(x-3y+ 3z=-42x+3y-z =15(4x-3y- z=19- What is the solution to the system?

Answer:x=5,y=1 and z=-2Step-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