Answer:
[tex]x=1, y=1, z=0[/tex]
Step-by-step explanation:
we have
[tex]-2x+2y+3z=0[/tex] ----> equation A
[tex]-2x-y+z=-3[/tex] -----> equation B
[tex]2x+3y+3z=5[/tex] ----> equation C
adds equation A and equation C
[tex]-2x+2y+3z=0\\2x+3y+3z=5\\-------\\2y+3y+3z+3z=0+5[/tex]
[tex]5y+6z=5[/tex] -----> equation D
adds equation B and equation C
[tex]-2x-y+z=-3\\2x+3y+3z=5\\--------\\-y+3y+z+3z=-3+5[/tex]
[tex]2y+4z=2[/tex] -----> equation E
we have the system
[tex]5y+6z=5[/tex] -----> equation D
[tex]2y+4z=2[/tex] -----> equation E
Multiply equation E by -1.5 both sides
[tex]-1.5(2y+4z)=-1.5(2)[/tex]
[tex]-3y-6z=-3[/tex] -----> equation F
adds equation D and equation F
[tex]5y+6z=5\\-3y-6z=-3\\---------\\5y-3y=5-3\\2y=2\\y=1[/tex]
Find the value of z
substitute the value of y in equation E (or substitute in equation D)
[tex]2(1)+4z=2[/tex]
[tex]2+4z=2[/tex]
[tex]4z=0[/tex]
[tex]z=0[/tex]
Find the value of x
substitute the value of y and z in equation A (or equation B or C)
[tex]-2x+2(1)+3(0)=0[/tex]
[tex]-2x+2=0[/tex]
[tex]-2x=-2[/tex]
[tex]x=1[/tex]
therefore
The solution of the system of equations is
[tex]x=1, y=1, z=0[/tex]
