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3 years ago

−14 − 8x = −2(−3x + 7)

Mathematics

2 answers:

marissa [1.9K]3 years ago

6 0

Answer:

-8x=6x+14-14

-8x-6x=0

-14x=0

X=0/14

X=0

?????

Kitty [74]3 years ago

3 0

Answer:

X= 0 or Zero

Step-by-step explanation:

Steps To solve this problem... (8 steps)

1) Rearrange terms:

−14 − 8x = −2(−3x + 7)

−8x − 14 = −2(−3x + 7)

2) Distribute:

−8x − 14 = −2−(3x + 7)

−8x − 14 = 6x - 14

3) Add 14 to both sides of the equation:

−8x − 14 = 6x - 14

−8x − 14 + 14 = 6x - 14 + 14

4) Simplify (Add the numbers):

-8x = 6x

5) Subtract 6x from both sides of the equation:

-8x = 6x

-8x - 6x = 6x - 6x

6) Simplify (Combine like terms):

-14x = 0

7) Divide both sides of the equation by the same term:

-14x = 0

-14x/-14 = 0/-14

8) Simplify (Cancel terms that are in both the numerator and denominator):

X= 0/-14

Hope this Helped! :)

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In matrix form, the system is given by

$\begin{bmatrix} -1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -20 \\ 29 \\ 29 \end{bmatrix}$

I'll use G-J elimination. Consider the augmented matrix

$\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]$

• Multiply through row 1 by -1.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]$

• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right]$

• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right]$

• Multiply row 3 by 1/3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]$

• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]$

• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]$

Then the solution to the system is

$\boxed{x=9, y=-3, z=8}$

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]$

The third row tells us that $z=8$ . Then in the second row,

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and in the first row,

$x-y+z=20 \implies x=20 + (-3) - 8 = 9$

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