Solve the System of Equations y=x^2-2x-3 y=x-3

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

Eliminate the equal sides of each equation and combine.

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

Solve $x^{2} - 2 x - 3 = x - 3$ for $x$ .

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

Subtract $x$ from both sides of the equation.

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

Subtract $x$ from $- 2 x$ .

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

Add $3$ to both sides of the equation.

$x^{2} - 3 x - 3 + 3 = 0$

Combine the opposite terms in $x^{2} - 3 x - 3 + 3$ .

Add $- 3$ and $3$ .

$x^{2} - 3 x + 0 = 0$

Add $x^{2} - 3 x$ and $0$ .

$x^{2} - 3 x = 0$

Factor $x$ out of $x^{2} - 3 x$ .

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

$x \cdot x - 3 x = 0$

Factor $x$ out of $- 3 x$ .

$x \cdot x + x \cdot - 3 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 3$ .

$x (x - 3) = 0$

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

$x = 0$

$x - 3 = 0 + y = x - 3$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $x (x - 3) = 0$ true.

$x = 0, 3$

Evaluate $y$ when $x = 0$ .

Substitute $0$ for $x$ .

$y = (0) - 3$

Substitute $0$ for $x$ in $y = (0) - 3$ and solve for $y$ .

Remove parentheses.

$y = 0 - 3$

Remove parentheses.

$y = (0) - 3$

Subtract $3$ from $0$ .

$y = - 3$

Evaluate $y$ when $x = 3$ .

Substitute $3$ for $x$ .

$y = (3) - 3$

Substitute $3$ for $x$ in $y = (3) - 3$ and solve for $y$ .

Remove parentheses.

$y = 3 - 3$

Remove parentheses.

$y = (3) - 3$

Subtract $3$ from $3$ .

$y = 0$

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

$(0, - 3)$

$(3, 0)$

The result can be shown in multiple forms.

Point Form:

$(0, - 3), (3, 0)$

Equation Form:

$x = 0, y = - 3$

$x = 3, y = 0$

Solve the System of Equations y=x^2-2x-3 y=x-3

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