MATH SOLVE

4 months ago

Q:
# x+2y-2z=-314y+2z=0-x+y-z=-2Solve for x,y,z in systems of equations

Accepted Solution

A:

Answer:[tex]x=-9\\ \\y=-\dfrac{11}{3}\\ \\z=\dfrac{22}{3}[/tex]Step-by-step explanation:Given the system of inequalities[tex]\left\{\begin{array}{r}x+2y-2z=-31\\ \\4y+2z=0\\ \\-x+y-z=-2\end{array}\right.[/tex]From the second equation,[tex]2z=-4y\\ \\z=-2y[/tex]Substitute it into the first and third equations:[tex]\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.[/tex]Add these two equations:[tex]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}[/tex]Substitute [tex]y=-\frac{11}{3}[/tex] into the third equation:[tex]-x-3\cdot \dfrac{11}{3}=-2\\ \\-x-11=-2\\ \\-x=-2+11\\ \\x=-9[/tex]