Answer:
x=5,y=1 and z=-2
Step-by-step explanation:
We are given that system of equation
[tex]x-3y+3z=-4[/tex] (I equation)
[tex]2x+3y-z=15[/tex] (II equation )
[tex]4x-3y-z=19[/tex] (III equation )
Equation II multiply by 3 then add with equation I
Then, we get
[tex]7x+6y=41[/tex] ....(Equation IV)
Subtract equation II from equation III then we get
[tex]2x-6y=4[/tex] (equation V)
Adding equation IV and equation V then, we get
[tex]9x=45[/tex]
[tex]x=5[/tex]
Substitute x=5 in equation V then, we get
[tex]2(5)-6y=4[/tex]
[tex]10-6y=4[/tex]
[tex]6y=10-4=6[/tex]
[tex]y=1[/tex]
Substitute x=5 and y=1 in equation then, we get
[tex]5-3(1)+3z=-4[/tex]
[tex]2+3z=-4[/tex]
[tex]3z=-4-2=-6[/tex]
[tex]z=\frac{-6}{3}=-2[/tex]
Hence, the solution for the given system of equation is given by
x=5,y=1 and z=-2
