Answer:
C. [tex]x=0,\ y=1,\ z=0[/tex]
Step-by-step explanation:
Consider the system of three equations:
[tex]\left\{\begin{array}{l}4x+3y+6z=3\\5x+5y+6z=5\\6x+3y+6z=3\end{array}\right.[/tex]
Multiply the first equation by 5, the second equation by 4 and subtract them:
[tex]\left\{\begin{array}{r}4x+3y+6z=3\\-5y+6z=-5\\6x+3y+6z=3\end{array}\right.[/tex]
Multiply the first equation by 3, the second equation by 2 and subtract them:
[tex]\left\{\begin{array}{r}4x+3y+6z=3\\-5y+6z=-5\\3y+6z=3\end{array}\right.[/tex]
Multiply the second equation by 3, the third equation by 5 and add the second and the third equations:
[tex]\left\{\begin{array}{r}4x+3y+6z=3\\-5y+6z=-5\\48z=0\end{array}\right.[/tex]
Fro mthe third equation
[tex]z=0[/tex]
Substitute it into the second equation:
[tex]-5y+6\cdot 0=-5\\ \\-5y=-5\\ \\y=1[/tex]
Substitute y=1 and z=0 into the first equation:
[tex]4x+3\cdot 1+6\cdot 0=3\\ \\4x+3=3\\ \\4x=0\\ \\x=0[/tex]
The solution is
[tex]x=0,\ y=1,\ z=0[/tex]
