Question

X+2y-2z=-31
4y+2z=0
-x+y-z=-2

Solve for x,y,z in systems of equations

Answer 1

`x=-9\\ \\y=-(11)/(3)\\ \\z=(22)/(3)`

Explanation:

Given the system of inequalities

`\left\{\begin{array}{r}x+2y-2z=-31\\ \\4y+2z=0\\ \\-x+y-z=-2\end{array}\right.`

From the second equation,

`2z=-4y\\ \\z=-2y`

Substitute it into the first and third equations:

`\left\{\begin{array}{r}x+2y-2(-2y)=-31\\ \\-x+y-(-2y)=-2\end{array}\right.\Rightarrow \left\{\begin{array}{r}x+6y=-31\\ \\-x+3y=-2\end{array}\right.`

Add these two equations:

`x+6y+(-x+3y)=-31+(-2)\\ \\x+6y-x+3y=-31-2\\ \\9y=-33\\ \\y=-(11)/(3)\\ \\z=2\cdot (11)/(3)=(22)/(3)`

Substitute
`y=-(11)/(3)` into the third equation:

`-x-3\cdot (11)/(3)=-2\\ \\-x-11=-2\\ \\-x=-2+11\\ \\x=-9`