Question

solve the system of equations y=-2x y=x2-8

Answer 1

Answer:

$x=-4\\y=8\\\\x=2\\y=-4$

Step-by-step explanation:

Make both equations equal to each other and solve for x:

$-2x=x^2-8\\0=x^2+2x-8$

Factor the quadratic equation. You need to find two number that when you add them you get 2 and when you multiply them you get -8.

These numbers are 4 and -2. Then:

$0=(x+4)(x-2)\\\\x+4=0\\x=-4\\\\x-2=0\\x=2$

Substitute the values of x calculated, into any of the original equations. Then:

 $y=-2(-4)\\y=8\\\\y=-2(2)\\y=-4$

Answer 2

For this case we must solve the following system of equations:

$y = -2x\\y = x ^ 2-8$

Matching we have:

$-2x = x ^ 2-8\\x ^ 2+ 2x-8 = 0$

The solution will be given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\a = 1\\b = 2\\c = -8$

Substituting:

$x = \frac {-2 \pm \sqrt {2 ^ 2-4 (1) (- 8)}} {2 (1)}\\x = \frac {-2 \pm \sqrt {4+32}} {2}\\x = \frac {-2 \pm \sqrt {36}} {2}\\x = \frac {-2 \pm6} {2}$

$x_ {1} = \frac {4} {2} = 2 \ then \ y_ {1} = - 2 (2) = - 4\\x_ {2} = \frac {-2-6} {2} = - 4 \ then \ y_ {2} = - 2 (-4) = 8$

Answer:

$x_ {1} = \frac {4} {2} = 2 \ then \ y_ {1} = - 2 (2) = - 4\\x_ {2} = \frac {-2-6} {2} = - 4 \ then \ y_ {2} = - 2 (-4) = 8$