What is the solution set for y = x^2 - 6y=x2−6 and y = -2x - 3y=−2x−3?

1 Answer
Stefan V.
Aug 11, 2015

{(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) :}

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