

$x = 3 - x^{2} + 2 x - 1$

$y = x - 1$

$x = 3 - x^{2} + 2 x - 1$

$y = x - 1$

$x = 3 - x^{2} + 2 x - 1$

$y = x - 1$

Subtract $1$ from $3$ .

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

$y = x - 1$

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

$y = x - 1$

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

$y = x - 1$

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

$y = x - 1$

Step 2

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

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

$y = x - 1$

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

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

$y = x - 1$

Subtract $x$ from $2 x$ .

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

$y = x - 1$

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

$y = x - 1$

Factor the left side of the equation.

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

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

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

$y = x - 1$

Factor $- 1$ out of $x$ .

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

$y = x - 1$

Rewrite $2$ as $- 1 (- 2)$ .

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

$y = x - 1$

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

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

$y = x - 1$

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

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

$y = x - 1$

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

$y = x - 1$

Factor.

Factor $x^{2} - x - 2$ 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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

$y = x - 1$

Write the factored form using these integers.

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

$y = x - 1$

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

$y = x - 1$

Remove unnecessary parentheses.

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

$y = x - 1$

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

$y = x - 1$

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

$y = x - 1$

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

$x - 2 = 0$

$x + 1 = 0$

$y = x - 1$

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

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

$x - 2 = 0$

$y = x - 1$

Add $2$ to both sides of the equation.

$x = 2$

$y = x - 1$

$x = 2$

$y = x - 1$

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

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

$x + 1 = 0$

$y = x - 1$

Subtract $1$ from both sides of the equation.

$x = - 1$

$y = x - 1$

$x = - 1$

$y = x - 1$

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

$x = 2, - 1$

$y = x - 1$

$x = 2, - 1$

$y = x - 1$

Step 3

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

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

$y = (2) - 1$

$x = 2$

Simplify the right side.

Subtract $1$ from $2$ .

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

Step 4

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

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

$y = (- 1) - 1$

$x = - 1$

Simplify the right side.

Subtract $1$ from $- 1$ .

$y = - 2$

$x = - 1$

$y = - 2$

$x = - 1$

$y = - 2$

$x = - 1$

Step 5

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

$(2, 1)$

$(- 1, - 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 2, y = 1$

$x = - 1, y = - 2$

Step 7

Find the Points of Intersection x=3-y^2 y=x-1

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