Question

What is the solution to the following system?

5x+2y+z=4
x+2z=4
2x+y-z=-1


A. x = 0, y = 1, z = 2
B. x = 1, y = –1, z = 2
C. x = 1, y = 0, z = –1
D. x = 2, y = 1, z = 0

Answer 1

A. x = 0, y = 1, z = 2 EDG 2021

Explanation:

i took the test

Answer 2

correct choice is A

Explanation:

From the second equation express x and substitute it into the first and third equations:

`\left\{\begin{array}{l}x=4-2z\\5(4-2z)+2y+z=4\\2(4-2z)+y-z=-1\end{array}\right..`

Then

`\left\{\begin{array}{l}x=4-2z\\20-10z+2y+z=4\\8-4z+y-z=-1\end{array}\right.\Rightarrow\left\{\begin{array}{l}x=4-2z\\2y-9z=-16\\y-5z=-9\end{array}\right..`

From the third equation
`y=-9+5z` and substituting it into the second, you get

`2(-9+5z)-9z=-16,\\ \\-18+10z-9z=-16,\\ \\z=-16+18,\\ \\z=2.`

If
`z=2,` then

`y=-9+5\cdot 2=1`

and

`x=4-2\cdot 2=0.`

The solution of the system is
`x=0,\ y=1,\ z=2.`