Answer :
Answer:
[tex]=\frac{774}{337},\:z=\frac{1225}{337},\:y=-\frac{968}{337}[/tex]
Step-by-step explanation:
Given:
[tex]\begin{bmatrix}5x+14y-2z=-36\\ 7x+3y+4z=22\\ 3x-2y-z=9\end{bmatrix}[/tex]
Solve:
[tex]\mathrm{Substitute\:}x=\frac{-36-14y+2z}{5}[/tex]
Thus,
[tex]\begin{bmatrix}7\cdot \frac{-36-14y+2z}{5}+3y+4z=22\\ 3\cdot \frac{-36-14y+2z}{5}-2y-z=9\end{bmatrix}[/tex]
Simplify
[tex]\begin{bmatrix}\frac{-83y+34z-252}{5}=22\\ \frac{-52y+z-108}{5}=9\end{bmatrix}[/tex]
[tex]\mathrm{Substitute\:}y=-\frac{-34z+362}{83}[/tex]
[tex]\begin{bmatrix}\frac{-52\left(-\frac{-34z+362}{83}\right)+z-108}{5}=9\end{bmatrix}[/tex]
Simplify
[tex]\begin{bmatrix}\frac{-337z+1972}{83}=9\end{bmatrix}[/tex]
[tex]\mathrm{For\:}y=-\frac{-34z+362}{83}[/tex]
[tex]\mathrm{Substitute\:}z=\frac{1225}{337}[/tex]
[tex]y=-\frac{-34\cdot \frac{1225}{337}+362}{83}[/tex]
[tex]y=-\frac{968}{337}[/tex]
[tex]\mathrm{For\:}x=\frac{-36-14y+2z}{5}[/tex]
[tex]\mathrm{Substitute\:}z=\frac{1225}{337},\:y=-\frac{968}{337}[/tex]
[tex]x=\frac{-36-14\left(-\frac{968}{337}\right)+2\cdot \frac{1225}{337}}{5}[/tex]
[tex]x=\frac{774}{337}[/tex]
Therefore, the solutions are:
[tex]=\frac{774}{337},\:z=\frac{1225}{337},\:y=-\frac{968}{337}[/tex]
~lenvy~