Answer:
C) x = 1, y = -2, z = 4
Step-by-step explanation:
Given system of equations:
[tex]\begin{cases}x - 4y-2z= 1\\x+ 5y + 2z= -1\\y+ 3z = 10\end{cases}[/tex]
To solve the given system of equations, begin by subtracting the first equation from the second equation to eliminate the term in x:
[tex]\begin{array}{lrrrrrrl}&x&+&5y&+&2z&=&-1\\\vphantom{\dfrac12}-&(x&-&4y&-&2z&=&\;\;1)\\\cline{2-8}\vphantom{\dfrac12}&&&9y&+&4z&=&-2\end{array}[/tex]
Rearrange the third equation to isolate y:
[tex]y=10-3z[/tex]
Substitute y = 10 - 3z into the equation 9y + 4z = -2, then solve for z:
[tex]\begin{aligned}9(10-3z)+4z&=-2\\90-27z+4z&=-2\\90-23z&=-2\\-23z&=-92\\z&=4\end{aligned}[/tex]
Therefore, z = 4.
Substitute the found value of z into the equation for y, and solve for y:
[tex]\begin{aligned}y&=10-3(4)\\y&=10-12\\y&=-2\end{aligned}[/tex]
Therefore, y = -2.
Finally, substitute z = 4 and y = -2 into one of the original equations and solve for x:
[tex]\begin{aligned}x+ 5(-2) + 2(4)&= -1\\x-10 + 8&= -1\\x-2&= -1\\x&= 1\end{aligned}[/tex]
Therefore, x = 1.
So, the solution to the given system of equations is:
[tex]\Large\boxed{\boxed{x = 1, \;y = -2, \;z = 4}}[/tex]
