Find the solutions to the system.
Y=x^2-2x-2
Y=4x+5
Setting the two expressions for y equal to each other, we get:
x^2 - 2x - 2 = 4x + 5
Moving all the terms to one side, we have:
x^2 - 6x - 7 = 0
Factoring, we get:
(x - 7)(x + 1) = 0
So x = 7 or x = -1. Substituting these values back into either expression for y, we get:
When x = 7, y = 4(7) + 5 = 33.
When x = -1, y = (-1)^2 - 2(-1) - 2 = 1 + 2 - 2 = 1.
Therefore, the solutions to the system are (7, 33) and (-1, 1).
To find the solutions to the system of equations, we can set the two equations equal to each other and solve for x:
x^2 - 2x - 2 = 4x + 5
Rearranging the equation to have it equal to zero:
x^2 - 6x - 7 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = -6, and c = -7. Substituting these values into the quadratic formula:
x = (-(-6) ± √((-6)^2 - 4(1)(-7))) / (2(1))
Simplifying:
x = (6 ± √(36 + 28)) / 2
x = (6 ± √64) / 2
x = (6 ± 8) / 2
x = (6 + 8) / 2 or x = (6 - 8) / 2
x = 14 / 2 or x = -2 / 2
x = 7 or x = -1
Therefore, the solutions to the system of equations are x = 7 and x = -1.
To find the corresponding values of y, we can substitute these x-values back into either of the original equations:
For x = 7:
y = (7)^2 - 2(7) - 2
y = 49 - 14 - 2
y = 33
So, one solution to the system of equations is x = 7 and y = 33.
For x = -1:
y = (-1)^2 - 2(-1) - 2
y = 1 + 2 - 2
y = 1
So, another solution to the system of equations is x = -1 and y = 1.