Answer:
C. (4, 3, 2)
Step-by-step explanation:
Given : [tex]3x+2y+z=20[/tex]
[tex]x-4y-z=-10[/tex]
[tex]2x+y+2z=15[/tex]
To Find: Solution:
[tex]3x+2y+z=20[/tex] -1
[tex]x-4y-z=-10[/tex] --2
[tex]2x+y+2z=15[/tex] --3
Substitute the value of z from 1 in 2 and 3
So in 2 , [tex]x-4y-(20-3x-2y)=-10[/tex]
[tex]x-4y-20+3x+2y=-10[/tex]
[tex]4x-2y=-10+20[/tex]
[tex]4x-2y=10[/tex] ---4
So, in 3 , [tex]2x+y+2(20-3x-2y)=15[/tex]
[tex]2x+y+40-6x-4y=15[/tex]
[tex]-4x-3y=15-40[/tex]
[tex]-4x-3y=-25[/tex] -5
Now solve 4 and 5
Substitute the value of x from 4 in 5
[tex]-4(\frac{10+2y}{4})-3y=-25[/tex]
[tex]-1(10+2y)-3y=-25[/tex]
[tex]-10-2y-3y=-25[/tex]
[tex]-10-5y=-25[/tex]
[tex]-5y=-15[/tex]
[tex]y=3[/tex]
Now substitute the value of y in 4
[tex]4x-2(3)=10[/tex]
[tex]4x-6=10[/tex]
[tex]4x=10+6[/tex]
[tex]4x=16[/tex]
[tex]x=4[/tex]
Now substitute the value of x and y in 1 to get value of z
[tex]3(4)+2(3)+z=20[/tex]
[tex]12+6+z=20[/tex]
[tex]18+z=20[/tex]
[tex]z=20-18[/tex]
[tex]z=2[/tex]
Thus The solution is (4,3,2)
Hence Option c is correct.
