It looks like you've provided two equations of lines:
1. y = x^2 + 3x + 3
2. y = -x - 1
To solve this system of equations, we need to find the value(s) of x where the two equations intersect, i.e., where the output y from the first equation equals the output y from the second equation.
We do this by setting the two equations equal to each other and solving for x:
x^2 + 3x + 3 = -x - 1
Combining terms and moving everything to one side of the equation gives us a quadratic equation:
x^2 + 3x + x + 3 + 1 = 0
x^2 + 4x + 4 = 0
This is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = 4, and c = 4.
The equation x^2 + 4x + 4 = 0 can be factored as:
(x + 2)(x + 2) = 0
Setting each factor equal to 0 gives the same solution for x:
x + 2 = 0
x = -2
So the two lines intersect at x = -2. To find the corresponding y-coordinate of the intersection point, we plug x = -2 into either of the original equations. It's easier to use the second equation, as it is already solved for y:
y = -(-2) - 1
y = 2 - 1
y = 1
Therefore, the solution to the system of equations -- the point where the two equations intersect -- is (x, y) = (-2, 1).
y=x^2+3x+3
y=-x-1
1 answer