Question

(x-3y+ 3z=-4 2x+3y-z =15 (4x-3y- z=19 - What is the solution to the system?

Answer 1

Answer:

x=5,y=1 and z=-2

Step-by-step explanation:

We are given that system of equation

$x-3y+3z=-4$ (I equation)

$2x+3y-z=15$  (II equation )

$4x-3y-z=19$  (III equation )

Equation II multiply by 3 then add  with equation I

Then, we get

$7x+6y=41$ ....(Equation IV)

Subtract equation II from equation III then we get

$2x-6y=4$  (equation V)

Adding equation IV and equation V then, we get

$9x=45$

$x=5$

Substitute x=5 in equation V then, we get

$2(5)-6y=4$

$10-6y=4$

$6y=10-4=6$

$y=1$

Substitute x=5 and y=1 in equation then, we get

$5-3(1)+3z=-4$

$2+3z=-4$

$3z=-4-2=-6$

$z=\frac{-6}{3}=-2$

Hence, the solution for the given system of equation is given by

x=5,y=1 and z=-2