Question

What is the solution set for `y = x^2 - 6` and `y = -2x - 3`?

Answer 1

`{(x = -3), (y = 3) :} " "` or `" "{(x=1), (y=-5) :}`

Explanation:

Notice that you were given two equations that deal with the value of `y`

`y = x^2 - 6" "` and `" "y = -2x-3`

In order for these equations to be true, you need to have

`x^2 - 6 = -2x-3`

Rearrange this equation into classic quadratic form

`x^2 + 2x -3 = 0`

You can use the quadratic formula to determine the two solutions

`x_(1,2) = (-2 +- sqrt(2^2 - 4 * 1 * (-3)))/(2 * 1)`

`x_(1,2) = (-2 +- sqrt(16))/2 = (-2 +- 4)/2 = {(x_1 = (-2-4)/2 = -3), (x_2 = (-2 + 4)/2 = 1) :}`

Now take these values of `x` to one of the orignal equations and find the corresponding values of `y`.

• when `x=-3`, you have

`y = (-3)^2 - 6 = 3`

• when `x=1`, you have

`y = 1^2 - 6 = -5`

So, the two possible solution sets are

`{(x = -3), (y = 3) :} " "` or `" "{(x=1), (y=-5) :}`