Solve the System of Equations y=x^2+6x+10 y=-2x-6

$y = x^{2} + 6 x + 10$ $y = - 2 x - 6$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} + 6 x + 10 = - 2 x - 6$

Step 2

Solve $x^{2} + 6 x + 10 = - 2 x - 6$ for $x$ .

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

Add $2 x$ to both sides of the equation.

$x^{2} + 6 x + 10 + 2 x = - 6$

Add $6 x$ and $2 x$ .

$x^{2} + 8 x + 10 = - 6$

Add $6$ to both sides of the equation.

$x^{2} + 8 x + 10 + 6 = 0$

Add $10$ and $6$ .

$x^{2} + 8 x + 16 = 0$

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$x^{2} + 8 x + 4^{2} = 0$

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

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 4$ .

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 3

Evaluate $y$ when $x = - 4$ .

Substitute $- 4$ for $x$ .

$y = - 2 \cdot - 4 - 6$

Substitute $- 4$ for $x$ in $y = - 2 \cdot - 4 - 6$ and solve for $y$ .

Remove parentheses.

$y = - 2 \cdot - 4 - 6$

Simplify $- 2 \cdot - 4 - 6$ .

Multiply $- 2$ by $- 4$ .

$y = 8 - 6$

Subtract $6$ from $8$ .

$y = 2$

Step 4

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

$(- 4, 2)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(- 4, 2)$

Equation Form:

$x = - 4, y = 2$

Step 6

Solve the System of Equations y=x^2+6x+10 y=-2x-6

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