Answer:
The solution is x=1,y=2,z=3
Step-by-step explanation:
The given system of equations is ;\
2x−3y+4z=8...(1)
3x+4y−5z=−4...(2)
4x−5y+6z=12...(3)
Make x the subject in equation (1)
[tex]x=\frac{8+3y-4z}{2}...(4)[/tex]
Put equation (4) into equation (2) and (3)
[tex]3(\frac{8+3y-4z}{2})+4y-5z=-4[/tex]
Multiply through by;
[tex]3(8+3y-4z)+8y-10z=-8[/tex]
Expand;
[tex]24+9y-12z+8y-10z=-8[/tex]
Simplify;
[tex]17y-22z=-32...(5)[/tex]
Equation (4) in (3)
[tex]4(\frac{8+3y-4z}{2})-5y+6z=12[/tex]
[tex]2(8+3y-4z)-5y+6z=12[/tex]
[tex]16+6y-8z-5y+6z=12[/tex]
[tex]y-2z=-4[/tex]
[tex]y=2z-4...(6)[/tex]
Put equation (6) into equation (5)
[tex]17(2z-4)-22z=-32[/tex]
[tex]34z-68-22z=-32[/tex]
[tex]34z-22z=-32+68[/tex]
[tex]12z=36[/tex]
z=3
Put z=3 into equation (6)
y=2(3)-4=2
Put y=2 and z=3 into equation 4
[tex]x=\frac{8+3(2)-4(3)}{2}=1[/tex]
The solution is x=1,y=2,z=3
