A system of equations is shown below.
5 +2y = -15
2x-2y=-6
What is the solution to the system of equations?
A
B
C
D
(-3,0)
(0,-3)
(-3,6)
(6,-3)

You can use the process elimination to solve. In other words, subtract or add the equations like normal.

5x + 2y = -15

+ 2x - 2y = -6

Because there is a 2y and -2y, we can add both equations to eliminate the y so that we can figure out what x is:

5x + 2x = 7x

2y + (-2y) = 0

-15 + (-6) = -21

We now have 7x = -21

Divide all by 7 to get x = -3

Plug in -3 to any of the two equations. I will use the second equation:

2(-3) - 2y = -6

-6 - 2y = -6

-2y = 0

y = 0

This means that the solution to the system of equations is (-3, 0) in the form of (x, y).

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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