What is the solution to this system of equations? x+3y−z=6 4x−2y+2z=−10 6x+z=−12 (−4, 0, 12) (0, −2, −12) (2, 1, −3) (−3, 5, 6)

Solution :  (− 3, 5, 6)

Step-by-step explanation:

We have the following system of equations that we have to solve for,

To solve this problem we can start by writing the matrix with their respective coefficients --- (1)

Now we can reduce this to row echelon form, receiving our solution --- (2)

Swap row 1 and 3,

Cancel leading coefficient in row 3,

Swap row 2 and 3

Cancel leading coefficient in row 3

At this point you can see that we have to cancel the leading coefficient in each row, to row echelon form. Continuing this pattern we have the following matrix,

As you can see, x = - 3, y = 5, and z = 6, giving us a solution of (− 3, 5, 6). This is the fourth option.

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

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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