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xxMikexx [17]
2 years ago
5

Solve the following system of equations for all three variables.

Mathematics
1 answer:
GrogVix [38]2 years ago
4 0

Answer:

(x,y,z) -> (2,4,1)

Step-by-step explanation:

-7x + y + z = -9

-7x + 5y - 9z = -3

7x - 6y + 4z = -6

Pick two pairs:

-7x + 5y - 9z = -3

7x - 6y + 4z = -6

and

-7x + y + z = -9

-7x + 5y - 9z = -3

Eliminate the same variable from each system:

-7x + 5y - 9z = -3

7x - 6y + 4z = -6

+ 5y - 9z = -3

- 6y + 4z = -6

<u><em>-1y - 5z = - 9</em></u>

-7x + y + z = -9

-7x + 5y - 9z = -3

-7x + y + z = -9

7x - 5y + 9z = 3

<u><em>-4y - 10z = -6</em></u>

Solve the system of the two new equations:

-1y - 5z = - 9           ->      -4 ( -1y - 5z = - 9)     ->   4y + 20z = 36  

-4y - 10z = -6          ->           -4y - 10z = -6      ->   -4y - 10z = -6

10z = 30

Thus, z = 3

-4y - 10z = -6

-4y - 10(3) = -6

-4y - 30 = -6

-4y = 24

Thus, y = -6

Substitute into one of the original equations:

-7x + y + z = -9

-7x + (-6) + (3) = -9

7x + -3  = -9

7x  = -6

x =

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Step-by-step explanation:

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2 years ago
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Lunna [17]

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Step-by-step explanation:

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6 0
3 years ago
Amy is solving the equation using the steps shown.
Delicious77 [7]

The correct next step Amy could take is to factor (x^2 - 16) on the left side of the equation to yield (x + 4)(x – 4)(x + 2) = 0.

The equation is given as:

\mathbf{(x^2 - 16)(x + 2) = 0}

The possible next steps are:

1. Expand the equation as follows

\mathbf{x^3+2x^2-16x-32 = 0}

2. Express x^2 -16 as a difference of two squares as follows

\mathbf{(x + 4)(x- 4)(x + 2) = 0}

However, the correct step in this case is to factor x^2 - 16 as a difference of two squares to give \mathbf{(x + 4)(x- 4)(x + 2) = 0}

This is so, because the question says Amy is trying to solve for variable x.

This step will ensure that she gets the possible values of x, unlike expanding the whole equation .

Hence, the correct next step is \mathbf{(x + 4)(x- 4)(x + 2) = 0}

Read more about equations at:

brainly.com/question/7674413

7 0
2 years ago
If f(x) = 2x^3 - 14x^2 + 38x -26 and x-1 is a factor of f(x) find all of the zeros of f(x) algebraically.
Anestetic [448]

Answer:

Step-by-step explanation:

First confirm that x = 1 is one of the zeros.

f(1) = 2(1)^3 - 14(1)^2 + 38(1) - 26

f(1) = 2 - 14 + 38 - 26

f(1) = -12 + 38 = + 26

f(1) = 26 - 26

f(1) = 0

=========================

next perform a long division

x -1  || 2x^3 - 14x^2 + 38x - 26 || 2x^2 - 12x + 26

          2x^3 - 2x^2

          ===========

                    -12x^2 + 28x

                     -12x^2 +12x

                     ==========

                                  26x -26

                                  26x - 26

                                 ========

                                      0

Now you can factor 2x^2 - 12x + 26

                                 2(x^2 - 6x + 13)

The discriminate of the quadratic is negative. (36 - 4*1*13) = - 16

So you are going to get a complex result.

x = -(-6) +/- sqrt(-16)

     =============

                 2

x  = 3 +/- 2i

f(x) = 2*(x - 1)*(x - 3 + 2i)*(x - 3 - 2i)

The zeros are

1

3 +/- 2i

8 0
3 years ago
Evaluate 17 + x<br> when x = 16
Vesnalui [34]
Answer: 17+ 16 = 33

Good luck !
3 0
3 years ago
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