Mathematics: (1 point) linear system - three variables solve the following system of equations using gaussian elimination method. if there are no solutions, type n for both x, y and z. if there are infinitely many solutions, type z for z, and expressions in terms of z for x and y. 6x + 8y + 72 = -3 - 3x + 6y + 6z = 5 2x – 9y – 32 = 7

From what I remember it looks correct....

(1 point) linear system - three variables solve

wolfsaway

05.01.2022

$x=\frac{-363}{70}$ ≈ -5.1857

$y=\frac{-192}{35}$ ≈ -5.4857

$z=\frac{313}{84}$ ≈ 3.7262

Step-by-step explanation:

Rewrite the equation system as:

6x+8y=-75

-3x+6y+6z=5

2x-9y=39

Now, write the system in its augmented matrix form:

$\left[\begin{array}{cccc}6&8&0&-75\\-3&6&6&5\\2&-9&0&39\end{array}\right]$

applying row reduction process to its associated augmented matrix:

Swap R1 and R3, and then Swap R1 and R2:

$\left[\begin{array}{cccc}-3&6&6&5\\2&-9&0&39\\6&8&0&-75\end{array}\right]$

R3+2R1

$\left[\begin{array}{cccc}-3&6&6&5\\2&-9&0&39\\0&20&12&-65\end{array}\right]$

3R2+2R1

$\left[\begin{array}{cccc}-3&6&6&5\\0&-15&12&127\\0&20&12&-65\end{array}\right]$

15R3+20R2

$\left[\begin{array}{cccc}-3&6&6&5\\0&-15&12&127\\0&0&420&1565\end{array}\right]$

Now we have a simplified system:

$-3x+6y+6z=5\\0-15y+12z=127\\0+0+420z=1565$

$-3x+6y+6z=5\hspace{5 mm}(1)\\0-15y+12z=127\hspace{3 mm}(2)\\0+0+420z=1565\hspace{3 mm}(3)$

From (3):

$z=\frac{313}{84}$ (4)

Replacing (4) in (2)

$y=\frac{-192}{35}$ (5)

Finally replacing (5) and (4) in (1)

$x=\frac{-363}{70}$

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