Question

What is the solution to the following system?

Answer 1

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

Answer 2

Answer:

C. (4, 3, 2)

Step-by-step 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

$-4(\frac{10+2y}{4})-3y=-25$

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

$-10-2y-3y=-25$

$-10-5y=-25$

$-5y=-15$

$y=3$

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.