What are the solutions to the system of equations? y=−x^2−5x−6 x+y=−3

Given:

y=−x^2−5x−6 

x+y=−3

Determine the solutions of the given equation.

Solution:

In order to determine the solutions of the given equation, we simply substitute the first equation to the second equation, then solve for y.

x + (-x^2 - 5x - 6) = -3

-x^2 - 4x - 6 + 3 = 0

-x^2 -4x - 3 = 0

a = -1

b = -4

c = -3

In order to solve for the roots of an equation, we simply must look at the 2nd and 3rd value. First we must think of two numbers that when added, the answer is -4, and when multiplied, the answer is -3. Hence, if we are unable to find the number, we will use the quadratic formula.

The Quadratic formula:

x = −b ± √(b^2 − 4ac)/2a

is used to solve quadratic equations where a ≠ 0, in the form
ax^2+bx+c=0

When b^2−4ac=0 there is one real root.

When b^2−4ac>0 there are two real roots.

When b^2−4ac<0 there are no real roots, only a complex number.

Substitute the given values of a, b and c to the quadratic formula. 
x = −b ± √(b^2 − 4ac)/2a
x = −(-4) ± √((-4)^2 − 4(-1)(-3))/2(-1)
x = 4 ± √(16 − 12))/-2
x = 4 ± √(4)/-2
x = 4 ± 2/-2

Solve for + -  separately.

x = 4 + 2 / -2
x = 6/-2
x = -3

x = 4 - 2 / - 2
x = 2/-2
x = -1

Final answer:

x = -3

x = -1