Answer:
x=1, y=1, z=0
Step-by-step explanation:
System of Equations
We are required to solve the system of equations by elimination:
[tex]\left\{\begin{matrix}-2x+2y+3z=0\qquad [1]\\-2x-y+z=-3 \qquad [2]\\2x+3y+3z=5 \qquad [3]\end{matrix}\right[/tex]
Adding [1] and [3], and subtracting [1] and [2]:
[tex]\left\{\begin{matrix}5y+6z=5\qquad [4]\\3y+2z=3\qquad [5] \end{matrix}\right.[/tex]
Multiplying [5] by -3 and adding to [4]:
[tex]-4y=-4[/tex]
Solving:
[tex]y = -4 / (-4) = 1[/tex]
[tex]y=1[/tex]
Substituting into [5]:
[tex]5(1)+6z=5[/tex]
Simplifying:
[tex]6z = 0[/tex]
z=0
Substituting into [3]
[tex]2x+3(1)+3(0)=5[/tex]
Operating:
[tex]2x+3=5[/tex]
[tex]2x = 2[/tex]
[tex]x = 2/2=1[/tex]
x=1
The solution is
x=1, y=1, z=0
