QZ wrote:

If \(2 x^2 - 2x - 12 = 0\) and \(y^2 - 5y + 6 = 0\) when \(x = -y\), then what is the value of \(x\) ?

a. -3

b. -2

c. 0

d. 2

e. 3

Dividing the first equation by 2 and solving for x, we have:

x^2 - x - 6 = 0

(x - 3)(x + 2) = 0

x = 3 or x = -2

Solving for y in the second equation, we have:

(y - 3)(y - 2) = 0

y = 3 or y = 2

Since x = -y, we see that x must be -2 (and y must be 2).

Alternate Solution:

Note that 2x^2 can also be expressed as 2x^2 = 2x + 12 using the first equation.

Since x = -y, let’s substitute y = -x in the second equation:

(-x)^2 -5(-x) + 6 = 0

x^2 + 5x + 6 = 0

Let’s multiply each side of this equality by 2:

2x^2 + 10x + 12 = 0

2x^2 = -10x - 12

Since we have two expressions for 2x^2, let’s set them equal to each other:

2x + 12 = -10x -12

12x = -24

x = -2

Answer: B

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