Question

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

Answer 1

`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 -6x -3`

`6 = -x^2 -6x -3`

Multiply both sides by `-1`

`-6 = x^2 +6x +3`

Add 6 to both sides:

`0 = x^2 +6x +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 = (x + 3) (x + 3 )`

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

`0 = (x + 3)`

`-3 = x`

To check, substitute `x and y` into the equation:

`y = -x^2 -6x -3`

`6 = -(-3)^2 -6(-3) -3`

`6 = -9 +18 -3`

`6 = -12 +18 = 6`

Answer 2

`x=-3`

Explanation:

`y=-x^2-6x-3` given `y=6`

substitute `y=6`

`:.-x^2-6x-3=6`

multiply both sides by `-1`

`:.x^2+6x+3=-6`

`:.x^2+6x+3+6=0`

`:.x^2+6x+9=0`

`:.(x+3)(x+3)=0`

`:.x+3=0`

`:.x=-3`

Check:

substitute `y=6,x=-3`

`:.6=-(-3)^2-6(-3)-3`

`:.6=-9+18-3`

`:.6=-9+15`

`:.6=6`