Question

Solve the system of equations.
4x+3y+6z=3
5x+5y+6z=5
6x+3y+6z=3

Answer 1

ageed its C

Explanation:

Answer 2

C.
`x=0,\ y=1,\ z=0`

Explanation:

Consider the system of three equations:

`\left\{\begin{array}{l}4x+3y+6z=3\\5x+5y+6z=5\\6x+3y+6z=3\end{array}\right.`

Multiply the first equation by 5, the second equation by 4 and subtract them:

`\left\{\begin{array}{r}4x+3y+6z=3\\-5y+6z=-5\\6x+3y+6z=3\end{array}\right.`

Multiply the first equation by 3, the second equation by 2 and subtract them:

`\left\{\begin{array}{r}4x+3y+6z=3\\-5y+6z=-5\\3y+6z=3\end{array}\right.`

Multiply the second equation by 3, the third equation by 5 and add the second and the third equations:

`\left\{\begin{array}{r}4x+3y+6z=3\\-5y+6z=-5\\48z=0\end{array}\right.`

Fro mthe third equation

`z=0`

Substitute it into the second equation:

`-5y+6\cdot 0=-5\\ \\-5y=-5\\ \\y=1`

Substitute y=1 and z=0 into the first equation:

`4x+3\cdot 1+6\cdot 0=3\\ \\4x+3=3\\ \\4x=0\\ \\x=0`

The solution is

`x=0,\ y=1,\ z=0`