Solve the system of equations by elimination. (2 marks)

y=2x^2-2x-3
y=-x^2-2x-3

To solve the system of equations by elimination, we need to eliminate one of the variables by adding or subtracting the equations. Let's start by multiplying the second equation by 2 to make the coefficients of x^2 the same:

2y = -2x^2 - 4x - 6

Now we have the following system of equations:

y = 2x^2 - 2x - 3
2y = -2x^2 - 4x - 6

Next, we can combine the two equations by adding them together:

y + 2y = (2x^2 - 2x - 3) + (-2x^2 - 4x - 6)

This simplifies to:

3y = 0

Divide both sides of the equation by 3 to solve for y:

y = 0

Now that we have the value of y, we can substitute it back into one of the original equations to solve for x. Let's substitute y = 0 into the first equation:

0 = 2x^2 - 2x - 3

To solve this quadratic equation, we can factor or use the quadratic formula. In this case, factoring may be a bit challenging, so let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

The coefficients for our quadratic equation are:
a = 2, b = -2, c = -3

Substituting these values into the quadratic formula, we get:

x = (-(-2) ± √((-2)^2 - 4 * 2 * -3)) / (2 * 2)

Simplifying further:

x = (2 ± √(4 + 24)) / 4

x = (2 ± √28) / 4

x = (2 ± 2√7) / 4

Now, we can simplify the expression:

x = (1 ± √7) / 2

So the solutions for the system of equations are:

x = (1 + √7) / 2
y = 0

x = (1 - √7) / 2
y = 0