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

1 Answer
Aug 11, 2015

Answer:

#{(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) :}#