Question

What is the solution to the system? X+y+z=2 2x+y-z=-1 X=5-2z

Answer 1

Answer:

x = 1, y = -1, z = 2 → (1, -1, 2)

Step-by-step explanation:

$\left\{\begin{array}{ccc}x+y+z=2&(1)\\2x+y-z=-1&(2)\\x=5-2z&(3)\end{array}\right\\\\\text{Substitute (3) to (1) and (2):}\\\\\left\{\begin{array}{ccc}(5-2z)+y+z=2\\2(5-2z)+y-z=-1&\text{use the distributive property}\end{array}\right\\\left\{\begin{array}{ccc}5-2z+y+z=2\\10-4z+y-z=-1\end{array}\right\qquad\text{combine like terms}\\\left\{\begin{array}{ccc}5+y-z=2&\text{subtract 5 from both sides}\\10+y-5z=-1&\text{subtract 10 from both sides}\end{array}\right$

$\left\{\begin{array}{ccc}y-z=-3\\y-5z=-11&\text{change the signs}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}y-z=-3\\-y+5z=11\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad4z=8\qquad\text{divide both sides by 4}\\.\qquad\qquad \boxed{z=2}\\\\\text{Put it to the first equation:}\\\\y-2=-3\qquad\text{add 2 to both sides}\\\boxed{y=-1}\\\\\text{Put the values of}\ z\\text{to (3):}\\\\x=5-2(2)\\x=5-4\\\boxed{x=1}$