Solve the System of Equations y=-x^2-3x+2 y=x+6

$y = - x^{2} - 3 x + 2$ $y = x + 6$

Eliminate the equal sides of each equation and combine.

$- x^{2} - 3 x + 2 = x + 6$

Solve $- x^{2} - 3 x + 2 = x + 6$ for $x$ .

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

Subtract $x$ from both sides of the equation.

$- x^{2} - 3 x + 2 - x = 6$

Subtract $x$ from $- 3 x$ .

$- x^{2} - 4 x + 2 = 6$

Subtract $6$ from both sides of the equation.

$- x^{2} - 4 x + 2 - 6 = 0$

Subtract $6$ from $2$ .

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

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} - 4 x - 4$ .

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

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

Factor $- 1$ out of $- 4 x$ .

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

Rewrite $- 4$ as $- 1 (4)$ .

$- (x^{2}) - (4 x) - 1 \cdot 4 = 0$

Factor $- 1$ out of $- (x^{2}) - (4 x)$ .

$- (x^{2} + 4 x) - 1 \cdot 4 = 0$

Factor $- 1$ out of $- (x^{2} + 4 x) - 1 (4)$ .

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

Factor using the perfect square rule.

Rewrite $4$ as $2^{2}$ .

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

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

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

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

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

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

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $0$ by $- 1$ .

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

Set the $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Evaluate $y$ when $x = - 2$ .

Substitute $- 2$ for $x$ .

$y = (- 2) + 6$

Substitute $- 2$ for $x$ in $y = (- 2) + 6$ and solve for $y$ .

Remove parentheses.

$y = - 2 + 6$

Remove parentheses.

$y = (- 2) + 6$

Add $- 2$ and $6$ .

$y = 4$

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

$(- 2, 4)$

The result can be shown in multiple forms.

Point Form:

$(- 2, 4)$

Equation Form:

$x = - 2, y = 4$

Solve the System of Equations y=-x^2-3x+2 y=x+6

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