Question

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

Solve for x,y,z in systems of equations

Answer 1

Answer:

$x=-9\\ \\y=-\dfrac{11}{3}\\ \\z=\dfrac{22}{3}$

Step-by-step 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:

$x+6y+(-x+3y)=-31+(-2)\\ \\x+6y-x+3y=-31-2\\ \\9y=-33\\ \\y=-\dfrac{11}{3}\\ \\z=2\cdot \dfrac{11}{3}=\dfrac{22}{3}$

Substitute $y=-\frac{11}{3}$ into the third equation:

$-x-3\cdot \dfrac{11}{3}=-2\\ \\-x-11=-2\\ \\-x=-2+11\\ \\x=-9$