Solve the System of Equations y=-x^2-3x-10 10=x-y

$y = - x^{2} - 3 x - 10$ $10 = x - y$

Replace all occurrences of $y$ with $- x^{2} - 3 x - 10$ in each equation.

Replace all occurrences of $y$ in $10 = x - y$ with $- x^{2} - 3 x - 10$ .

$10 = x - (- x^{2} - 3 x - 10)$

$y = - x^{2} - 3 x - 10$

Simplify the right side.

Simplify $x - (- x^{2} - 3 x - 10)$ .

Simplify each term.

Apply the distributive property.

$10 = x + x^{2} - (- 3 x) + 10$

$y = - x^{2} - 3 x - 10$

Simplify.

Multiply $- - x^{2}$ .

Multiply $- 1$ by $- 1$ .

$10 = x + 1 x^{2} - (- 3 x) + 10$

$y = - x^{2} - 3 x - 10$

Multiply $x^{2}$ by $1$ .

$10 = x + x^{2} - (- 3 x) + 10$

$y = - x^{2} - 3 x - 10$

$10 = x + x^{2} - (- 3 x) + 10$

$y = - x^{2} - 3 x - 10$

Multiply $- 3$ by $- 1$ .

$10 = x + x^{2} + 3 x + 10$

$y = - x^{2} - 3 x - 10$

Multiply $- 1$ by $- 10$ .

$10 = x + x^{2} + 3 x + 10$

$y = - x^{2} - 3 x - 10$

$10 = x + x^{2} + 3 x + 10$

$y = - x^{2} - 3 x - 10$

$10 = x + x^{2} + 3 x + 10$

$y = - x^{2} - 3 x - 10$

Add $x$ and $3 x$ .

$10 = x^{2} + 4 x + 10$

$y = - x^{2} - 3 x - 10$

$10 = x^{2} + 4 x + 10$

$y = - x^{2} - 3 x - 10$

$10 = x^{2} + 4 x + 10$

$y = - x^{2} - 3 x - 10$

$10 = x^{2} + 4 x + 10$

$y = - x^{2} - 3 x - 10$

Solve for $x$ in $10 = x^{2} + 4 x + 10$ .

Rewrite the equation as $x^{2} + 4 x + 10 = 10$ .

$x^{2} + 4 x + 10 = 10$

$y = - x^{2} - 3 x - 10$

Subtract $10$ from both sides of the equation.

$x^{2} + 4 x + 10 - 10 = 0$

$y = - x^{2} - 3 x - 10$

Combine the opposite terms in $x^{2} + 4 x + 10 - 10$ .

Subtract $10$ from $10$ .

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

$y = - x^{2} - 3 x - 10$

Add $x^{2} + 4 x$ and $0$ .

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

$y = - x^{2} - 3 x - 10$

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

$y = - x^{2} - 3 x - 10$

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

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

$x \cdot x + 4 x = 0$

$y = - x^{2} - 3 x - 10$

Factor $x$ out of $4 x$ .

$x \cdot x + x \cdot 4 = 0$

$y = - x^{2} - 3 x - 10$

Factor $x$ out of $x \cdot x + x \cdot 4$ .

$x (x + 4) = 0$

$y = - x^{2} - 3 x - 10$

$x (x + 4) = 0$

$y = - x^{2} - 3 x - 10$

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

$x = 0$

$x + 4 = 0$

$y = - x^{2} - 3 x - 10$

Set $x$ equal to $0$ .

$x = 0$

$y = - x^{2} - 3 x - 10$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

$y = - x^{2} - 3 x - 10$

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = - x^{2} - 3 x - 10$

$x = - 4$

$y = - x^{2} - 3 x - 10$

The final solution is all the values that make $x (x + 4) = 0$ true.

$x = 0, - 4$

$y = - x^{2} - 3 x - 10$

$x = 0, - 4$

$y = - x^{2} - 3 x - 10$

Replace all occurrences of $x$ with $0$ in each equation.

Replace all occurrences of $x$ in $y = - x^{2} - 3 x - 10$ with $0$ .

$y = - {(0)}^{2} - 3 \cdot 0 - 10$

$x = 0$

Simplify the right side.

Simplify $- {(0)}^{2} - 3 \cdot 0 - 10$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = - 0 - 3 \cdot 0 - 10$

$x = 0$

Multiply $- 1$ by $0$ .

$y = 0 - 3 \cdot 0 - 10$

$x = 0$

Multiply $- 3$ by $0$ .

$y = 0 + 0 - 10$

$x = 0$

$y = 0 + 0 - 10$

$x = 0$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 10$

$x = 0$

Subtract $10$ from $0$ .

$y = - 10$

$x = 0$

$y = - 10$

$x = 0$

$y = - 10$

$x = 0$

$y = - 10$

$x = 0$

$y = - 10$

$x = 0$

Replace all occurrences of $x$ with $- 4$ in each equation.

Replace all occurrences of $x$ in $y = - x^{2} - 3 x - 10$ with $- 4$ .

$y = - {(- 4)}^{2} - 3 \cdot - 4 - 10$

$x = - 4$

Simplify the right side.

Simplify $- {(- 4)}^{2} - 3 \cdot - 4 - 10$ .

Simplify each term.

Raise $- 4$ to the power of $2$ .

$y = - 1 \cdot 16 - 3 \cdot - 4 - 10$

$x = - 4$

Multiply $- 1$ by $16$ .

$y = - 16 - 3 \cdot - 4 - 10$

$x = - 4$

Multiply $- 3$ by $- 4$ .

$y = - 16 + 12 - 10$

$x = - 4$

$y = - 16 + 12 - 10$

$x = - 4$

Simplify by adding and subtracting.

Add $- 16$ and $12$ .

$y = - 4 - 10$

$x = - 4$

Subtract $10$ from $- 4$ .

$y = - 14$

$x = - 4$

$y = - 14$

$x = - 4$

$y = - 14$

$x = - 4$

$y = - 14$

$x = - 4$

$y = - 14$

$x = - 4$

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

$(0, - 10)$

$(- 4, - 14)$

The result can be shown in multiple forms.

Point Form:

$(0, - 10), (- 4, - 14)$

Equation Form:

$x = 0, y = - 10$

$x = - 4, y = - 14$

Solve the System of Equations y=-x^2-3x-10 10=x-y

Related Questions

Support LCMGCF.com by disabling your adblocker.