Question

what is the solution to this system of equations?\\(\

$$\begin{cases}\\ \\ \\ \\ x + 3y - z = 6\\\\4x - 2y + 2z = -10\\\\\\ \\ \\ \\ 6x + z = -12\\end{cases}$$

\\)\\(\circ\\ (-3, 5, 6)\\)\\(\circ\\ (0, -2, -12)\\)\\(\circ\\ (2, 1, -3)\\)\\(\circ\\ (-4, 0, 12)\\)

Explanation:

Step1: Isolate z from third equation

$z = -12 - 6x$

Step2: Substitute z into first equation

$x + 3y - (-12 - 6x) = 6$
Simplify: $x + 3y + 12 + 6x = 6 \implies 7x + 3y = -6$

Step3: Substitute z into second equation

$4x - 2y + 2(-12 - 6x) = -10$
Simplify: $4x - 2y -24 -12x = -10 \implies -8x -2y = 14 \implies 4x + y = -7$

Step4: Solve for y from Step3

$y = -7 - 4x$

Step5: Substitute y into Step2 equation

$7x + 3(-7 - 4x) = -6$
Simplify: $7x -21 -12x = -6 \implies -5x = 15 \implies x = -3$

Step6: Find y using x=-3

$y = -7 - 4(-3) = -7 +12 = 5$

Step7: Find z using x=-3

$z = -12 -6(-3) = -12 +18 = 6$

Answer:

$(-3, 5, 6)$