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

##### 1 Answer

Aug 11, 2015

#### Answer:

*or*

#### Explanation:

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

#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

*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

*or*