Solve the system by elimination (show all work)

-2x +2y+3z=0,

-2x-y+z=-3,

2x+3y+3z=5

Step-by-step explanation:

{−2x + 2y + 3z = 0

{−2x - y + z = −3 ←

{2x + 3y + 3z = 5 ←

2y + 4z = 2 ↷

{−2x + 2y + 3z = 0 ←

{−2x - y + z = −3

{2x + 3y + 3z = 5 ←

5y + 6z = 5 ↷

{5y + 6z = 5

{2y + 4z = 2

−⅖[5y + 6z = 5]

{−2y - 2⅖z = −2 >> New Temporary Equation

{2y + 4z = 2

____________

1⅗z = 0

___ _

1⅗ 1⅗

[Plug this back into both equations above to get the y-term of 1, then plugging both z-term and y-term into all three equations at the very top, will give you the x-term of 1]; ;

I am joyous to assist you anytime.

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.

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step-by-step explanation:

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