To find the solutions of the system algebraically, we need to set the two equations equal to each other:
x^2 - 2x - 6 = 4x + 10
Now, we can solve for x by rearranging the equation and simplifying:
x^2 - 2x - 6 = 4x + 10
x^2 - 6x - 16 = 0
Now, we have a quadratic equation that we can solve using the quadratic formula:
x = (-(-6) ± √((-6)^2 - 4(1)(-16))) / 2(1)
x = (6 ± √(36 + 64)) / 2
x = (6 ± √100) / 2
x = (6 ± 10) / 2
So, the solutions for x are:
x1 = (6 + 10) / 2 = 16 / 2 = 8
x2 = (6 - 10) / 2 = -4 / 2 = -2
Now that we have the values for x, we can find the corresponding y values by plugging them back into the second equation:
For x = 8:
y = 4(8) + 10 = 32 + 10 = 42
For x = -2:
y = 4(-2) + 10 = -8 + 10 = 2
Therefore, the solutions to the system algebraically are:
(8, 42) and (-2, 2)
Find the solutions of the system algebraically.
y= x^2-2x-6
y=4x＋10
1 answer