Question

What are the solutions to the following system of equations? x + y = 3 y = x2 − 9 (3, 0) and (1, 2) (−3, 0) and (1, 2) (3, 0) and (−4, 7) (−3, 0) and (−4, 7)

Answer 1

The solutions for the given system of equations are:

(3, 0), (-4, 7).

How to solve the given system of equations:

Here we have the system:

x + y = 3

y = x² - 9

To solve this, we can replace the second equation into the first one, so we get:

x + (x² - 9) = 3

Now we can solve this quadratic equation for x, we need to solve:

x² + x - 12 = 0.

The solutions are given by Bhaskara's formula:

$x = \frac{-1 \pm \sqrt{1^2 - 4*1*(-12)} }{2*1} \\\\x = \frac{-1 \pm 7 }{2}$

Then the two solutions are:

x = (-1 + 7)/2 = 3

x = (-1 - 7)/2 = -4

To get the y-values correspondent, we can evaluate the linear equation in these two values:

y = 3 - x.

For x = 3:

y = 3 - 3 = 0

For x = -4:

y = 3 + 4 = 7

Then the two solutions are: (3, 0), (-4, 7).

If you want to learn more about systems of equations:

brainly.com/question/13729904

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