Find the solutions to the system of equations. select all that apply.

y=x^2-4
y=-2x-5

a. (-2,0)
b. (-5,0)
c. (2,0)
d. (-1,-3)

Step-by-step explanation:

Hello there!

The system of equation given is:

y = x^2-4

y = -2x-5

And we want to solve this system:

the first step we can do is replace y in the first equation by the given in the second equation:

y = x^2-4 = -2x - 5

now we can solve it for x:

x^2-4 +2x + 5 = 0

x^2 +2x + 1 = 0

and now we can use the cuadratic equation forumla:

this is, if we have a function of the form:

ax^2 + bx + c = 0

the solutions are:

in our equation we have a=1, b = 2 and c= 1

we replace it in the formula given, and we obtain:

Now we got the value of x, and we could replace it in one of the equations and get the value of y:

y = -2x-5

y = -2*(-1) - 5 = 2 -5 = -3

then the pair is (-1, -3)

and the correct option is D.

step-by-step explanation:

i hate black

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3