+1836 Solve the system of equations
If there is a solution, write it as an ordered triple of integers or common fractions in simplest form. For example, if , , and were a solution, you would enter (1,3/2,-4).
If there is no solution, enter the word "none" as your answer.\(Solve the system of equations \begin{align*} x+y+2z & = 1, \\ -2x +4y + 2z & = 1, \\ y +z &= 1. \end{align*} If there is a solution, write it as an ordered triple of integers or common fractions in simplest form. For example, if $x=1$, $y=\frac64$, and $z=-4$ were a solution, you would enter (1,3/2,-4). If there is no solution, enter the word "none" as your answer. \)
+129852 x + y + 2z = 1 (1)
-2x + 4y + 2z = 1 (2)
y + z = 1 (3)
Multiply (3) by 2 and rearrange → 2z = 2 - 2y (4)
Sub (4) into (1) and simplify
x + y + [2 - 2 y] = 1
x - y = -1 (5)
Multiply (2) by -1 and add to (1)
3x - 3y = 0 divide both sides by 3
x - y = 0 (6)
But.....we have a contradiction between (5) and (6)......x - y cannot take on two values.....thus.....this system has no solution
