Question

What is the solution to the system of equations?

Answer 1

Answer:

(-10, 2, 6)

Step-by-step explanation:

$\left\{\begin{array}{ccc}x+3y+2z=8&(1)\\3x+y+3z=-10&(2)\\-2x-2y-z=10&(3)\end{array}\right\qquad\text{subtract both sides of the equations (1) from (2)}\\\\\underline{-\left\{\begin{array}{ccc}3x+y+3z=-10\\x+3y+2z=8\end{array}\right }\\.\qquad2x-2y+z=-18\qquad(4)\qquad\text{add both sides of the equations (3) and (4)}\\\\\underline{+\left\{\begin{array}{ccc}-2x-2y-z=10\\2x-2y+z=-18\end{array}\right}\\.\qquad-4y=-8\qquad\text{divide both sides by (-4)}\\.\qquad\qquad y=2\qquad\text{put the value of y to (1) and (3)}$

$\left\{\begin{array}{ccc}x+3(2)+2z=8\\-2x-2(2)-z=10\end{array}\right\\\left\{\begin{array}{ccc}x+6+2z=8&\text{subtract 6 from both sides}\\-2x-4-z=10&\text{add 4 to both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+2z=2&\text{multiply both sides by 2}\\-2x-z=14\end{array}\right\\\underline{+\left\{\begin{array}{ccc}2x+4z=4\\-2x-z=14\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad3z=18\qquad\text{divide both sides by 3}\\.\qquad\qquad z=6\qquad\text{put the value of z to the first equation}$

$x+2(6)=2\\x+12=2\qquad\text{subtract 10 from both sides}\\x=-10$