Question

What is the sum of the three solutions? (find the values for x, y, and z, then add the answers)

Answer 1

Answer: x + y + z = 4
-1 + 2 + 3 = 4

Explanation:

2x + 3y - z = 5 (1)
x - 3y + 2z= -6 (2)
3x + y - 4z = -8 (3)
————————-
2x + 3y - z = 5 (Add (1) and (2)
X - 3y + 2z = -6
————————
3x + z = -1 (4)

3x + y - 4z = -8 Add (3) and (2)
x - 3y + 2z = -6
————————
3(3x + y - 4z = -8)
x - 3y + 2z = -6
—————————
9x + 3y - 12z = -24
x -3y + 2z = -6
—————————-
10x - 10z = -30
x - z = -3 (5)

Add (4) and (5)

3x - z = -1
x - z = -3
—————
4x = -4
x = -1

Plug x = -1 in (5)
x - z = -3
-1 - z = -3
-z = -2
z = 2

Plug x and z in (2):
x - 3y + 2z = -6
-1 - 3y + 2(2) = -6
-1 - 3y + 4 = -6
-3y + 3 = -6
-3y = -9
y = -9/-3
y = 3

Therefore, y + z + x = 3 + 2 -1 = 4

Answer 2

Answer:

Once we got

$x=-1$

$y=3$

$z=2$

$\boxed{\text{The sum is 4}}$

Step-by-step explanation:

Given the linear system:

$\begin{cases} 2x + 3y-z = 5 \\ x- 3y + 2z = -6 \\ 3x + y - 4z = -8 \end{cases}$

Let's solve it using matrices. I will use Cramer's rule

$M=\left[\begin{array}{ccc}2&3&-1\\1&-3&2\\3&1&-4\end{array}\right]$

Considering determinant as D.

$D=\begin{vmatrix}2&3&-1\\1&-3&2\\3&1&-4\\\end{vmatrix}=40$

$M_x = \left[\begin{array}{ccc}5&3&-1\\-6&-3&2\\-8&1&-4\end{array}\right] \implies D_x = \begin{vmatrix}5&3&-1\\-6&-3&2\\-8&1&-4\\\end{vmatrix}=-40$

$M_y = \left[\begin{array}{ccc}2&5&-1\\1&-6&2\\3&-8&-4\end{array}\right] \implies D_y = \begin{vmatrix}2&5&-1\\1&-6&2\\3&-8&-4\\\end{vmatrix}=120$

$M_z = \left[\begin{array}{ccc}2&3&5\\1&-3&-6\\3&1&-8\end{array}\right] \implies D_z= \begin{vmatrix}2&3&5\\1&-3&-6\\3&1&-8\\\end{vmatrix}=80$

So, we have

$x=\frac{D_x}{D} =\frac{-40}{40}=-1$

$y=\frac{D_y}{D} =\frac{120}{40}=3$

$z=\frac{D_z}{D} =\frac{80}{40}=2$