Question

What is the solution to the following system?

3x+2y+z=20

x-4y-z=-10

2x+y+2z=15


A. (2, 3, 4)

B. (4, 3, –2)

C. (4, 3, 2)

D. (6, 7, –2)

Answer 1

it is C i believe you plug in the answers to get it

Answer 2

C. (4, 3, 2)

Explanation:

Given :
`3x+2y+z=20`

`x-4y-z=-10`

`2x+y+2z=15`

To Find: Solution:

`3x+2y+z=20` -1

`x-4y-z=-10` --2

`2x+y+2z=15` --3

Substitute the value of z from 1 in 2 and 3

So in 2 ,
`x-4y-(20-3x-2y)=-10`

`x-4y-20+3x+2y=-10`

`4x-2y=-10+20`

`4x-2y=10` ---4

So, in 3 ,
`2x+y+2(20-3x-2y)=15`

`2x+y+40-6x-4y=15`

`-4x-3y=15-40`

`-4x-3y=-25` -5

Now solve 4 and 5

Substitute the value of x from 4 in 5

`-4((10+2y)/(4))-3y=-25`

`-1(10+2y)-3y=-25`

`-10-2y-3y=-25`

`-10-5y=-25`

`-5y=-15`

`y=3`

Now substitute the value of y in 4

`4x-2(3)=10`

`4x-6=10`

`4x=10+6`

`4x=16`

`x=4`

Now substitute the value of x and y in 1 to get value of z

`3(4)+2(3)+z=20`

`12+6+z=20`

`18+z=20`

`z=20-18`

`z=2`

Thus The solution is (4,3,2)

Hence Option c is correct.