Question

x - 3y + 6z = 21

3x + 2y - 5z = - 30

2x - 5y + 2z = -6

The sum of values of x

x, y

y, and z

z of the solution is

Answer 1

Let's solve the system of linear equations using the Cramer's rule.

$\Delta=\begin{vmatrix} 1& - 3&6\\ 3 & 2&-5\\ 2&-5&2 \end{vmatrix}=\\ =1\cdot2\cdot2+(-3)\cdot(-5)\cdot2+6\cdot3\cdot(-5)-\\-6\cdot2\cdot2-1\cdot(-5)\cdot(-5)-(-3)\cdot3\cdot2=\\ =4+30-90-24-25+18=-87;$

$\Delta_1=\begin{vmatrix} 21& - 3&6\\ -30 & 2&-5\\ -6&-5&2 \end{vmatrix}=\\ =21\cdot2\cdot2+(-3)\cdot(-5)\cdot(-6)+6\cdot(-30)\cdot(-5)-\\-6\cdot2\cdot(-6)-21\cdot(-5)\cdot(-5)-(-3)\cdot(-30)\cdot2=\\ =84-90+900+72-525-180=261;$

$\Delta_2=\begin{vmatrix} 1& 21&6\\ 3 & - 30&-5\\ 2&-6&2 \end{vmatrix}=\\ =1\cdot(-30)\cdot2+21\cdot(-5)\cdot2+6\cdot3\cdot(-6)-\\-6\cdot(-30)\cdot2-1\cdot(-5)\cdot(-6)-21\cdot3\cdot2=\\ =-60-210-108+360-30-126=-174;$

$\Delta_3=\begin{vmatrix} 1& - 3&21\\ 3 & 2&-30\\ 2&-5&-6 \end{vmatrix}=\\ =1\cdot2\cdot(-6)+(-3)\cdot(-30)\cdot2+21\cdot3\cdot(-5)-\\-21\cdot2\cdot2-1\cdot(-30)\cdot(-5)-(-3)\cdot3\cdot(-6)=\\ =-12+180-315-84-150-54=-435;$

$x=\cfrac{\Delta_1} {\Delta} =\cfrac{261}{-87}-3;\\ y=\cfrac{\Delta_2} {\Delta} =\cfrac{-174}{-87}=2;\\ z=\cfrac{\Delta_3} {\Delta} =\cfrac{-435}{-87}=5;\\ x+y+z=-3+2+5=4.$