# How do you solve y=-x^2-6x-3, y=6 using substitution?

##### 2 Answers
Feb 17, 2017

$x = - 3$; $y = 6$

#### Explanation:

$y = 6$ is given in the question, but it still is part of the answer.

To solve for $x$, substitute $y = 6$ into the equation:

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

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

Multiply both sides by $- 1$

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

Add 6 to both sides:

$0 = {x}^{2} + 6 x + 9$

Factor the equation by calculating factors of the first and last numbers that will add or subtract to arrive at the inner number.

$0 = \left(x + 3\right) \left(x + 3\right)$

In this case both factors will provide the same answer for $x$:

$0 = \left(x + 3\right)$

$- 3 = x$

To check, substitute $x \mathmr{and} y$ into the equation:

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

$6 = - {\left(- 3\right)}^{2} - 6 \left(- 3\right) - 3$

$6 = - 9 + 18 - 3$

$6 = - 12 + 18 = 6$

Feb 17, 2017

$x = - 3$

#### Explanation:

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

substitute $y = 6$

$\therefore - {x}^{2} - 6 x - 3 = 6$

multiply both sides by $- 1$

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

$\therefore {x}^{2} + 6 x + 3 + 6 = 0$

$\therefore {x}^{2} + 6 x + 9 = 0$

$\therefore \left(x + 3\right) \left(x + 3\right) = 0$

$\therefore x + 3 = 0$

$\therefore x = - 3$

Check:

substitute $y = 6 , x = - 3$

$\therefore 6 = - {\left(- 3\right)}^{2} - 6 \left(- 3\right) - 3$

$\therefore 6 = - 9 + 18 - 3$

$\therefore 6 = - 9 + 15$

$\therefore 6 = 6$