Answer:
B. [tex]x=3,\ y=20,\ z=-14[/tex]
Step-by-step explanation:
Consider the system of three equations:
[tex]\left\{\begin{array}{l}2x+2y+3z=4\\5x+3y+5z=5\\3x+4y+6z=5\end{array}\right.[/tex]
Multiply the first equation by 5, the second equation by 2 and subtract them:
[tex]\left\{\begin{array}{r}2x+2y+3z=4\\4y+5z=10\\3x+4y+6z=5\end{array}\right.[/tex]
Multiply the first equation by 3, the second equation by 2 and subtract them:
[tex]\left\{\begin{array}{r}2x+2y+3z=4\\4y+5z=10\\-2y-3z=2\end{array}\right.[/tex]
Multiply the third equation by 2 and add the second and the third equations:
[tex]\left\{\begin{array}{r}2x+2y+3z=4\\4y+5z=10\\-z=14\end{array}\right.[/tex]
Fro mthe third equation
[tex]z=-14[/tex]
Substitute it into the second equation:
[tex]4y+5\cdot (-14)=10\\ \\4y=10+70\\ \\ 4y=80\\ \\y=20[/tex]
Substitute y=20 and z=-14 into the first equation:
[tex]2x+2\cdot 20+3\cdot (-14)=4\\ \\2x+40-42=4\\ \\2x-2=4\\ \\2x=6\\ \\x=3[/tex]
The solution is
[tex]x=3,\ y=20,\ z=-14[/tex]
