Solve the following system of equations algebraically:y, equals, x, squared, plus, 6, x, minus, 6y=x 2 +6x−6y, equals, 3, x, minus, 2y=3x−2

The system of equations is:

1. y = x^2 + 6x - 6
2. y = 3x - 2

To solve this system algebraically, we can set the two equations equal to each other because they both equal y:

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

Next, we can simplify this equation by subtracting 3x and adding 2 to both sides:

x^2 + 3x - 4 = 0

This is a quadratic equation, which we can solve by factoring:

(x + 4)(x - 1) = 0

Setting each factor equal to zero gives the solutions x = -4 and x = 1.

We can substitute these values back into the original equations to find the corresponding y values.

For x = -4:

y = (-4)^2 + 6*(-4) - 6 = 16 - 24 - 6 = -14

For x = 1:

y = (1)^2 + 6*1 - 6 = 1 + 6 - 6 = 1

So the solutions to the system of equations are (-4, -14) and (1, 1).