Question

Solve the system by elimination. − 2 x + 2 y + 3 z = 0 − 2 x − y + z = − 3 2 x + 3 y + 3 z = 5

Answer 1

Answer:

$x=1, y=1, z=0$

Step-by-step explanation:

we have

$-2x+2y+3z=0$ ----> equation A

$-2x-y+z=-3$  -----> equation B

$2x+3y+3z=5$ ----> equation C  

adds equation A and equation C

$-2x+2y+3z=0\\2x+3y+3z=5\\-------\\2y+3y+3z+3z=0+5$

$5y+6z=5$ -----> equation D

adds equation B and equation C

$-2x-y+z=-3\\2x+3y+3z=5\\--------\\-y+3y+z+3z=-3+5$

$2y+4z=2$ -----> equation E

we have the system

$5y+6z=5$ -----> equation D

$2y+4z=2$ -----> equation E

Multiply equation E by -1.5 both sides

$-1.5(2y+4z)=-1.5(2)$

$-3y-6z=-3$ -----> equation F

adds equation D and equation F

$5y+6z=5\\-3y-6z=-3\\---------\\5y-3y=5-3\\2y=2\\y=1$

Find the value of z

substitute the value of y in equation E (or substitute in equation D)

$2(1)+4z=2$

$2+4z=2$

$4z=0$

$z=0$

Find the value of x

substitute the value of y and z in equation A (or equation B or C)

$-2x+2(1)+3(0)=0$

$-2x+2=0$

$-2x=-2$

$x=1$

therefore

The solution of the system of equations is

$x=1, y=1, z=0$