Solve the System of Equations y=x^2 y=x+2

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

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} = x + 2$

Step 2

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

Subtract $x$ from both sides of the equation.

$x^{2} - x = 2$

Subtract $2$ from both sides of the equation.

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

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 $- 1$ .

$- 2, 1 + y = x + 2$

Write the factored form using these integers.

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

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

$x - 2 = 0$

$x + 1 = 0 + y = x + 2$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = 2, - 1$

Step 3

Evaluate $y$ when $x = 2$ .

Substitute $2$ for $x$ .

$y = (2) + 2$

Substitute $2$ for $x$ in $y = (2) + 2$ and solve for $y$ .

Remove parentheses.

$y = 2 + 2$

Remove parentheses.

$y = (2) + 2$

Add $2$ and $2$ .

$y = 4$

Step 4

Evaluate $y$ when $x = - 1$ .

Substitute $- 1$ for $x$ .

$y = (- 1) + 2$

Substitute $- 1$ for $x$ in $y = (- 1) + 2$ and solve for $y$ .

Remove parentheses.

$y = - 1 + 2$

Remove parentheses.

$y = (- 1) + 2$

Add $- 1$ and $2$ .

$y = 1$

Step 5

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

$(2, 4)$

$(- 1, 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 2, y = 4$

$x = - 1, y = 1$

Step 7

Solve the System of Equations y=x^2 y=x+2

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