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Sonja [21]
3 years ago
15

What is the solution to the following system?

Mathematics
2 answers:
VladimirAG [237]3 years ago
7 0
It is C i believe you plug in the answers to get it
Yuki888 [10]3 years ago
7 0

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.

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